MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Frobenius method
Question Bank - Set 2
Liberty University
Question 1
Question
Consider the differential equation (x2−6)x2y′′ + 2xy′+ (1 −x2)y= 0. Use the
Frobenius method to find a general solution about x= 0.
Solution
Step 1: Assume a solution of the form y=∑∞
n=0 anxn+r.
Step 2: Find the derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation.
(x2−6)x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+2x(∞
∑
n=0
an(n+r)xn+r−1)+(1−x2)(∞
∑
n=0
anxn+r)= 0
Step 4: Simplify the equation by expanding and rearranging terms.
∞
∑
n=0 [an(n+r)(n+r−1)xn+r−6an(n+r)(n+r−1)xn+r+ 2an(n+r)xn+r+anxn+r−anxn+r+2]= 0
Step 5: Combine like terms and set the coefficient of each power of xto zero.
(n+r)(n+r−1−6(n+r)(n+r−1) + 2(n+r) + 1)anxn+r−anxn+r+2 = 0
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
Step 6: Equate the coefficient of xn+rto zero and solve for rto find the
indicial equation.
r2−5r= 0 =⇒r= 0,5
Step 7: Find the recurrence relation by substituting r= 0 into the equation.
(0 + 0)(0 + 0 −1)an−6(0 + 0)(0 + 0 −1)an+ 2(0 + 0)an+ 1an= 0
Step 8: Solve the recurrence relation.
Question 2
Question
Find the Frobenius series solution to the differential equation
x2y′′ + 2xy′−(x2+ 1)y= 0
about x= 0.
Solution
To find the Frobenius series solution to the given differential equation, we will
assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where anare constants to
be determined and ris the initial guess for the root of the indicial equation.
Step 1: Find the indicial equation.
The indicial equation is given by the characteristic equation of the differ-
ential equation. For the given differential equation, the characteristic equation
is obtained by substituting y(x) = xr∑∞
n=0 anxninto the differential equation
and collecting like powers of x.
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+2x
∞
∑
n=0
an(n+r)xn+r−1−(x2+1)
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
an−2xn+r−
∞
∑
n=2
an−2xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
(an−2+an)xn+r= 0
2
From the equation above, we can find the indicial equation by setting the
coefficient of xr−2to zero:
a0r(r−1) = 0
This gives us the indicial equation r(r−1) = 0 with roots r= 0,1.
So, we have two possible solutions: y1(x) = ∑∞
n=0 anxnand y2(x) = ∑∞
n=0 anxn+1.
Step 2: Finding the recurrence relation for y1(x).
For y1(x) = ∑∞
n=0 anxn, substituting r= 0 into the original differential
equation gives:
2a0−a0= 0 =⇒a0= 0
2a1−a1= 0 =⇒a1= 0
And, for n≥2:
n(n−1)an+ 2nan−an−2= 0 =⇒an=an−2
n(n+ 1)
So, the coefficients anfor y1(x)can be written as a function of a0and a1.
Step 3: Finding the recurrence relation for y2(x).
For y2(x) = ∑∞
n=0 anxn+1, substituting r= 1 into the
Question 3
Question
Use the Frobenius method to find a series solution for the differential equation:
x2y′′ +xy′−(x+ 1)y= 0
at x= 0, with an indicial equation yielding the roots r1= 1 and r2= 0.
Solution
To solve this differential equation using the Frobenius method, we assume a
series solution of the form:
y(x) =
∞
∑
n=0
anxr+n
where ris the smaller root of the indicial equation. Substituting this into
the differential equation gives:
x2
∞
∑
n=0
an(r+n)(r+n−1)xr+n−2+x
∞
∑
n=0
an(r+n)xr+n−1−(x+1)
∞
∑
n=0
anxr+n= 0
We expand each term and simplify:
∞
∑
n=0
an(r+n)(r+n−1)xr+n+
∞
∑
n=0
an(r+n)xr+n−
∞
∑
n=0
anxr+n+1 −
∞
∑
n=0
anxr+n= 0
3
Now, we will collect terms with the same powers of x:
(r(r−1)a0+ra1)xr+((r+ 1)ra1+ (r+ 1)ra2−a0)xr+1+
∞
∑
n=2
((r+n)(r+n−1)an+ (r+n)an−an−2)xr+n= 0
Since we know the roots of the indicial equation to be r1= 1 and r2= 0, we
can start solving for the coefficients anone by one.
Question 4
Question
Consider the differential equation
x2y′′ +x(1 −x)y′−y= 0
Find the Frobenius series solution about the ordinary point x= 0.
Solution
To find the Frobenius series solution about the ordinary point x= 0, we assume
the solution has the form
y(x) =
∞
∑
n=0
anxn+r
Substitute this into the differential equation and solve for r:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Grouping terms by powers of xgives
[a0r(r−1) + a0r−a0]xr+
∞
∑
n=1
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
4
Now, for the series to be a solution, the coefficient of each power of xmust
vanish. This leads to the indicial equation:
r(r−1) + r= 0
r2−r= 0
r(r−1) = 0
This gives us r= 0 or r= 1.
Step 1: Let’s first consider the case when r= 0. This leads to the solution
of the form y1(x) = ∑∞
n=0 anxn.
Step 2: Now, let’s consider the case when r= 1. This leads to the solution
of the form y2(x) = ∑∞
n=0 anxn+1.
Therefore, the Frobenius series solution about the ordinary point x= 0 is
given by y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxnand y2(x) =
∑∞
n=0 anxn+1.
Question 5
Question
Find the Frobenius series solution for the differential equation
x2y′′ +xy′+ (x2−1)y= 0
near the regular singular point x= 1.
Solution
To find the Frobenius series solution, we assume a solution of the form y(x) =
∑∞
n=0 an(x−1)n+r, where ris the order of the regular singular point. In this
case, r= 1 because the coefficient of y′is singular at x= 1.
Step 1: Find y′and y′′
y′(x) =
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
=
∞
∑
n=0
[(n+r)an(x−1)n+r−1]
y′′ (x) =
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]
5
Step 2: Substitute y,y′, and y′′ into the differential equation
x2y′′ +xy′+ (x2−1)y= 0
x2
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2] + x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+(x2−1)
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Simplify the equation by combining like terms
∞
∑
n=0
[(n+r)(n+r−1)anxn+r]+
∞
∑
n=0
[(n+r)anxn+r]+
∞
∑
n=0
anxn+r+2−
∞
∑
n=0
anxn+r= 0
Step 4: Solve for the recurrence relation
(n+r)(n+r−1)an+ (n+r)an+an−an= 0
(n+r)(n+r−1)an+ (n+r)an= 0
(n+r)(n+ 2r−1)an= 0
an= 0 or n=−2r+ 1
Step 5: Write down the Frobenius series solution The Frobenius
series solution near x= 1 is given by
y(x) =
∞
∑
n=0
an(x−1)n+1 =
∞
∑
n=0
an+2r−1(x−1)n+1
Therefore, the solution to the differential equation near x= 1 is
y(x) =
∞
∑
n=0
an+1(x−1)n+1
Question 6
Question
Find the Frobenius series solution for the differential equation (1 −x2)y′′ −
2xy′+ 2y= 0 centered at x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y. First derivative:
y′=
∞
∑
n=0
an(n+r)xn+r−1
Second derivative:
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
6
Step 3: Substitute y,y′, and y′′ into the differential equation.
(1−x2)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by collecting terms with the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2−2
∞
∑
n=0
an(n+r)xn+r+2
∞
∑
n=0
anxn+r= 0
Step 5: Combine series terms with the same power of xand simplify the
equation further.
∞
∑
n=0
[an(n+r)(n+r−1) −2an(n+r)] xn+r+
∞
∑
n=0
[2an−an(n+r)(n+r−1)] xn+r+2 = 0
Step 6: Equate the coefficients of the terms with the same power of xto
zero. This leads to the indicial equation:
a0r(r−1) = 0
and the recurrence relation:
(n+r+ 2)(n+r+ 1)an+2 −(n+r)(n+r−1−2(n+r))an= 0
Step 7: Solve the indicial equation to find the values of r. Since a0= 0, we
have r(9r−1) = 0. This gives r= 0 or r= 1.
Step 8: Use the first few terms of the recurrence relation to find the coeffi-
cients a1and a2in terms of a0. When r= 0:
2a2−a0= 0 =⇒a2=a0
2
When r= 1:
6a3−2a1= 0 =⇒a3=a1
3
Therefore, the Frobenius series solution for the given differential equation
centered at x= 0 is:
y(x) = a0(1 + a1
2x+a2
3x2+· · · )+a0x(1 + a3
2x+· · · )
Question 7
Question
Consider the differential equation:
x2y′′ −x(y′−1) −2y= 0
Use the Frobenius method to find a general solution near the regular singular
point x= 0.
7
Solution
Step 1: Identify the singular point
The given differential equation can be rewritten as:
y′′ −y′−1
x−2
x2y= 0
The point x= 0 is a regular singular point, as the coefficient of y′′ is singular
at x= 0.
Step 2: Assume a Frobenius series solution
Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the singular
point, anare constants to be determined, and y′and y′′ are calculated accord-
ingly.
Step 3: Find y′and y′′
Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation
Substitute y,y′, and y′′ into the differential equation and simplify to collect
terms with the same power of x.
Step 5: Solve for the recurrence relation
Set the coefficient of each term of xin the differential equation to zero to
obtain a recurrence relation for an.
Step 6: Find the indicial equation
The indicial equation is found by setting the coefficient of xrin the recurrence
relation to zero.
Step 7: Solve for the first few terms
Using the recurrence relation and indicial equation, solve for the first few
terms of the series solution y=∑∞
n=0 anxn+r.
Step 8: General solution
The general solution near the regular singular point x= 0 is y=c1y1(x) +
c2y2(x), where y1(x)and y2(x)are linearly independent series solutions obtained
using the Frobenius method.
Question 8
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +xy′−y= 0
8
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Find the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums into a single series:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−an]xn+r= 0
Step 5: Equate the coefficients of like powers of xto find the indicial equa-
tion:
an(n+r)(n+r−1) + an(n+r)−an= 0
an[(n+r)(n+r−1) + (n+r)−1] = 0
Step 6: Solve the indicial equation r2−1=0to find the roots r1= 1 and
r2=−1.
Step 7: The solutions to the indicial equation are not equal, so the general
solution is:
y(x) = c1x+c2x−1
Question 9
Question
Consider the differential equation
x2y′′ +xy′+ (x2−1)y= 0.
Use the Frobenius method to find the indicial roots and at least two linearly
independent solutions near the regular singular point x= 0.
9
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+rand compute y′
and y′′ .
Step 2: Substitute y,y′, and y′′ into the differential equation and collect like
terms.
Step 3: Set the coefficient of each power of xto zero to get indicial equations
for determining the values of r.
Step 4: Solve the indicial equations to find the indicial roots r1and r2.
Step 5: If the indicial roots are different, proceed to find two linearly inde-
pendent solutions. If the roots are equal, the method needs to be adjusted.
Step 6: Write out the first few terms of the series solution based on the roots
found in Step 4.
Step 7: Use the recurrence relation to find a general formula for the coeffi-
cients an.
Step 8: Repeat the process to find a second linearly independent solution.
Step 9: Ensure both solutions are linearly independent, and the general
solution is a linear combination of these solutions.
This process will lead us to the desired solutions near the regular singular
point.
Question 10
Question
Consider the differential equation (x2−1)y′′ +2xy′−2y= 0. Find the Frobenius
series solution about x= 0.
Solution
Step 1: Write the differential equation in the form x2y′′ +p(x)xy′+q(x)y= 0:
y′′ +2x
x2−1y′−2
x2−1y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equa-
tion.
Step 3: Find the indicial equation by substituting y=xrinto the differential
equation. The indicial equation is given by:
r(r−1) + 2r
−1= 0
r2−r= 0
r(r−1) = 0
10
which gives the roots r= 0 and r= 1.
Step 4: Write the Frobenius series solution utilizing the two roots r= 0 and
r= 1 as:
y=a0x0+a1x1+a2x2+· · ·
y=a0+a1x+a2x2+· · ·
Question 11
Question
Consider the following differential equation:
(x2−2)x2y′′ −4xy′+ 6y= 0
Use the Frobenius method to find at least two solutions near x= 0.
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions are of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation. Let’s substitute y(x)into the dif-
ferential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+r)(n+r−1)xn+r−2]
−4x[∞
∑
n=0
an(n+r)xn+r−1]+ 6
∞
∑
n=0
anxn+r= 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
Now, equate the terms multiplied by xn+rto zero and collect like terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
an(n+r)xn+r+6
∞
∑
n=0
anxn+r= 0
This equation should hold for all values of n, so we can equate the coefficients
of xr+nto zero. This will give us the indicial equation.
Indicial equation: r2−1 = 0
11
⇒r=±1
There are two roots for the indicial equation: r1= 1 and r2=−1.
Step 2: Find the recurrence relation for r= 1.Assuming r= 1, we
substitute r= 1 into the general form of the solution and simplify:
y(x) =
∞
∑
n=0
anxn+1
Next, we substitute y(x)and r= 1 into the differential equation and simplify:
(x2−2)x2y′′ −4xy′+ 6y= 0
⇒(x2−2)x2[∞
∑
n=0
an(n+ 1)(n)xn]−4x[∞
∑
n=0
an(n+ 1)xn]+ 6
∞
∑
n=0
anxn+1 = 0
Expanding and simplifying, we get:
∞
∑
n=0
an(n+1)(n)xn+2−2
∞
∑
n=0
an(n+1)xn+2−4
∞
∑
n=0
an(n+1)xn+1+6
∞
∑
n=0
anxn+2 = 0
Equate the terms multiplied by xn+2 to zero and collect like terms. This
gives the recurrence relation for r= 1.
Step 3: Find the recurrence relation for r=−1.By following
Question 12
Question
Find a Frobenius series solution for the differential equation:
x2y′′ +xy′−x2y= 0
about the ordinary point x= 0.
Solution
To find a Frobenius series solution, we assume a solution of the form y(x) =
∞
∑
n=0
anxn+r, where anare constants to be determined and ris a constant to be
found by substitution.
Step 1: Find the indicial equation by substituting the assumed
form into the differential equation.
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1,
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2.
12
Substitute these into the differential equation and equate coefficients of like
powers of x:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−x2
∞
∑
n=0
anxn+r= 0.
Simplify the equation and set each coefficient to zero to get the indicial equation.
Step 2: Solve the Indicial Equation. The indicial equation will be
obtained by setting the coefficient of xrto zero. This will give the possible
values of r.
Step 3: Determining the Recurrence Relationship. Once we have
found the values of r, substitute back into the original differential equation and
determine the recurrence relationship for the coefficients an.
Step 4: Finding the First Few Terms. Using the recurrence relation-
ship, find the values of the coefficients for the first few terms of the series.
Continue the process until you have a few terms of the series solution.
Question 13
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′−p(p+
1)y= 0 at x= 0 when p= 3.
Solution
Step 1: Write the general form of the Frobenius series solution as y=∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y.
y′=
∞
∑
n=0
(n+r)anxn+r−1and y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain the
indicial equation.
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−p(p+1)
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by combining terms with the same power of x.
Then set the coefficient of each power of xto zero to determine the recurrence
relation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−p(p+ 1)
∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of xrto zero to obtain the indicial equation.
r2= 0 ⇒r= 0
13
Step 6: Determine the recurrence relation for the coefficients an.
(n+r)(n+r−1)an+ (n+r)an−p(p+ 1)an= 0
Solving for anand simplifying gives:
an=p(p+ 1) −n(n−1)
n(2n+2+p)
Step 7: Substitute p= 3 into the formula for anto obtain the specific form
of the coefficients.
Step 8: Write the Frobenius series solution with p= 3 as y=a0+a1x+
a2x2+. . . and simplify further if needed.
Question 14
Question
Use the Frobenius method to find the general solution of the differential equation
(2x2+ 1)y′′ +xy′−y= 0 near the ordinary point x= 0.
Solution
Step 1: Write the equation in standard form.
The given differential equation is (2x2+ 1)y′′ +xy′−y= 0. To write this
equation in standard form, divide through by 2x2+ 1:
(2x2+ 1)y′′ +xy′−y= 0
⇒y′′ +x
2x2+ 1y′−1
2x2+ 1y= 0
Step 2: Assume a power series solution.
Assume that the solution is in the form of a power series: y(x) = ∑∞
n=0 anxn+r,
where anare constants to be determined and ris the root of the indicial equation
given by the equation r2+r−1
2= 0.
Step 3: Find the first few terms of y(x)near x= 0.
Substitute y(x)into the differential equation and express the derivatives y′and
y′′ using the power series form. Then, equate coefficients of like powers of xto
determine the values of an.
Step 4: Determine the recursion relation.
After finding the values of an, write down the first few terms of the solution
y(x). From this, determine a recursion relation for the coefficients an.
Step 5: Find the general solution.
Using the determined values of anand the recursion relation, write down the
general solution of the differential equation near x= 0 in terms of the power
series.
14
Therefore, the general solution of the differential equation (2x2+ 1)y′′ +
xy′−y= 0 near x= 0 using the Frobenius method involves assuming a power
series solution, determining the values of the coefficients anthrough equating
coefficients of like powers of x, finding the recursion relation, and then writing
down the general form of the solution.
Question 15
Question
Consider the differential equation (1 −x)y′′ −2xy′+ 2y= 0 near x= 0. Use
the Frobenius method to find the first three nonzero terms of the solution.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x).
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and combine
like terms.
(1−x)
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−2x
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation by expanding the terms and reindexing the
sums.
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2−
∞
∑
n=0
an(n+r)(n+r−1)xn+r−1−2
∞
∑
n=0
an(n+r)xn+r−1+2
∞
∑
n=0
anxn+r= 0
Step 5: Match the powers of xfor each term and write down the ODE for
each power.
n= 0 : a0r(r−1) = 0 ⇒r= 0,1
n= 1 : a1(1 + r)(r)−a0r(r−1) −2a0(r) = 0
n= 2 : a2(2 + r)(1 + r)−a1(1 + r)(r)−2a1(1 + r)+2a0= 0
Step 6: Solve the ODE for n= 0 to find rand check the second solution.
For n= 0, we get the indicial equation r(r−1) = 0, which gives r= 0 or r= 1.
Let’s assume r= 0 and find the second solution.
15
Step 7: Substitute r= 0 into the ODE for n= 1 and solve for a1.
a1= 0
Step 8: Substitute r= 0 into the ODE for n= 2 and solve for a2.
2a2= 0 ⇒a2= 0
Step 9: The first three non-zero terms of the solution are y(x) = a0+a1x+
a2x2=a0. Therefore, the solution of the differential equation is y(x) = a0.
Question 16
Question
Consider the differential equation:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions near the
regular singular point x= 0.
Solution
Let’s assume a solution of the form y(x) = ∑∞
n=0 anxn+r. The derivatives of
y(x)are:
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Substitute y(x),y′(x), and y′′ (x)back into the differential equation:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Group terms by powers of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
Now, set the exponents of xto be equal in each term:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
16
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2 −
∞
∑
n=0
anxn+r= 0
We would then collect like terms with nindexes.
Question 17
Question
Use the Frobenius method to find the general solution to the following differen-
tial equation:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the given differential equation in the form x2y′′ +p(x)y′+q(x)y= 0
where p(x)and q(x)are functions of x.
x2y′′ +xy′−4y= 0
Step 2: Assume a solution of the form y=∑∞
n=0 anxn+r, where ris the
root of the indicial equation.
Step 3: Find the first and second derivatives of y.
y′=
∞
∑
n=0
an(n+r)xn+r−1
y′′ =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 4: Substitute y,y′, and y′′ back into the differential equation.
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(∞
∑
n=0
an(n+r)xn+r−1)−4(∞
∑
n=0
anxn+r)= 0
Step 5: Simplify the equation and collect like terms.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 6: Combine the sums into a single series equation.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
17
∞
∑
n=0
[an(n+r)(n+r−1) + an(n+r)−4an]xn+r= 0
Step 7: Set the coefficients of xn+requal to zero.
an(n+r)(n+r−1) + an(n+r)−4an= 0
Step 8: Solve the indicial equation r(r−1) + r−4=0to find the roots r1
and r2.
Step 9: Once you find the roots r1and r2, determine the corresponding
recurrence relation for the equation.
Step 10: Use the recurrence relation and the initial conditions to find the
coefficients an.
Step 11: Write the general solution using the calculated coefficients.
Question 19
Question
Find the Frobenius series solution of the differential equation (x2−3x+ 2)y′′ −
xy′+y= 0 about the regular singular point x= 1.
Solution
Step 1: Find the indicial equation.
Assume a Frobenius series solution of the form y=∑∞
n=0 an(x−1)n+r.
Substituting this into the differential equation, we get:
(x2−3x+2)
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r−2]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 2: Simplify the equation.
Rearranging the terms, we have:
∞
∑
n=0
[(n+r)(n+r−1)an(x−1)n+r]−3
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+2
∞
∑
n=0
[an(x−1)n+r]−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1]+
∞
∑
n=0
an(x−1)n+r= 0
Step 3: Combine terms with the same power of (x−1).
Expanding out the sums and collecting like terms, we can write the resulting
equation in index form:
∞
∑
n=0
[(n+r)(n+r−1)−3(n+r)+2]an(x−1)n+r−x
∞
∑
n=0
[(n+r)an(x−1)n+r−1] = 0
Step 4: Find the indicial roots.
18
The indicial equation is obtained by equating the coefficient of a0(x−1)rto
zero:
(r)(r−1) −3r+ 2 = 0
Solving this quadratic equation gives the roots r= 1 and r= 2.
Therefore, the indicial roots are r1= 1 and r2= 2.
Question 20
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Calculate the first and second derivatives of y(x):
y′(x) =
∞
∑
n=0
an(n+r)xn+r−1
y′′ (x) =
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 3: Substitute y(x),y′(x), and y′′ (x)into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
Step 4: Simplify the equation and reindex the sums to have a single sum:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 3
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
anxn+r= 0
becomes
∞
∑
n=0
(an(n+r)(n+r−1) + 3an(n+r)−an)xn+r= 0
Step 5: Set the coefficient of each power of xto zero to form indicial equation:
an(n+r)(n+r−1) + 3an(n+r)−an= 0
n(n+ 2r−1)an= 0
19
Step 6: Solve for the roots of nto determine the values of r:
n(n+ 2r−1) = 0
This gives the roots n= 0 and n= 1 −2r.
Step 7: Case 1: When n= 0, we have r= 1/2.
Step 8: Case 2: When n= 1 −2r, we have r=−(1/2).
Step 9: General solution: The general solution will be of the form:
y(x) = c1x1/2+c2x−1/2
where c1and c2are arbitrary constants.
Question 21
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +xy′−4y= 0
Solution
Step 1: Write the differential equation in standard form around the regular
singular point x= 0.
The given equation is: x2y′′ +xy′−4y= 0
Step 2: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the power series into the differential equation to get:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
Step 3: Multiply out the terms and combine like powers of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−4
∞
∑
n=0
anxn+r= 0
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r−
∞
∑
n=0
4anxn+r= 0
Step 4: Set the coefficients of xn+rto zero to find the indicial equation.
a0r(r−1) = 0 (coefficients of xr)
20
a1(r+ 1)r+a0r= 0 (coefficients of x1+r)
Step 5: Solve the indicial equation to find the roots r1and r2.
r1= 2, r2=−2
Step 6: Write the general solution with r1and r2.
y(x) = c1x2+c2x−2
Question 22
Question
Find the Frobenius series solution to the differential equation:
x2y′′ +xy′−(2x2+ 1)y= 0
at the regular singular point x= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ +1
xy′−(2
x2+1
x2)y= 0
y′′ +1
xy′−2
x2y−1
xy= 0
Step 2: Assume the solution is in the form of a power series:
y=
∞
∑
n=0
anxn+r
Step 3: Differentiate yto find y′and y′′ :
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y,y′, and y′′ into the differential equation and combine
terms:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1−2
x2
∞
∑
n=0
anxn+r−1
x
∞
∑
n=0
anxn+r= 0
21
Step 5: Combine the series and factor out terms with the same power of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an]xn+r−2−
∞
∑
n=0
2anxn+r−2−
∞
∑
n=0
anxn+r−1= 0
Step 6: Equate all coefficients of xn+r−2,xn+r−1, and xn+rto zero to find
the recurrence relation for an:
n(n+ 2r−1)an+ (n+r)an−2an= 0
r(r−1)an+ (2r−1)an−2an= 0
(n+r+ 1)an+1 = (2 −n)an
Step 7: Use the recurrence relation to determine the coefficients anto obtain
the Frobenius solution.
Question 23
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′−y= 0
Solution
Step 1: Write the differential equation as a power series. Let y=∑∞
n=0 anxn+r.
Then, we have:
y′=
∞
∑
n=0
(n+r)anxn+r−1
y′′ =
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Substitute y,y′, and y′′ into the differential equation:
x2
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x
∞
∑
n=0
(n+r)anxn+r−1−
∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+
∞
∑
n=0
(n+r)anxn+r−
∞
∑
n=0
anxn+r= 0
22
Step 3: Adjust the indices of summation. To combine all terms into one
summation, set n=m−rin the first two summations:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an−an]xn+r= 0
Step 4: Combine terms and simplify.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 5: Set the coefficient of anxn+rto zero.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 6: Find the roots of the indicial equation. The roots are found by
solving n2+ (2r+ 1)n+ (r2+r−1) = 0. The roots are n=−rand n=r−1.
Step 7: Determine the recurrence relation. Substitute n=−rinto the
equation:
a−r= 0
Substitute n=r−1into the equation:
ar−1=−1
2r(r+ 1)ar−1
Step 8: General solution. The general solution is given by:
y(x) = c1x−r+c2xr−1
Question 24
Question
Find two linearly independent power series solutions for the differential equation
x2y′′ +xy′+ (x2−2)y= 0 at x= 0 using the Frobenius method.
Solution
To find the power series solutions at x= 0, we assume solutions of the form y=
∑∞
n=0 anxn+r. Let’s substitute this assumption into the differential equation
and solve for rand the coefficients an.
Step 1: Find the indicial equation Substitute y=∑∞
n=0 anxn+rinto
the differential equation:
x2
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+x
∞
∑
n=0
an(n+r)xn+r−1+(x2−2)
∞
∑
n=0
anxn+r= 0
23
Step 2: Simplify the equation Rearrange terms and combine like powers
of x:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
anxn+r+2−2
∞
∑
n=0
anxn+r= 0
Step 3: Group terms by powers of xGroup terms with the same power
of x:
∞
∑
n=0
(an(n+r)(n+r−1) + an(n+r)−2an)xn+r+
∞
∑
n=0
anxn+r+2 = 0
Step 4: Set the coefficients of each power of xto zero For the first
sum to be zero, the coefficient of xn+rmust be zero:
an(n+r)(n+r−1) + an(n+r)−2an= 0
n(n+ 2r−1)an= 0
Step 5: Find the roots of the indicial equation Setting n(n+2r−1) =
0, we find two possible roots: r= 0 and r=1
2.
Therefore, the two linearly independent solutions are y1(x) = ∑∞
n=0 anxn
and y2(x) = ∑∞
n=0 anxn+1
2where anare undetermined coefficients.
Question 25
Question
Find the Frobenius series solution to the differential equation:
x2y′′ + 2xy′+ (x2+ 1)y= 0,where y=
∞
∑
n=0
anxn+r.
Solution
Step 1: Find rand the indicial equation.
The given differential equation can be written in the form:
x2y′′ + 2xy′+ (x2+ 1)y= 0
Substitute y=∑∞
n=0 anxn+rback into the differential equation and equate
coefficients of like powers of xto get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+ 2
∞
∑
n=0
an(n+r)xn+r+
∞
∑
n=0
an(xn+2 +xn) = 0
24
Simplify and rewrite in terms of a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 2an(n+r) + an−2]xn+r= 0
Equate the coefficients of xr−2to zero to find r:
[(r)(r−1) + 2(r)]a0= 0
r2+r+ 2r= 0
r2+ 3r= 0
r(r+ 3) = 0
r= 0,−3
Therefore, the possible values for rare r= 0 and r=−3.
Step 2: Find a solution for r= 0.
For r= 0, the recursive relation becomes:
an=−an−2
n(n−1)
Starting with a0and a1:
a0=C1, a1=−a−1
1=−C2(let C1, C2be constants)
Therefore, the series solution for r= 0 is:
y1(x) = C1+C2x
Step 3: Find a solution for r=−3.
For r=−3, the recursive relation becomes:
an=−an−2
(n−3)(n−2),starting with a0=C3, a1=C4, a2=C5
Therefore, the series solution for r=−3is:
y2(x) = C3x−3+C4x−2+C5x−1
The general solution of the differential equation is:
y(x) = c1x−3+c2x−2+c3x+c4x3+c5x4+· · ·
25
