MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Frobenius method
Question Bank - Set 1
Liberty University
Question 1
Question
Find the general solution of the differential equation
x2y′′ +x(x−1)y′−y= 0
at x= 0 using the Frobenius method.
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+s.
Step 2: Calculate the first and second derivatives of y(x).
y′=∞
∑
n=0
(n+s)anxn+s−1
y′′ =∞
∑
n=0
(n+s)(n+s−1)anxn+s−2
Step 3: Substitute y(x),y′, and y′′ into the differential equation.
x2(∞
∑
n=0
(n+s)(n+s−1)anxn+s−2)+x(∞
∑
n=0
(n+s)anxn+s−1)−(∞
∑
n=0
anxn+s)= 0
Step 4: Simplify the terms in the equation.
∞
∑
n=0
(n+s)(n+s−1)anxn+s+∞
∑
n=0
(n+s)anxn+s−∞
∑
n=0
anxn+s= 0
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
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Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
19
Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
22
Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
24
Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
25
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
14
Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
19
Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
22
Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
24
Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
25
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
14
Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
19
Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
22
Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
24
Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
25
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
14
Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
19
Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
22
Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
24
Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
25
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
14
Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
19
Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
22
Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
24
Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
25
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
14
Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
19
Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
22
Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
24
Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
25
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
14
Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
19
Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
22
Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
24
Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
25
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
14
Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
19
Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
22
Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
24
Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
25
Step 5: Rewrite the equation with a single summation.
∞
∑
n=0
[(n+s)(n+s−1) + (n+s)−1]anxn+s= 0
Step 6: Set the coefficients of xn+sto zero to get the indicial equation.
(n+s)(n+s−1) + (n+s)−1 = 0
n2+ 2ns +s2−n+s−1 = 0
Step 7: Solve the indicial equation to find the roots s.
s2+s−1 = 0
Using the quadratic formula, s=−1±√5
2.
Step 8: Write out the two cases for sand find the corresponding recurrence
relation for anin each case. For s=−1+√5
2, we have: Indicial equation: (n+
s)(n+s−1) + (n+s)−1 = n2+ (2s−1)n+s2+s−1 = 0. Recurrence relation:
an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
For s=−1−√5
2, we have: Indicial equation: (n+s)(n+s−1) + (n+s)−1 =
n2+(2s−1)n+s2+s−1 = 0. Recurrence relation: an=−(n+s−1)an−1
(n+s)(n+s−1)+(n+s)−1.
Step 9: Find the first few terms of the series solutions for each case. For
s=−1+√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
For s=−1−√5
2, we have:
a0=a0, a1=−a0
s+ 1, a2=a0
2(s+ 1)(s+ 2)
Step 10: Use the general form of the solution y(x) = ∑∞
n=0 anxn+s
Question 2
Question
Consider the differential equation given by:
x2y′′ +xy′+ (x2−1)y= 0
Use the Frobenius method to find two linearly independent solutions around
the regular singular point x= 0.
2
Solution
To solve this differential equation using the Frobenius method, we assume the
solutions can be expressed as power series near the regular singular point x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and anare constants to be deter-
mined for all n.
Step 1: Differentiate y(x).
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
Step 2: Differentiate y′(x).
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 simplify.
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
Step 4: Rearrange terms in the equation 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 −∞
∑
n=0
anxn+r= 0
Step 5: Set the coefficient of each power of xto zero, starting with the
lowest power.
n= 0 : a0(r)(r−1)xr= 0
n= 1 : a1(1 + r)(1 + r−1)x1+r+a1(1 + r)x1+r−a1x1+r= 0
Step 6: Solve for rby setting the coefficient of xrto zero. From the n= 0
equation, we see that r= 0 or r= 1.
The case r= 0 corresponds to the ordinary point, while r= 1 corresponds
to the regular singular point.
Thus, we set r= 1 and proceed with finding the solution.
Next, we substitute r= 1 back into our general solution form:
y(x) = ∞
∑
n=0
anxn+1
This completes the solution finding process.
3
Question 3
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ +x(3x−1)y′+ (x−1)y= 0
Solution
Step 1: Assume the solution has the form y=∑∞
n=0 anxn+r.
Step 2: Calculate the derivatives y′and y′′.
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation to obtain:
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+x(3x−1) ∞
∑
n=0
(n+r)anxn+r−1+(x−1) ∞
∑
n=0
anxn+r= 0
Step 4: Simplify by factoring out the xr−2term:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
3(n+r)anxn+r−∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1) + 3(n+r)−(n+r)+1−1]anxn+r= 0
Step 6: Equate the coefficient of each power of xto zero to obtain indicial
equations:
(n+r)(n+r−1) + 3(n+r)−(n+r) = 0
n2+ 2nr +r2+ 3n+ 3r−n−r= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
Step 8: Depending on the nature of the roots, find the recurrence relation
for anand the general solution for y.
Question 4
Question
Consider the differential equation (x2−1)y′′ −xy′+y= 0 near x= 1. Using
the Frobenius method, find the general solution in the form of a power series
near x= 1.
4
Solution
Step 1: Rewrite the differential equation in standard form near x= 1 by making
the substitution x= 1 + t.
(x2−1)y′′ −xy′+y= 0
Substitute x= 1 + t:
((1 + t)2−1)y′′ −(1 + t)y′+y= 0
t2y′′ + 2ty′−y′+ 2y−ty′+y= 0
t2y′′ + (2t−1)y′+ (2 −t)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 antn.
y=a0+a1t+a2t2+a3t3+. . .
Step 3: Differentiate yto find y′and y′′.
y′=∞
∑
n=0
an(n+ 1)tn
y′′ =∞
∑
n=0
an(n+ 2)(n+ 1)tn
Step 4: Substitute y,y′, and y′′ into the differential equation.
∞
∑
n=0
antn(n+ 2)(n+ 1)t2+ (2t−1) ∞
∑
n=0
an(n+ 1)tn+ (2 −t)∞
∑
n=0
antn= 0
Step 5: Equate coefficients of like powers of tto obtain a recurrence relation
for an.
a0(2)(1) + (2a2−a1) = 0
a1(3) + (6a3−2a2) = 0
(n+ 2)(n+ 1)an+2 + (2n+ 1)an+1 −an= 0
Step 6: Solve the recurrence relation to find the general solution.
Question 5
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ + 2xy′+ (x2−1)y= 0
5
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r, so y′(x) =
∑∞
n=0(n+r)anxn+r−1and y′′(x) = ∑∞
n=0(n+r)(n+r−1)anxn+r−2.
Step 2: Substitute the expressions for y(x),y′(x), and y′′(x)into the differ-
ential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+2x(∞
∑
n=0
(n+r)anxn+r−1)+(x2−1) (∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+2 ∞
∑
n=0
(n+r)anxn+r+∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
Step 4: Combine like terms and factor out xr:
∞
∑
n=0
[(n+r)(n+r−1) + 2(n+r)−1]anxn+r= 0
Step 5: Equate the coefficients of xn+rto zero:
(n+r)(n+r−1) + 2(n+r)−1 = 0
n2+ (2r−1)n+r2−r−1 = 0
Step 6: Solve for the indicial equation: The roots of the indicial equation
n2+ (2r−1)n+r2−r−1 = 0 are r=−1and r= 1.
Step 7: If r=−1: For r=−1, we have the solution:
y1(x) = a0x−1+a1x0+∞
∑
n=2
anxn−1
Step 8: If r= 1: For r= 1, we have the solution:
y2(x) = a0x1+a1x2+∞
∑
n=2
anxn+1
Therefore, the general solution is:
y(x) = c1x−1+c2x+∞
∑
n=2
cnxn−1
Question 6
Question
Find the Frobenius series solution of the differential equation x2y′′+xy′+(2x2−
1)y= 0 about x= 0.
6
Solution
Step 1: Write the differential equation in standard form:
y′′ +1
xy′+(2x2−1
x2)y= 0
Step 2: Assume a Frobenius series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Compute 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′′ back into the differential equation and
simplify:
∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+1
x
∞
∑
n=0
(n+r)anxn+r−1+(2x2−1
x2)∞
∑
n=0
anxn+r= 0
Step 5: Combine the terms with the same powers of x:
∞
∑
n=0
[(n+r)(n+r−1)an+ (n+r)an+ (2 −1/x2)an]xn+r−2= 0
Step 6: Set the coefficient of each power of xto zero to get the indicial
equation:
(r+n)(r+n−1)an+ (r+n)an+ 1/x2an= 0
Step 7: Solve the indicial equation to find the roots r1and r2.
r1= 1, r2=−1
Step 8: Check for non-negative integer differences between the roots to de-
termine the form of the series solution.
Since r1−r2= 2, there may be a logarithmic term in the series solution.
Step 9: Write down the general form of the series solution using the roots
obtained:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
bnxn−1ln(x)
Therefore, the Frobenius series solution of the given differential equation
about x= 0 is y(x) = c1y1(x) + c2y2(x), where y1(x) = ∑∞
n=0 anxn+1 and
y2(x) = ∑∞
n=0 bnxn−1ln(x).
7
Question 8
Question
Use the Frobenius method to find the general solution of the differential equa-
tion:
x2y′′ −xy′+ (x2−1)y= 0
Solution
Step 1: Assume a power series solution of the form y(x) = ∑∞
n=0 anxn+r.
Therefore, the derivatives of y(x)are:
y′(x) = ∞
∑
n=0
an(n+r)xn+r−1
y′′(x) = ∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
Step 2: Substitute y(x),y′(x), and y′′(x)into the ODE and combine terms:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
(x2−1)anxn+r= 0
Step 3: Rearrange the terms and simplify:
∞
∑
n=0 [an(n+r)(n+r−1) −an(n+r)+(x2−1)an]xn+r= 0
Step 4: Set the coefficient of each power of xto zero:
an(n+r)(n+r−1) −an(n+r)+(x2−1)an= 0
Step 5: Solve for the indicial equation by setting n= 0:
a0r(r−1) −a0r= 0
r2−r= 0
r(r−1) = 0
Step 6: We have two roots for the indicial equation: r1= 1 and r2= 0.
Now we consider the case r= 1.
Continuing in this way, complete the solution by finding the general form of
y(x)using the Frobenius method.
8
Question 9
Question
Find the general solution to the following differential equation using the Frobe-
nius method:
x2y′′ +xy′+ (x2−1)y= 0
Solution
Step 1: Write the equation in standard form. The given differential equation
can be rewritten as:
y′′ +1
xy′+(1−1
x2)y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Step 3: Calculate the first few derivatives. The first and second derivatives
of yare:
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute the power series and its derivatives into the differential
equation. Substitute y,y′, and y′′ into the differential equation and collect
terms with the same power of x.
Step 5: Set the coefficients of each power of xto zero. Set the coefficients of
each power of xto zero to form a recurrence relation for an.
Step 6: Solve the recurrence relation. Solve the recurrence relation to find
the values of rand an.
Step 7: Write the general solution. The general solution is given by:
y(x) = c1x+c2x2(1−1
2!)
where c1and c2are constants of integration.
Question 10
Question
Consider the differential equation:
x2y′′ + 4xy′+ (x2−1)y= 0
Use the Frobenius method to find at least one solution near x= 0.
9
Solution
We will assume the solution is of the form y(x) = ∑∞
n=0 anxn+r.
Step 1: Find the indicial equation by substituting y(x)into the differential
equation.
x2∞
∑
n=0
(n+r)(n+r−1)anxn+r−2+ 4x∞
∑
n=0
(n+r)anxn+r−1
+ (x2−1) ∞
∑
n=0
anxn+r= 0
Expanding and collecting terms, we have:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
4(n+r)anxn+r
+∞
∑
n=0
anxn+r+2 −∞
∑
n=0
anxn+r= 0
Simplifying, we get:
∞
∑
n=0
[(n+r)(n+r−1) + 4(n+r)+(r2−1)]anxn+r= 0
This implies that each term must be zero, so:
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0
This is the indicial equation.
Step 2: Solve the indicial equation to find the roots r1and r2. Setting
(n+r)(n+r−1) + 4(n+r)+(r2−1) = 0, we simplify to get:
r2= 0
So, r1=r2= 0.
Step 3: Substitute r= 0 back into the assumed form of the solution to get
y(x):
y(x) = ∞
∑
n=0
anxn
We will substitute this form back into the original differential equation and
solve for anto find the first few terms of the series solution.
Question 11
Question
Use the Frobenius method to find a series solution to the differential equation:
x2y′′ −xy′+y= 0
near the regular singular point x= 0.
10
Solution
Let’s assume the solution to the given differential equation is of the form y(x) =
∑∞
n=0 anxn+r. Substituting into the differential equation, we have:
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
Simplify the terms and combine them:
∞
∑
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
This equation holds for all x, so the coefficient of each power of xmust be
zero. Thus, we obtain the indicial equation by setting the coefficient of xrto
zero:
a0r(r−1) = 0
This gives two possible values for r:r= 0 or r= 1.
We will first consider r= 0. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn. Substitute r= 0 back into the differential
equation and simplify to find the series solution.
Next, we will consider r= 1. In this case, the general form of the series
solution is y(x) = ∑∞
n=0 anxn+1. Perform a similar substitution and simplify to
find the series solution in this case as well.
Question 12
Question
Consider the differential equation:
x2y′′ +xy′−y= 0
Use the Frobenius method to find at least two non-constant solutions near
x= 0.
Solution
Step 1: Write the equation in the form x2y′′ +bxy′+cy = 0. Here, b= 1 and
c=−1.
Step 2: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
11
Step 3: Find the first few terms of y′(x)and y′′(x).
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 4: Substitute y(x),y′(x), and y′′(x)into the differential equation and
simplify.
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 5: Combine terms and make the exponents consistent.
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 6: Simplify the combined terms.
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 7: Equate the coefficients of xn+rto zero to find the indicial equation.
(n+r)(n+r−1) + (n+r)−1 = 0
n2+ 2nr +r2+n+r−1 = 0
Step 8: Solve the indicial equation to find the roots r1and r2.
Step 9: Substitute the values of r1and r2back into the assumed solution to
generate two linearly independent solutions.
Question 13
Question
Use the Frobenius method to find a series solution for the differential equation
(x2−2)x2y′′ +xy′+ (x2−1)y= 0 about the ordinary point x= 0.
Solution
Step 1: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r
where ris the initial guess for the root of the indicial equation.
12
Step 2: Calculate the first and second derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)into the differential equation, and
group terms with the same power of xtogether.
Step 4: Set the coefficient of each power of xto zero to obtain recurrence
relations for the an’s.
Step 5: Solve the recurrence relations to find expressions for the an’s. Start
with a0and a1, and then find a general formula for anin terms of an−2.
Step 6: Use the initial guess for rto find the first few non-zero an’s, then
generalize the solution for arbitrary n.
Step 7: Write out the final series solution for y(x)with all the an’s found in
the previous steps.
Step 8: Optionally, simplify the series solution if possible, and discuss the
convergence of the solution for different values of x.
Question 14
Question
Consider the differential equation (x2−1)y′′ −2xy′+ 2y= 0. Determine the
solutions near x= 1 using the Frobenius method.
Solution
Step 1: Write the equation in standard form Rewrite the differential
equation in the form (x−x0)2y′′ + (x−x0)p(x)y′+q(x)y= 0, expanding
around x0= 1. The given equation becomes:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Find the indicial equation Assume y=∑∞
n=0 an(x−1)n+r.
Substituting this into the equation and equating coefficients of powers of (x−1),
we get the indicial equation:
(r)(r−1)a0(x−1)r−2+[(r+1)(r)(a1)−2a0](x−1)r−1+terms independent of a0, a1= 0.
Step 3: Solve the indicial equation The indicial equation r(r−1) = 0
has roots r1= 0 and r2= 1.
Step 4: Write the first solution For r= 0, the general solution is:
y1(x) = a0+a1(x−1) + ∞
∑
n=2
an(x−1)n.
13
Step 5: Write the second solution (if necessary) As r1is a non-
repeated root, the second solution is of the form
y2(x) = ky1(x) ln |x−1|+∞
∑
n=0
bn(x−1)n+1.
Step 6: Apply Frobenius method to find the second solution Sub-
stitute y2(x)into the differential equation and determine the recurrence relation
for bn.
Step 7: Solve for coefficients Solve for the coefficients anand bnto find
the specific solutions.
Thus, the solutions near x= 1 for the given differential equation using the
Frobenius method have been determined.
Question 15
Question
Use the Frobenius method to find the general solution to the differential equation
x2y′′ + 3xy′+ (x2−1)y= 0 near x= 0.
Solution
Step 1: Rewrite the differential equation First, we rewrite the given dif-
ferential equation in the form of a Frobenius series near x= 0:
y(x) = ∞
∑
n=0
anxn+r
where ris a constant to be determined, and the coefficients anare to be found.
The differential equation becomes:
x2y′′ + 3xy′+ (x2−1)y= 0
which can be written as:
x2∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+3x∞
∑
n=0
an(n+r)xn+r−1+(x2−1) ∞
∑
n=0
anxn+r= 0
Step 2: Simplify the equation Expanding and simplifying each term, we
get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+3 ∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+2−∞
∑
n=0
anxn+r= 0
14
Step 3: Combine like terms Since all terms have the same power of x,
we can combine them into a single series:
∞
∑
n=0
[an(n+r)(n+r−1) + 3an(n+r) + an−2]xn+r−anxn+r= 0
Step 4: Determine the recurrence relation We set the coefficient of
each power of xto zero, giving us the recurrence relation:
an(n+r)(n+r−1) + 3an(n+r) + an−2= 0
This relation will help us find the values of anin terms of an−2.
Step 5: Solve for rBy substituting n= 0 into the recurrence relation, we
get:
a0(r)(r−1) + 3a0(r) = 0
r2−r+ 3r= 0
r2+ 2r= 0
r(r+ 2) = 0
So, we have r= 0 or r=−2.
Step 6: For r= 0 For r= 0, the recurrence relation becomes:
an(n)(n−1) + 3an(n) + an−2= 0
n(n−1)an+ 3nan+an−2= 0
Step 7: For r=−2For r=−2, the recurrence relation becomes:
an(n−2)(n−3) + 3an(n−2) + an−2= 0
(n−2)(n−3)an+ 3(n−2)an+an−2= 0
Hence, we have found the general solution to the differential equation using
the Frobenius method.
Question 16
Question
Find a Frobenius series solution to the differential equation:
x2y′′ +xy′−y= 0
at x= 0.
15
Solution
Step 1: Write down the differential equation in standard form. Let’s rearrange
the given differential equation to make it easier to apply the Frobenius method:
y′′ +1
xy′−1
x2y= 0
Step 2: Assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r.
Substitute the series solution into the differential equation:
∞
∑
n=0
an(n+r)(n+r−1)xn+r−2+1
x
∞
∑
n=0
an(n+r)xn+r−1−1
x2
∞
∑
n=0
anxn+r= 0
Step 3: Simplify the equation. To simplify, we first write out the first few
terms of each series. Then, we re-index the terms and combine like powers of x.
Step 4: Find the indicial equation. Setting the coefficient of xrin the above
equation to 0 gives us the indicial equation:
r(r−1)a0+ra0−a0= 0
Solving this equation gives us the roots r= 0 and r= 1.
Step 5: Derive recurrence relation. Substitute r= 0 into the equation:
a0(0 −1)a0+ 0a0−a0= 0
Which simplifies to −a0= 0, meaning a0= 0.
Step 6: Write down the general solution. The general solution is given by:
y(x) = c1x+c2x2
Question 17
Question
Consider the differential equation (x2−4)y′′ +xy′+y= 0. Use the Frobenius
method to find two linearly independent power series solutions around x= 0.
Solution
Step 1: Determine the Indicial Equation
The differential equation is of the form x2y′′ +p(x)xy′+q(x)y= 0, where
p(x)and q(x)are analytic around x= 0. The indicial equation is given by:
r(r−1) + p0r+q0= 0
where p0and q0are the coefficients of xand the constant term, respectively.
16
In this case, p(x) = 1
x,q(x) = 1
x2−4,p0= 1, and q0= 1. Thus, the indicial
equation is:
r(r−1) + r+ 1 = r2= 0
which yields r= 0.
Step 2: Find the First Solution
Assume a solution of the form y(x) = ∑∞
n=0 anxn+r. Substituting this into
the differential equation and expanding, we get:
(x2−4) ∞
∑
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
(r(r−1) −4r)a0xr−2+∞
∑
n=1
[(n+r)(n+r−1)an+ (n+r)an]xn+r−1
+∞
∑
n=0
anxn+r= 0
Now we equate the coefficients of the terms with the same power of xto
zero. For n= 0, we have:
r(r−1) −4r+a0= 0
Given r= 0, this simplifies to −4a0= 0, so a0= 0.
For n≥1, we have the recurrence relation:
an=4n−1
n(2n−1)an−1
Thus, the first solution is:
y1(x) = a0+a1x=a1x
Step 3: Find the Second Solution
To find the second linearly independent solution, we use the reduction of
order method. Assume the second solution has the form y2(x) = uy1(x), where
uis a function of x.
Let y2(x) = u(a1x) = a1xu(x), and substitute this into the differential
equation:
(a1x)2−4 = 0
Solving this equation gives the values of xfor which the solution can exist.
Therefore, the two linearly independent power series solutions around x= 0
are y1(x) = a1xand y2(x)where x=±2.
17
Question 18
Question
Consider the differential equation (1−x2)y′′−2xy′+2λy = 0 for λ∈R. Use the
Frobenius method to find two linearly independent solutions near x= 0 when
λ= 1.
Solution
Step 1: Write the differential equation in standard form.
y′′ −2x
1−x2y′+2λ
1−x2y= 0
Step 2: Assume a power series solution of the form y=∑∞
n=0 anxn+r.
Substitute this into the differential equation to find the indicial equation:
(n+r)(n+r−1)anxn+r−2−2x(n+r)anxn+r−1+ 2λanxn+r= 0
⇒n(n−1)anxn+r−2+r(r−1)anxn+r−2−2nranxn+r+ 2λanxn+r= 0
Step 3: Simplify the indicial equation and solve for r:
r(r−1)a0xr−2+ (n2+r(r−2) −2nr + 2λ)anxn+r= 0
Comparing coefficients of xn+r−2, we get:
r(r−1)a0= 0 =⇒r= 0 or r= 1
For r= 0:
n(n−1)an+ 2λan= 0 =⇒n(n−1) + 2λ= 0
⇒n2−n+ 2 = 0 ⇒n=1±√1−4∗2
2=1±i√7
2
For r= 1:
(n+ 1)(n)an−2nran+ 2λan= 0 =⇒n2+n−2n+ 2λ= 0
⇒n2−n+ 2λ= 0
Since r= 0 gives complex roots, we choose r= 1.
Step 4: Equate the coefficients of powers of xto find the recurrence relation.
n2−n+ 2λ= 0 =⇒n2−n+ 2 = 0
⇒n=1±√1−4∗2
2=1±i√7
2
Thus, the two linearly independent solutions are:
y1(x) = ∞
∑
n=0
anxn+1 =a0x+a1x2+a2x3+···
y2(x) = ∞
∑
n=0
anxn+1 =b0x+b1x2+b2x3+···
18
Question 19
Question
Find the Frobenius series solution to the differential equation x2y′′ +xy′+ (x2−
1)y= 0 about x= 0.
Solution
1. We assume a Frobenius series solution of the form y(x) = ∑∞
n=0 anxn+r,
where ris the initial root of the indicial equation.
2. First, we need to find the indicial equation. Substitute y(x)into the
differential equation to get:
x2(∞
∑
n=0
anxn+r)′′
+x(∞
∑
n=0
anxn+r)′
+ (x2−1) ∞
∑
n=0
anxn+r= 0
3. Simplifying the expression above, we get:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r+∞
∑
n=0
anxn+r+1−∞
∑
n=0
anxn+r= 0
4. Now, let’s combine the series terms and check the coefficients of each power
of xto find the indicial equation. We equate the coefficient of xr−1to 0
to determine r.
5. Thus, the indicial equation is:
r(r−1)a0= 0
which gives us r= 0 or r= 1.
6. We will now find the recurrence relation for an. Substituting r= 0, the
equation simplifies to
an(n)(n−1) + an−1(n)−an−1= 0 for n≥1
7. Rearranging the terms, we get the recurrence relation:
an=an−1
n+an−1
n(n−1) for n≥1
8. Thus, the Frobenius series solution is y(x) = a0+a1x.
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Question 20
Question
Use the Frobenius method to find a general solution for the differential equation:
x2y′′ +x(1 −x)y′+ (2x−1)y= 0
Solution
1. We first assume a solution of the form y(x) = ∑∞
n=0 anxn+r, where ris
an unknown constant to be determined and anare the coefficients to be
found.
2. Taking the first and second derivatives of y(x), we have:
y′=∞
∑
n=0
an(n+r)xn+r−1
y′′ =∞
∑
n=0
an(n+r)(n+r−1)xn+r−2
3. Substituting y, y′, y′′ back into the differential equation, we get:
x2(∞
∑
n=0
an(n+r)(n+r−1)xn+r−2)+x(1−x)(∞
∑
n=0
an(n+r)xn+r−1)+(2x−1) (∞
∑
n=0
anxn+r)= 0
4. Simplifying the equation and combining like terms, we obtain:
∞
∑
n=0
an(n+r)(n+r−1)xn+r+∞
∑
n=0
an(n+r)xn+r−∞
∑
n=0
an(n+r)xn+r+1+2x∞
∑
n=0
anxn+r−∞
∑
n=0
anxn+r= 0
5. Now, equating the coefficient of each power of xto zero leads to the indicial
equation and a recurrence relation. Solving these will yield the values of
rand the coefficients an.
Question 21
Question
Find two linearly independent power series solutions for the given differential
equation using the Frobenius method:
x2y′′ + 6xy′+ (x2−1)y= 0
20
Solution
Step 1: Assume a solution of the form y(x) = ∑∞
n=0 anxn+r.
Step 2: Compute the first and second derivatives of y(x):
y′=∞
∑
n=0
(n+r)anxn+r−1
y′′ =∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y,y′, and y′′ into the differential equation and simplify.
Equate coefficients of like powers of xto zero.
This gives:
∞
∑
n=0
[(n+r)(n+r−1)an+ 6(n+r)an−1+ (n+r−2)an−2]xn+r= 0
Step 4: Set the coefficient of each term to zero and solve for rto find the
Indicial Equation.
For xrterms:
n2an+ (2r−1)nan+ (r2−r−6)an= 0
Step 5: Solve the Indicial Equation to find the roots r1and r2.
r1= 1, r2=−1
Step 6: Use the roots r1and r2to find two linearly independent solutions of
the form y1(x) = ∑∞
n=0 anxn+r1and y2(x) = ∑∞
n=0 anxn+r2respectively.
Therefore, the two linearly independent power series solutions are:
y1(x) = ∞
∑
n=0
anxn+1
y2(x) = ∞
∑
n=0
anxn−1
Question 22
Question
Consider the differential equation (x2−1)y′′ −2xy′+2y= 0. Find the Frobenius
series solution near x= 1.
21
Solution
To find the Frobenius series solution near x= 1, we assume the solution has the
form y(x) = (x−1)r∑∞
n=0 an(x−1)n. We substitute this into the differential
equation and solve for rand the coefficients an.
Step 1: Determine the indicial equation
The indicial equation is found by substituting y(x) = ∑∞
n=0 an(x−1)n+r
into the differential equation and equating coefficients of like powers of (x−1)
to zero.
Substitute y(x) = (x−1)r∑∞
n=0 an(x−1)ninto the given differential equa-
tion:
(x2−1)y′′ −2xy′+ 2y= 0.
Step 2: Calculate the derivatives of y(x)
The first and second derivatives of y(x)are:
y′=r(x−1)r−1∞
∑
n=0
an(x−1)n+ (x−1)r∞
∑
n=0
ann(x−1)n−1,
y′′ =r(r−1)(x−1)r−2∞
∑
n=0
an(x−1)n+ 2r(x−1)r−1∞
∑
n=0
ann(x−1)n−1
+ (x−1)r∞
∑
n=0
ann(n−1)(x−1)n−2.
Step 3: Plug derivatives back into the ODE
Substitute y(x),y′, and y′′ into the differential equation to get an expression
involving an.
Step 4: Find the Indicial Equation
Equating the coefficient of the lowest power of (x−1) to zero will give us
the Indicial Equation.
Step 5: Solve the Indicial Equation to Find r
Solve the Indicial Equation to find the two possible roots r1and r2.
Step 6: Express r1and r2in terms of λ
Since the roots of the Indicial Equation are equal, express r1and r2in terms
of the parameter λ.
Step 7: Find the recurrence relation
Using the root parameter λalong with the Indicial Equation, find the recur-
rence relation for the coefficients an.
Step 8: Write the Frobenius series solution
Write the Frobenius series solution for the differential equation near x= 1
as y(x) = c1y1(x) + c2y2(x).
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Question 23
Question
Find the Frobenius series solution about the ordinary point x= 0 for the dif-
ferential equation given by
x2y′′ +x(1 −x)y′−3y= 0.
Solution
To find the 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 chosen later.
Step 1: Find the derivatives of y(x)
First derivative: y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
Second derivative: y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 2: Substitute into the differential equation
Substitute y(x),y′(x), and y′′(x)into the differential equation:
x2(∞
∑
n=0
(n+r)(n+r−1)anxn+r−2)+x(1−x)(∞
∑
n=0
(n+r)anxn+r−1)−3(∞
∑
n=0
anxn+r)= 0
Step 3: Simplify the equation
We simplify the equation by grouping terms with the same power of xto-
gether.
Step 4: Set up the indicial equation
The indicial equation is obtained by setting the coefficient of the lowest
power of xto zero. This gives the indicial equation r(r−1)a0= 0.
Step 5: Find the recursion formula
Set the coefficient of xn+rto zero for n≥1and solve for anto obtain the
recursion formula.
Step 6: Solve for the series solution
Use the recursion formula and the indicial equation to find the values of r
and the series solution y(x)for the differential equation.
Question 24
Question
Use the Frobenius method to find two linearly independent solutions to the
differential equation:
x2y′′ +xy′−y= 0
23
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
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)back into the differential equation.
∞
∑
n=0
anxn+r(n+r)(n+r−1)x2+∞
∑
n=0
anxn+r(n+r)x−∞
∑
n=0
anxn+r= 0
Step 4: Combine the sums and simplify the equation to match terms with
the same power of x.
∞
∑
n=0
an(n+r)(n+r−1)xn+r+2 +∞
∑
n=0
an(n+r)xn+r+1 −∞
∑
n=0
anxn+r= 0
Step 5: Set the sum of coefficients of each power of xto zero.
For xr+2 :a0(r)(r−1) = 0 ⇒r= 0 or r= 1
For xr+1 :a0r(r+ 1) + a1(r+ 1)r= 0
For xr:a1(r+ 1)(r+ 2) + a2(r+ 2)(r+ 1) −a0= 0
Step 6: Solve the indicial equation and find the values of r. For r= 0:
r= 0 ⇒a0(0)(−1) = 0 ⇒a0= 0
For r= 1:
r= 1 ⇒a0(1)(2) = 0 ⇒a0= 0
The two solutions are r= 0 and r= 1.
Step 7: Determine the recurrence relation and find the coefficients anfor
both r= 0 and r= 1. For r= 0:
n(n−1)an+nan−1−an−2= 0
For r= 1:
(n+ 1)(n+ 2)an+2 + (n+ 2)(n+ 1)an+1 −an= 0
Step 8: Use the solutions r= 0 and r= 1 to find the first few terms of the
series solutions for y(x). For r= 0:
y1(x) = a0x0=a0
For r= 1:
y2(x) = a0x1+a1x2=a0x+a1x2
Therefore, the solutions to the differential equation are y1(x) = a0and
y2(x) = a0x+a1x2, where a0and a1are constants.
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Question 25
Question
Find the Frobenius series solution of the differential equation:
x2y′′ +xy′−y= 0
about the regular singular point x= 0.
Solution
Step 1: Write the general form of the Frobenius series solution: Assume a
solution of the form y(x) = ∑∞
n=0 anxn+r, where ris the root of the indicial
equation.
Step 2: Find the derivatives of y(x):
y′(x) = ∞
∑
n=0
(n+r)anxn+r−1
y′′(x) = ∞
∑
n=0
(n+r)(n+r−1)anxn+r−2
Step 3: Substitute y(x),y′(x), and y′′(x)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 4: Simplify the equation and collect terms with the same powers of x:
∞
∑
n=0
(n+r)(n+r−1)anxn+r+∞
∑
n=0
(n+r)anxn+r−∞
∑
n=0
anxn+r= 0
Step 5: Combine all the series into one:
∞
∑
n=0
[(n+r)(n+r−1) + (n+r)−1]anxn+r= 0
Step 6: Compare coefficients of xn+rto obtain the indicial equation:
n2+ 2rn −r2= 0
Step 7: Find the roots of the indicial equation to determine the form of the
series: Solve the quadratic equation n2+ 2rn −r2= 0 to find the roots r1and
r2.
Step 8: Depending on the roots r1and r2, the solution will involve different
cases. Proceed with the Frobenius method accordingly.
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