MATH 334 - DIFFERENTIAL
EQUATIONS - Cauchy-Euler equations
Question Bank - Set 3
Liberty University
Question 1
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xm.
Solution
Step 1: Begin by finding the derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the derivatives into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m(m−1) −3m+ 4 = 0
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
Step 3: Solve for the roots of the characteristic equation.
m−2=0
m= 2
Step 4: Write down the general solution using the roots found.
∴y=c1x2+c2x2ln(x)
Question 2
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using a substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y=xr⇒y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute these derivatives into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0 ⇒x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0 ⇒r2−4r+ 4 = 0
Step 4: Solve the characteristic equation for r.
r2−4r+ 4 = 0 ⇒(r−2)2= 0 ⇒r= 2
Step 5: Write down the general solution.
y=c1x2+c2x2ln x
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln x.
Question 3
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Write the characteristic equation by assuming a solution of the form
y=xr.
The characteristic equation is: r(r−1) −3r+ 4 = 0
⇒r2−4r+ 4 = 0
⇒(r−2)2= 0
2
⇒r= 2
(Repeated root)
Step 2: Write down the general solution for the Cauchy-Euler equation.
y(x) = c1x2+c2x2ln |x|
where c1and c2are arbitrary constants.
Step 3: Therefore, the solution to the Cauchy-Euler equation is:
y(x) = c1x2+c2x2ln |x|
Question 4
Question
Solve the Cauchy-Euler equation (x2−3x+ 2)y′′ −2xy′+ 2y= 0 using the
substitution y=xr.
Solution
Step 1: Find the derivatives y′and y′′.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
(x2−3x+ 2)y′′ −2xy′+ 2y= 0
(x2−3x+ 2)(r(r−1)xr−2)−2x(rxr−1)+2xr= 0
r(r−1)xr−3r(r−1)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3r(r−1)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
r2xr−r(r−3)(r−2)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
Step 4: Factor out common terms.
r2xr−r(r−3)(r−2)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
r2xr−r(r−3)(r−2)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
r2xr−r(r2−5r+ 6)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
3
Step 5: Set the expression equal to zero and solve for r.
r2−r(r2−5r+ 6) + 2r(r−1) = 0
r2−r3+ 5r2−6r+ 2r2−2r= 0
r2+ 5r2+ 2r2−r3−6r−2r= 0
8r2−r3−8r= 0
r(8r−r2−8) = 0
r(r2−8r+ 8) = 0
r(r−4)(r−2) = 0
Step 6: Write down the roots of the characteristic equation. The roots are
r1= 4 and r2= 2.
Step 7: Write down the general solution using the roots.
y(x) = C1x4+C2x2
Question 5
Question
Solve the Cauchy-Euler equation:
x2y′′ −4xy′+ 6y= 0
Solution
Step 1: Assume the solution has the form y=xr. Then find the derivatives of
y.
Step 2: Calculate the first and second derivative of ywith respect to x.
Step 3: Substitute y,y′, and y′′ back into the original differential equation
and simplify.
Step 4: Solve the resulting equation by factoring out common terms and
finding the roots of the characteristic equation.
Step 5: Write down the general solution of the Cauchy-Euler equation.
Step 6: Express the general solution in terms of x.
Step 7: Write the final solution after substituting back for y.
Question 6
Question
Solve the Cauchy-Euler differential equation:
x2y′′ −5xy′+ 6y=x2
4
Solution
Step 1: Write the characteristic equation associated with the Cauchy-Euler
equation using the auxiliary equation m(m−1). For the given differential
equation, the characteristic equation is:
m(m−1) −5m+ 6 = 0
Step 2: Simplify the characteristic equation:
m2−m−5m+ 6 = 0
m2−6m+ 6 = 0
Step 3: Solve the characteristic equation to find the roots m1and m2:
m=6±√62−4∗1∗6
2
m=6±√36 −24
2
m=6±√12
2
m=6±2√3
2
m= 3 ±√3
Step 4: The general solution to the differential equation is then given by:
y(x) = c1xm1+c2xm2
Step 5: Substitute the roots m1= 3 + √3 and m2= 3 −√3 into the general
solution:
y(x) = c1x3+√3+c2x3−√3
Step 6: Therefore, the solution to the Cauchy-Euler differential equation is:
y(x) = c1x3+√3+c2x3−√3
Question 7
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
5
Solution
Step 1: Determine the derivatives of ywith respect to x.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the Cauchy-Euler equation.
x2·r(r−1)xr−2−5x·rxr−1+ 6xr= 0
Step 3: Simplify the equation.
r(r−1)xr−5rxr+ 6xr= 0
Step 4: Factor out xr.
xr(r2−r−5r+ 6) = 0
Step 5: Simplify the expression within the parentheses.
xr(r2−6r+ 6) = 0
Step 6: Factor the quadratic expression r2−6r+ 6 = 0 to solve for r.
(r−3)(r−2) = 0
Step 7: Solve for r.
r= 2 or r= 3
Step 8: Write down the general solution using the roots r1= 2 and r2= 3.
y(x) = c1x2+c2x3
where c1and c2are arbitrary constants.
Question 8
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
6
Solution
To solve the given Cauchy-Euler equation, we will make a substitution y=xm
and then find the values of mthat satisfy the equation.
Step 1: Find the first and second derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m(m−1) −3m+ 4 = 0
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
Step 3: Write down the general solution after finding the roots. Since
(m−2)2= 0, the repeated root is m= 2. Therefore, the general solution is
given by:
y=c1x2+c2x2ln |x|
where c1and c2are arbitrary constants.
Question 9
Question
Solve the Cauchy-Euler equation
x2y′′ −xy′+ 4y= 0.
Solution
Step 1: Let’s assume a solution of the form y=xr. We will differentiate y
accordingly.
Step 2: Compute the first and second derivatives of ywith respect to x.
y′=rxr−1
y′′ =r(r−1)xr−2
7
Step 3: Substitute y,y′, and y′′ into the differential equation.
x2(r(r−1)xr−2)−x(rxr−1)+4xr= 0.
Step 4: Simplify the equation.
r(r−1)xr−rxr+ 4xr= 0.
Step 5: Combine like terms.
r2xr−rxr+ 4xr= 0.
Step 6: Factor out xr.
xr(r2−r+ 4) = 0.
Step 7: Set the expression equal to zero to find possible values of r.
r2−r+ 4 = 0.
Step 8: Solve the quadratic equation r2−r+ 4 = 0 to find the roots.
r=1±p1−4(1)(4)
2(1)
r=1±√−15
2
r=1±i√15
2
Step 9: Thus, the general solution to the Cauchy-Euler equation is
y=c1x1+i√15
2+c2x1−i√15
2,
where c1and c2are arbitrary constants.
Question 10
Question
Find the general solution to the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: Write down the characteristic equation by assuming y=xr. The
characteristic equation is r(r−1)−5r+6 = 0, which simplifies to r2−6r+6 = 0.
Step 2: Solve the characteristic equation. Factoring the quadratic equation
r2−6r+6 = 0 gives (r−3)2−3 = 0, which further simplifies to (r−3+√3)(r−
3−√3) = 0. Thus, we have r= 3 ±√3.
Step 3: Determine the general solution. Since the roots are distinct and real,
the general solution is given by y(x) = c1x3+√3+c2x3−√3, where c1and c2are
arbitrary constants.
8
Question 11
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 given that y1(x) = xis
a solution.
Solution
Step 1: Find the auxiliary equation by assuming y=xr. Step 2: Substitute
y=xrinto the Cauchy-Euler equation. Step 3: Solve the resulting auxiliary
equation to find the roots. Step 4: Form the general solution using the roots
and the original solution y1(x) = x.
Step 1: Assume y=xr. Then, the derivatives are: y′=rxr−1and y′′ =
r(r−1)xr−2.
Step 2: Substitute y=xr,y′=rxr−1, and y′′ =r(r−1)xr−2into the
Cauchy-Euler equation: x2r(r−1)xr−2−3xrxr−1+ 4xr= 0
Step 3: Simplify the equation: r(r−1)xr−3rxr+ 4xr= 0
Factor out xr:xr(r2−4r+ 4) = 0
Solve the auxiliary equation: r2−4r+ 4 = (r−2)2= 0
Thus, we have a repeated root r= 2.
Step 4: The general solution is given by: y(x) = c1xr1+c2xr2, where r1= 2,
r2= 2, and r1=r2= 2.
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is: y(x) = c1x2+c2x2= (c1+c2)x2
Given that y1(x) = xis a solution, we have c1+c2= 1. Therefore, c1=1
2
and c2=1
2.
So, the solution to the Cauchy-Euler equation is y(x) = 1
2x2+1
2x2=x2.
Question 12
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 2y= 0
Solution
To solve the Cauchy-Euler equation, we assume a solution of the form y=xm.
We then substitute yand its derivatives into the differential equation.
Step 1: Find the first and second derivatives of y
y=xm
y′=mxm−1
9
y′′ =m(m−1)xm−2
Step 2: Substitute yand its derivatives into the differential equa-
tion Substitute y,y′, and y′′ into the differential equation:
x2(m(m−1)xm−2)−3x(mxm−1)+2xm= 0
Step 3: Simplify the equation
m(m−1)xm−3mxm+ 2xm= 0
m(m−1) −3m+ 2 = 0
Step 4: Solve for the roots of the characteristic equation
m2−m−3m+ 2 = 0
m2−4m+ 2 = 0
m=4±p(−4)2−4∗1∗2
2
m=4±√16 −8
2
m=4±√8
2
m=4±2√2
2
m= 2 ±√2
Step 5: Write the general solution The general solution is given by:
y(x) = c1x2+√2+c2x2−√2
where c1and c2are constants.
Question 13
Question
Solve the Cauchy-Euler equation
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand find the derivatives of
y(x). Step 2: Substitute y(x) and its derivatives into the differential equation.
Step 3: Simplify the equation and solve for r. Step 4: Determine the general
solution for the Cauchy-Euler equation.
10
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 18y= 0
Solution
Step 1: Write down the characteristic equation by substituting y=xminto the
differential equation.
m(m−1)xm−5mxm+ 18xm= 0
Step 2: Simplify the characteristic equation.
m2−6m+ 18 = 0
Step 3: Find the roots of the characteristic equation using the quadratic
formula.
m=6±p(−6)2−4∗1∗18
2=6±√36 −72
2=6±i√36
2= 3 ±3i
Step 4: Write down the general solution using the roots found in Step 3.
y(x) = C1x3cos(3 ln(x)) + C2x3sin(3 ln(x))
where C1and C2are arbitrary constants.
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: We start by substituting y=xrinto the Cauchy-Euler equation x2y′′ −
5xy′+ 6y= 0:
x2(r)(r−1)xr−2−5x(r)xr−1+ 6xr= 0
Step 2: Simplify the equation by combining like terms:
r(r−1)xr−5rxr+ 6xr= 0
Step 3: Factor out xrfrom the equation:
11
xr[r(r−1) −5r+ 6] = 0
Step 4: Simplify the equation further:
xr[r2−r−5r+ 6] = 0
xr[r2−6r+ 6] = 0
Step 5: Set the expression inside the brackets equal to zero to find the roots:
r2−6r+ 6 = 0
Step 6: Solve the quadratic equation using the quadratic formula:
r=6±√62−4·1·6
2=6±√12
2= 3 ±√3
Step 7: The two roots are r= 3 + √3 and r= 3 −√3.
Step 8: Therefore, the general solution to the Cauchy-Euler equation x2y′′ −
5xy′+ 6y= 0 is:
y(x) = c1x3+√3+c2x3−√3
Question 16
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 9y= 0.
Solution
We first assume that the solution is in the form of y=xr. We will differentiate
this and substitute back into the equation to determine the possible values for
r.
Step 1: Find the derivatives of y
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
12
Step 2: Substitute derivatives back into the equation
x2y′′ −3xy′+ 9y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+9xr= 0
r(r−1)xr−3rxr+ 9xr= 0
r(r−1) −3r+ 9 = 0
r2−r−3r+ 9 = 0
r2−4r+ 9 = 0
(r−3)(r−3) = 0
r= 3
Step 3: Form the general solution Since r= 3, the solution in the form
of y=Cxris y=Cx3.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
9y= 0 is y=Cx3, where Cis a constant.
Question 17
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Find the first and second derivatives of y(x).
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2
Step 3: Substitute y(x), y′(x), and y′′(x) into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
Step 5: Factor out xrfrom the equation.
xr(r2−4r+ 4) = 0
Step 6: Solve for rby setting the equation equal to zero.
r2−4r+ 4 = 0
13
Step 7: Find the roots of the characteristic equation.
r=4±√42−4·4
2=4±0
2= 2 (Repeated root)
Step 8: Write the general solution using the roots of the characteristic equa-
tion.
y(x) = c1x2+c2x2ln |x|
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y(x) = c1x2+c2x2ln |x|, where c1and c2are arbitrary constants.
Question 18
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xm. Then
the derivatives of ycan be found as follows:
y′=mxm−1and y′′ =m(m−1)xm−2
Substitute y,y′, and y′′ into the Cauchy-Euler equation:
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m2−4m+ 4 = 0
Step 1: Solve the characteristic equation m2−4m+ 4 = 0 to find the roots.
The characteristic equation factors as (m−2)2= 0, so m= 2 is a repeated root.
Step 2: Write out the general solution based on the roots of the character-
istic equation. Since m= 2 is a repeated root, the general solution is:
y(x) = c1x2+c2x2ln |x|
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is y(x) = c1x2+c2x2ln |x|.
Question 19
Question
Solve the Cauchy-Euler differential equation: x2y′′ + 5xy′+ 6y= 0.
14
Solution
Step 1: Determine the characteristic equation by assuming a solution in the
form y=xr. The characteristic equation is obtained by substituting y=xr
into the differential equation:
x2(r)(r−1)xr−2+ 5x(r)xr−1+ 6xr= 0
Step 2: Simplify the characteristic equation.
r(r−1)xr+ 5rxr+ 6xr= 0
r2−r+ 5r+ 6 = 0
r2+ 4r+ 6 = 0
Step 3: Solve the characteristic equation to find the roots of r. Using the
quadratic formula: r=−b±√b2−4ac
2a, where a= 1, b= 4, c= 6.
r=−4±p42−4(1)(6)
2(1)
r=−4±√16 −24
2
r=−4±√−8
2
r=−4±2i√2
2
r=−2±i√2
Step 4: Write the general solution based on the roots obtained. Since the
roots are complex, the general solution is:
y(x) = c1x−2cos√2 ln(x)+c2x−2sin√2 ln(x)
where c1and c2are arbitrary constants.
Question 20
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−y= 0.
15
Solution
Step 1: Assume a solution of the form y(x) = xrand find the derivatives.
Step 2: Compute the first and second derivatives:
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2.
Step 3: Substitute yand its derivatives into the differential equation:
x2(r(r−1)xr−2)+3x(rxr−1)−xr= 0.
Step 4: Simplify the equation:
r(r−1)xr+ 3rxr−xr= 0.
Step 5: Combine like terms:
r2xr−rxr+ 3rxr−xr= 0.
Step 6: Factor out xr:
(r2−r+ 3r−1)xr= 0.
Step 7: Simplify the expression:
(r2+ 2r−1)xr= 0.
Step 8: Set the equation equal to zero:
r2+ 2r−1 = 0.
Step 9: Solve for rusing the quadratic formula:
r=−2±p22−4·1·(−1)
2·1.
Step 10: Simplify the expression:
r=−2±√4+4
2=−2±√8
2.
Step 11: Further simplify the expression:
r=−2±2√2
2=−1±√2.
Step 12: The general solution is given by:
y(x) = c1x−1+√2+c2x−1−√2,
where c1and c2are arbitrary constants.
16
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ + 3xy′−4y= 0 using the substitution
y=xr.
Solution
Step 1: Find the first and second derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′=rxr−1, and y′′ =r(r−1)xr−2into the
Cauchy-Euler equation.
x2y′′ + 3xy′−4y= 0
x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
r(r−1)xr+ 3rxr−4xr= 0
Step 3: Simplify the equation.
r(r−1) + 3r−4=0
r2−r+ 3r−4=0
r2+ 2r−4=0
(r+ 2)(r−2) = 0
Step 4: Find the roots of the characteristic equation.
r+ 2 = 0 or r−2=0
r=−2 or r= 2
Step 5: Write the general solution.
y=c1x2+c2x−2
Therefore, the solution to the Cauchy-Euler equation x2y′′ + 3xy′−4y= 0
is y=c1x2+c2x−2, where c1and c2are arbitrary constants.
Question 22
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y= 0 by making the substitution
y=xm. Hence, find the general solution of the equation.
17
Solution
Step 1: Differentiate the substitution y=xmto find y′and y′′.
y′=mxm−1,
y′′ =m(m−1)xm−2.
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation x2y′′ −xy′+
y= 0 and simplify.
x2(m(m−1)xm−2)−x(mxm−1) + xm= 0,
m(m−1)xm−mxm+xm= 0,
m(m−1) −m+ 1 = 0,
m2−2m+ 1 = 0,
(m−1)2= 0.
Step 3: Since we have a repeated root at m= 1, the general solution is given
by y=c1x+c2xln(x), where c1and c2are arbitrary constants.
Question 23
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
18
Step 4: Find the roots of the characteristic equation.
r= 2
Step 5: Write down the general solution.
y=c1x2+c2x2ln(x)
Question 24
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y=x3
Solution
To solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y=x3, we will first assume
a solution of the form y=xr.
Step 1: Find the derivatives Calculate the first and second derivatives
of y=xr:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute into the Cauchy-Euler equation Substitute y,y′,
and y′′ into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr=x3
Step 3: Simplify the equation Simplify the equation by multiplying and
combining like terms:
r(r−1)xr−5rxr+ 6xr=x3
r2xr−rxr−5rxr+ 6xr=x3
r2−6r+ 6 = x3−r
Step 4: Find the characteristic equation Based on the equation above,
we get the characteristic equation as:
r2−6r+ 6 = 0
Step 5: Solve the characteristic equation Solve the characteristic equa-
tion by factoring or using the quadratic formula:
r=6±p62−4(1)(6)
2(1)
19
r=6±√12
2
r= 3 ±i√3
Step 6: Find the general solution Since the roots are complex, we have
the general solution in the form:
y=c1x3cos√3 ln(x)+c2x3sin√3 ln(x)
where c1and c2are arbitrary constants.
Question 25
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′−y= 0.
Solution
Step 1: Assume a solution of the form y=xrand compute the first and second
derivatives.
Step 2: Substitute yand its derivatives into the Cauchy-Euler equation and
simplify.
Step 3: Solve the resulting characteristic equation r(r−1) −r+ 1 = 0 to
find the roots r1and r2.
Step 4: Write down the general solution using the roots found in step 3:
y(x) = c1xr1+c2xr2
Step 5: Determine the specific solution based on the initial conditions or
boundary constraints provided in the problem.
Step 6: If there are repeated roots in step 3, then multiply by natural logs
of x.
Step 7: Complete the solution by plugging in the initial conditions or bound-
ary constraints to find the values of c1and c2.
Question 26
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 8y= 0.
20
Solution
Step 1: Write the characteristic equation by substituting y=xrinto the differ-
ential equation.
x2r(r−1)xr−2−5xr(xr−1)+8xr= 0
r(r−1) −5r+ 8 = 0
r2−6r+ 8 = 0
(r−2)(r−4) = 0
Step 2: Find the roots of the characteristic equation to determine the form
of the general solution. The roots are r1= 2 and r2= 4.
Step 3: Write down the general solution to the Cauchy-Euler equation using
the roots obtained in Step 2.
y(x) = c1x2+c2x4
Thus, the general solution to the given Cauchy-Euler equation x2y′′ −5xy′+
8y= 0 is y(x) = c1x2+c2x4, where c1and c2are arbitrary constants.
Question 27
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y= 0
Solution
Step 1: Assume the solution is of the form y=xr.
Step 2: Find the derivatives of y.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −5xy′+ 6y=x2[r(r−1)xr−2]−5x[rxr−1]+6xr
=r(r−1)xr−5rxr+ 6xr
Step 4: Simplify the expression.
r(r−1)xr−5rxr+ 6xr= 0
Step 5: Factor out xr.
xr[r(r−1) −5r+ 6] = 0
21
Step 6: We get the characteristic equation from r(r−1) −5r+ 6 = 0.
r2−r−5r+ 6 = r2−6r+ 6 = 0
Step 7: Solve the characteristic equation:
For r1and r2such that r1×r2= 6 and r1+r2= 6:
r1= 2 and r2= 3
Step 8: The solutions are then y1=x2and y2=x3.
Step 9: Therefore, the general solution to the Cauchy-Euler equation is:
y=c1x2+c2x3
where c1and c2are arbitrary constants.
Question 28
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: We first express the equation in standard form by dividing through by
x2:
y′′ −3
xy′+4
x2y= 0
Step 2: Now, we assume a solution of the form y=xmand differentiate to
find y′and y′′.
y′=mxm−1
y′′ =m(m−1)xm−2
Step 3: Substitute the expressions for y,y′, and y′′ into the differential
equation:
m(m−1)xm−2−3
xmxm−1+4
x2xm= 0
Step 4: Simplify the equation by multiplying through by x2:
m(m−1)xm−3mxm+ 4xm= 0
Step 5: Combine the terms involving xm:
(m2−3m+ 4)xm= 0
Step 6: The characteristic equation is m2−3m+ 4 = 0. Solve for musing
the quadratic formula:
m=3±p(−3)2−4∗1∗4
2=3±√−7
2=3±i√7
2
22
Step 7: Since the roots are complex, the general solution is given by:
y=c1x3+i√7
2+c2x3−i√7
2
Question 29
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xmand then
solve for the characteristic equation.
Step 1: Find the first and second derivatives of y.
Let y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
Step 3: Simplify the equation.
m(m−1) −3m+ 4 = 0
Step 4: Solve the characteristic equation m2−4m+ 4 = 0.
m2−4m+ 4 = 0
(m−2)2= 0
m= 2 (double root)
Step 5: Find the general solution using the root m= 2.
y(x) = c1x2+c2x2ln x
where c1and c2are arbitrary constants.
Therefore, the general solution to the Cauchy-Euler equation is y(x) = c1x2+
c2x2ln x.
23
Question 30
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
We will solve the given Cauchy-Euler equation using the substitution y=xm.
Step 1: Find the first and second derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the derivatives of yinto the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m(m−1) −3m+ 4 = 0 (dividing by xm)
m2−4m+ 4 = 0
(m−2)2= 0
m= 2 (repeated root)
Step 3: Write the general solution using the repeated root m= 2.
y=c1xm1+c2xm2
=c1x2+c2x2
= (c1+c2)x2
Step 4: Therefore, the general solution to the Cauchy-Euler equation x2y′′−
3xy′+ 4y= 0 is y= (c1+c2)x2, where c1and c2are arbitrary constants.
Question 31
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xr.
24
Solution
Step 1: Find the derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′, and y′′ into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
r2xr−rxr−3rxr+ 4xr= 0
(r2−r−3r+ 4)xr= 0
(r2−4r+ 4)xr= 0
(r−2)2xr= 0
Step 3: Set the characteristic equation (r−2)2= 0 equal to zero and solve
for r.
(r−2)2= 0
r−2=0
r= 2
Step 4: Write down the general solution using the roots of the auxiliary
equation. Since we have a double root at r= 2, the general solution is given by
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 32
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
To solve a Cauchy-Euler equation, we make the substitution y=xrfor some
real number r. Then we differentiate ywith respect to xto find y′and y′′, and
substitute them into the differential equation.
25
Step 1: Find y′and y′′. Let y=xr. Then:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
r(r−1) −3r+ 4 = 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 3: Write the general solution using the found value of r. Since r= 2,
one of the solutions is y1=x2. The general solution is then:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 33
Question
Solve the Cauchy-Euler equation: x2y′′ + 3xy′+ 4y= 0.
Solution
Step 1: We first assume a solution of the form y=xrand find the derivatives.
Step 2: Compute the first and second derivatives of y. Step 3: Substitute the
derivatives of yinto the differential equation x2y′′ + 3xy′+ 4y= 0. Step 4:
Simplify the equation obtained by substituting y, y′,and y′′. Step 5: Divide
the entire equation by xrto simplify further. Step 6: Set the resulting equation
equal to zero. Step 7: Solve the characteristic equation to find the roots. Step
8: Write down the general solution using the roots found in the previous step.
Step 9: Write down the final solution by multiplying the general solution by xr.
Question 34
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y=x3.
26
Solution
Let’s first assume that the solution has the form y=xr. We can then find the
characteristic equation by substituting yand its derivatives into the Cauchy-
Euler equation.
Step 1: Find the derivatives Let y=xr. Then,
y′=rxr−1and y′′ =r(r−1)xr−2.
Step 2: Substitute into the equation Substitute y,y′, and y′′ into the
equation x2y′′ −xy′+y=x3:
x2(r(r−1)xr−2)−x(rxr−1) + xr=x3.
Step 3: Simplify and solve for rSimplify the equation:
r(r−1)xr−rxr+xr=x3.
r(r−1)xr−rxr+xr=x3.
Factor out xr:
xr(r(r−1) −r+ 1) = x3.
Solve for r:
r2−r−r+ 1 = 0
r2−2r+ 1 = 0
(r−1)2= 0
r= 1.
Step 4: General solution With r= 1, we get a single root. Therefore,
the general solution is given by
y=c1x1+c2x1ln(x),
where c1and c2are arbitrary constants.
Thus, the solution to the Cauchy-Euler equation x2y′′ −xy′+y=x3is
y=c1x+c2xln(x).
Question 35
Question
Solve the Cauchy-Euler differential equation: x2y′′ +xy′−4y= 0.
27
Question 2
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using a substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y=xr⇒y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute these derivatives into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0 ⇒x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0 ⇒r2−4r+ 4 = 0
Step 4: Solve the characteristic equation for r.
r2−4r+ 4 = 0 ⇒(r−2)2= 0 ⇒r= 2
Step 5: Write down the general solution.
y=c1x2+c2x2ln x
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln x.
Question 3
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Write the characteristic equation by assuming a solution of the form
y=xr.
The characteristic equation is: r(r−1) −3r+ 4 = 0
⇒r2−4r+ 4 = 0
⇒(r−2)2= 0
2
⇒r= 2
(Repeated root)
Step 2: Write down the general solution for the Cauchy-Euler equation.
y(x) = c1x2+c2x2ln |x|
where c1and c2are arbitrary constants.
Step 3: Therefore, the solution to the Cauchy-Euler equation is:
y(x) = c1x2+c2x2ln |x|
Question 4
Question
Solve the Cauchy-Euler equation (x2−3x+ 2)y′′ −2xy′+ 2y= 0 using the
substitution y=xr.
Solution
Step 1: Find the derivatives y′and y′′.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
(x2−3x+ 2)y′′ −2xy′+ 2y= 0
(x2−3x+ 2)(r(r−1)xr−2)−2x(rxr−1)+2xr= 0
r(r−1)xr−3r(r−1)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3r(r−1)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
r2xr−r(r−3)(r−2)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
Step 4: Factor out common terms.
r2xr−r(r−3)(r−2)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
r2xr−r(r−3)(r−2)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
r2xr−r(r2−5r+ 6)xr−1+ 2r(r−1)xr−2−2rxr+ 2xr= 0
3
Step 5: Set the expression equal to zero and solve for r.
r2−r(r2−5r+ 6) + 2r(r−1) = 0
r2−r3+ 5r2−6r+ 2r2−2r= 0
r2+ 5r2+ 2r2−r3−6r−2r= 0
8r2−r3−8r= 0
r(8r−r2−8) = 0
r(r2−8r+ 8) = 0
r(r−4)(r−2) = 0
Step 6: Write down the roots of the characteristic equation. The roots are
r1= 4 and r2= 2.
Step 7: Write down the general solution using the roots.
y(x) = C1x4+C2x2
Question 5
Question
Solve the Cauchy-Euler equation:
x2y′′ −4xy′+ 6y= 0
Solution
Step 1: Assume the solution has the form y=xr. Then find the derivatives of
y.
Step 2: Calculate the first and second derivative of ywith respect to x.
Step 3: Substitute y,y′, and y′′ back into the original differential equation
and simplify.
Step 4: Solve the resulting equation by factoring out common terms and
finding the roots of the characteristic equation.
Step 5: Write down the general solution of the Cauchy-Euler equation.
Step 6: Express the general solution in terms of x.
Step 7: Write the final solution after substituting back for y.
Question 6
Question
Solve the Cauchy-Euler differential equation:
x2y′′ −5xy′+ 6y=x2
4
Solution
Step 1: Write the characteristic equation associated with the Cauchy-Euler
equation using the auxiliary equation m(m−1). For the given differential
equation, the characteristic equation is:
m(m−1) −5m+ 6 = 0
Step 2: Simplify the characteristic equation:
m2−m−5m+ 6 = 0
m2−6m+ 6 = 0
Step 3: Solve the characteristic equation to find the roots m1and m2:
m=6±√62−4∗1∗6
2
m=6±√36 −24
2
m=6±√12
2
m=6±2√3
2
m= 3 ±√3
Step 4: The general solution to the differential equation is then given by:
y(x) = c1xm1+c2xm2
Step 5: Substitute the roots m1= 3 + √3 and m2= 3 −√3 into the general
solution:
y(x) = c1x3+√3+c2x3−√3
Step 6: Therefore, the solution to the Cauchy-Euler differential equation is:
y(x) = c1x3+√3+c2x3−√3
Question 7
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
5
Solution
Step 1: Determine the derivatives of ywith respect to x.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the Cauchy-Euler equation.
x2·r(r−1)xr−2−5x·rxr−1+ 6xr= 0
Step 3: Simplify the equation.
r(r−1)xr−5rxr+ 6xr= 0
Step 4: Factor out xr.
xr(r2−r−5r+ 6) = 0
Step 5: Simplify the expression within the parentheses.
xr(r2−6r+ 6) = 0
Step 6: Factor the quadratic expression r2−6r+ 6 = 0 to solve for r.
(r−3)(r−2) = 0
Step 7: Solve for r.
r= 2 or r= 3
Step 8: Write down the general solution using the roots r1= 2 and r2= 3.
y(x) = c1x2+c2x3
where c1and c2are arbitrary constants.
Question 8
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
6
Solution
To solve the given Cauchy-Euler equation, we will make a substitution y=xm
and then find the values of mthat satisfy the equation.
Step 1: Find the first and second derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m(m−1) −3m+ 4 = 0
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
Step 3: Write down the general solution after finding the roots. Since
(m−2)2= 0, the repeated root is m= 2. Therefore, the general solution is
given by:
y=c1x2+c2x2ln |x|
where c1and c2are arbitrary constants.
Question 9
Question
Solve the Cauchy-Euler equation
x2y′′ −xy′+ 4y= 0.
Solution
Step 1: Let’s assume a solution of the form y=xr. We will differentiate y
accordingly.
Step 2: Compute the first and second derivatives of ywith respect to x.
y′=rxr−1
y′′ =r(r−1)xr−2
7
Step 3: Substitute y,y′, and y′′ into the differential equation.
x2(r(r−1)xr−2)−x(rxr−1)+4xr= 0.
Step 4: Simplify the equation.
r(r−1)xr−rxr+ 4xr= 0.
Step 5: Combine like terms.
r2xr−rxr+ 4xr= 0.
Step 6: Factor out xr.
xr(r2−r+ 4) = 0.
Step 7: Set the expression equal to zero to find possible values of r.
r2−r+ 4 = 0.
Step 8: Solve the quadratic equation r2−r+ 4 = 0 to find the roots.
r=1±p1−4(1)(4)
2(1)
r=1±√−15
2
r=1±i√15
2
Step 9: Thus, the general solution to the Cauchy-Euler equation is
y=c1x1+i√15
2+c2x1−i√15
2,
where c1and c2are arbitrary constants.
Question 10
Question
Find the general solution to the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: Write down the characteristic equation by assuming y=xr. The
characteristic equation is r(r−1)−5r+6 = 0, which simplifies to r2−6r+6 = 0.
Step 2: Solve the characteristic equation. Factoring the quadratic equation
r2−6r+6 = 0 gives (r−3)2−3 = 0, which further simplifies to (r−3+√3)(r−
3−√3) = 0. Thus, we have r= 3 ±√3.
Step 3: Determine the general solution. Since the roots are distinct and real,
the general solution is given by y(x) = c1x3+√3+c2x3−√3, where c1and c2are
arbitrary constants.
8
Question 11
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 given that y1(x) = xis
a solution.
Solution
Step 1: Find the auxiliary equation by assuming y=xr. Step 2: Substitute
y=xrinto the Cauchy-Euler equation. Step 3: Solve the resulting auxiliary
equation to find the roots. Step 4: Form the general solution using the roots
and the original solution y1(x) = x.
Step 1: Assume y=xr. Then, the derivatives are: y′=rxr−1and y′′ =
r(r−1)xr−2.
Step 2: Substitute y=xr,y′=rxr−1, and y′′ =r(r−1)xr−2into the
Cauchy-Euler equation: x2r(r−1)xr−2−3xrxr−1+ 4xr= 0
Step 3: Simplify the equation: r(r−1)xr−3rxr+ 4xr= 0
Factor out xr:xr(r2−4r+ 4) = 0
Solve the auxiliary equation: r2−4r+ 4 = (r−2)2= 0
Thus, we have a repeated root r= 2.
Step 4: The general solution is given by: y(x) = c1xr1+c2xr2, where r1= 2,
r2= 2, and r1=r2= 2.
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is: y(x) = c1x2+c2x2= (c1+c2)x2
Given that y1(x) = xis a solution, we have c1+c2= 1. Therefore, c1=1
2
and c2=1
2.
So, the solution to the Cauchy-Euler equation is y(x) = 1
2x2+1
2x2=x2.
Question 12
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 2y= 0
Solution
To solve the Cauchy-Euler equation, we assume a solution of the form y=xm.
We then substitute yand its derivatives into the differential equation.
Step 1: Find the first and second derivatives of y
y=xm
y′=mxm−1
9
y′′ =m(m−1)xm−2
Step 2: Substitute yand its derivatives into the differential equa-
tion Substitute y,y′, and y′′ into the differential equation:
x2(m(m−1)xm−2)−3x(mxm−1)+2xm= 0
Step 3: Simplify the equation
m(m−1)xm−3mxm+ 2xm= 0
m(m−1) −3m+ 2 = 0
Step 4: Solve for the roots of the characteristic equation
m2−m−3m+ 2 = 0
m2−4m+ 2 = 0
m=4±p(−4)2−4∗1∗2
2
m=4±√16 −8
2
m=4±√8
2
m=4±2√2
2
m= 2 ±√2
Step 5: Write the general solution The general solution is given by:
y(x) = c1x2+√2+c2x2−√2
where c1and c2are constants.
Question 13
Question
Solve the Cauchy-Euler equation
x2y′′ + 3xy′−y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand find the derivatives of
y(x). Step 2: Substitute y(x) and its derivatives into the differential equation.
Step 3: Simplify the equation and solve for r. Step 4: Determine the general
solution for the Cauchy-Euler equation.
10
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 18y= 0
Solution
Step 1: Write down the characteristic equation by substituting y=xminto the
differential equation.
m(m−1)xm−5mxm+ 18xm= 0
Step 2: Simplify the characteristic equation.
m2−6m+ 18 = 0
Step 3: Find the roots of the characteristic equation using the quadratic
formula.
m=6±p(−6)2−4∗1∗18
2=6±√36 −72
2=6±i√36
2= 3 ±3i
Step 4: Write down the general solution using the roots found in Step 3.
y(x) = C1x3cos(3 ln(x)) + C2x3sin(3 ln(x))
where C1and C2are arbitrary constants.
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: We start by substituting y=xrinto the Cauchy-Euler equation x2y′′ −
5xy′+ 6y= 0:
x2(r)(r−1)xr−2−5x(r)xr−1+ 6xr= 0
Step 2: Simplify the equation by combining like terms:
r(r−1)xr−5rxr+ 6xr= 0
Step 3: Factor out xrfrom the equation:
11
xr[r(r−1) −5r+ 6] = 0
Step 4: Simplify the equation further:
xr[r2−r−5r+ 6] = 0
xr[r2−6r+ 6] = 0
Step 5: Set the expression inside the brackets equal to zero to find the roots:
r2−6r+ 6 = 0
Step 6: Solve the quadratic equation using the quadratic formula:
r=6±√62−4·1·6
2=6±√12
2= 3 ±√3
Step 7: The two roots are r= 3 + √3 and r= 3 −√3.
Step 8: Therefore, the general solution to the Cauchy-Euler equation x2y′′ −
5xy′+ 6y= 0 is:
y(x) = c1x3+√3+c2x3−√3
Question 16
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 9y= 0.
Solution
We first assume that the solution is in the form of y=xr. We will differentiate
this and substitute back into the equation to determine the possible values for
r.
Step 1: Find the derivatives of y
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
12
Step 2: Substitute derivatives back into the equation
x2y′′ −3xy′+ 9y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+9xr= 0
r(r−1)xr−3rxr+ 9xr= 0
r(r−1) −3r+ 9 = 0
r2−r−3r+ 9 = 0
r2−4r+ 9 = 0
(r−3)(r−3) = 0
r= 3
Step 3: Form the general solution Since r= 3, the solution in the form
of y=Cxris y=Cx3.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
9y= 0 is y=Cx3, where Cis a constant.
Question 17
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Find the first and second derivatives of y(x).
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2
Step 3: Substitute y(x), y′(x), and y′′(x) into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
Step 5: Factor out xrfrom the equation.
xr(r2−4r+ 4) = 0
Step 6: Solve for rby setting the equation equal to zero.
r2−4r+ 4 = 0
13
Step 7: Find the roots of the characteristic equation.
r=4±√42−4·4
2=4±0
2= 2 (Repeated root)
Step 8: Write the general solution using the roots of the characteristic equa-
tion.
y(x) = c1x2+c2x2ln |x|
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y(x) = c1x2+c2x2ln |x|, where c1and c2are arbitrary constants.
Question 18
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xm. Then
the derivatives of ycan be found as follows:
y′=mxm−1and y′′ =m(m−1)xm−2
Substitute y,y′, and y′′ into the Cauchy-Euler equation:
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m2−4m+ 4 = 0
Step 1: Solve the characteristic equation m2−4m+ 4 = 0 to find the roots.
The characteristic equation factors as (m−2)2= 0, so m= 2 is a repeated root.
Step 2: Write out the general solution based on the roots of the character-
istic equation. Since m= 2 is a repeated root, the general solution is:
y(x) = c1x2+c2x2ln |x|
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is y(x) = c1x2+c2x2ln |x|.
Question 19
Question
Solve the Cauchy-Euler differential equation: x2y′′ + 5xy′+ 6y= 0.
14
Solution
Step 1: Determine the characteristic equation by assuming a solution in the
form y=xr. The characteristic equation is obtained by substituting y=xr
into the differential equation:
x2(r)(r−1)xr−2+ 5x(r)xr−1+ 6xr= 0
Step 2: Simplify the characteristic equation.
r(r−1)xr+ 5rxr+ 6xr= 0
r2−r+ 5r+ 6 = 0
r2+ 4r+ 6 = 0
Step 3: Solve the characteristic equation to find the roots of r. Using the
quadratic formula: r=−b±√b2−4ac
2a, where a= 1, b= 4, c= 6.
r=−4±p42−4(1)(6)
2(1)
r=−4±√16 −24
2
r=−4±√−8
2
r=−4±2i√2
2
r=−2±i√2
Step 4: Write the general solution based on the roots obtained. Since the
roots are complex, the general solution is:
y(x) = c1x−2cos√2 ln(x)+c2x−2sin√2 ln(x)
where c1and c2are arbitrary constants.
Question 20
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−y= 0.
15
Solution
Step 1: Assume a solution of the form y(x) = xrand find the derivatives.
Step 2: Compute the first and second derivatives:
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2.
Step 3: Substitute yand its derivatives into the differential equation:
x2(r(r−1)xr−2)+3x(rxr−1)−xr= 0.
Step 4: Simplify the equation:
r(r−1)xr+ 3rxr−xr= 0.
Step 5: Combine like terms:
r2xr−rxr+ 3rxr−xr= 0.
Step 6: Factor out xr:
(r2−r+ 3r−1)xr= 0.
Step 7: Simplify the expression:
(r2+ 2r−1)xr= 0.
Step 8: Set the equation equal to zero:
r2+ 2r−1 = 0.
Step 9: Solve for rusing the quadratic formula:
r=−2±p22−4·1·(−1)
2·1.
Step 10: Simplify the expression:
r=−2±√4+4
2=−2±√8
2.
Step 11: Further simplify the expression:
r=−2±2√2
2=−1±√2.
Step 12: The general solution is given by:
y(x) = c1x−1+√2+c2x−1−√2,
where c1and c2are arbitrary constants.
16
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ + 3xy′−4y= 0 using the substitution
y=xr.
Solution
Step 1: Find the first and second derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′=rxr−1, and y′′ =r(r−1)xr−2into the
Cauchy-Euler equation.
x2y′′ + 3xy′−4y= 0
x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
r(r−1)xr+ 3rxr−4xr= 0
Step 3: Simplify the equation.
r(r−1) + 3r−4=0
r2−r+ 3r−4=0
r2+ 2r−4=0
(r+ 2)(r−2) = 0
Step 4: Find the roots of the characteristic equation.
r+ 2 = 0 or r−2=0
r=−2 or r= 2
Step 5: Write the general solution.
y=c1x2+c2x−2
Therefore, the solution to the Cauchy-Euler equation x2y′′ + 3xy′−4y= 0
is y=c1x2+c2x−2, where c1and c2are arbitrary constants.
Question 22
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y= 0 by making the substitution
y=xm. Hence, find the general solution of the equation.
17
Solution
Step 1: Differentiate the substitution y=xmto find y′and y′′.
y′=mxm−1,
y′′ =m(m−1)xm−2.
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation x2y′′ −xy′+
y= 0 and simplify.
x2(m(m−1)xm−2)−x(mxm−1) + xm= 0,
m(m−1)xm−mxm+xm= 0,
m(m−1) −m+ 1 = 0,
m2−2m+ 1 = 0,
(m−1)2= 0.
Step 3: Since we have a repeated root at m= 1, the general solution is given
by y=c1x+c2xln(x), where c1and c2are arbitrary constants.
Question 23
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
18
Step 4: Find the roots of the characteristic equation.
r= 2
Step 5: Write down the general solution.
y=c1x2+c2x2ln(x)
Question 24
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y=x3
Solution
To solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y=x3, we will first assume
a solution of the form y=xr.
Step 1: Find the derivatives Calculate the first and second derivatives
of y=xr:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute into the Cauchy-Euler equation Substitute y,y′,
and y′′ into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr=x3
Step 3: Simplify the equation Simplify the equation by multiplying and
combining like terms:
r(r−1)xr−5rxr+ 6xr=x3
r2xr−rxr−5rxr+ 6xr=x3
r2−6r+ 6 = x3−r
Step 4: Find the characteristic equation Based on the equation above,
we get the characteristic equation as:
r2−6r+ 6 = 0
Step 5: Solve the characteristic equation Solve the characteristic equa-
tion by factoring or using the quadratic formula:
r=6±p62−4(1)(6)
2(1)
19
r=6±√12
2
r= 3 ±i√3
Step 6: Find the general solution Since the roots are complex, we have
the general solution in the form:
y=c1x3cos√3 ln(x)+c2x3sin√3 ln(x)
where c1and c2are arbitrary constants.
Question 25
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′−y= 0.
Solution
Step 1: Assume a solution of the form y=xrand compute the first and second
derivatives.
Step 2: Substitute yand its derivatives into the Cauchy-Euler equation and
simplify.
Step 3: Solve the resulting characteristic equation r(r−1) −r+ 1 = 0 to
find the roots r1and r2.
Step 4: Write down the general solution using the roots found in step 3:
y(x) = c1xr1+c2xr2
Step 5: Determine the specific solution based on the initial conditions or
boundary constraints provided in the problem.
Step 6: If there are repeated roots in step 3, then multiply by natural logs
of x.
Step 7: Complete the solution by plugging in the initial conditions or bound-
ary constraints to find the values of c1and c2.
Question 26
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 8y= 0.
20
Solution
Step 1: Write the characteristic equation by substituting y=xrinto the differ-
ential equation.
x2r(r−1)xr−2−5xr(xr−1)+8xr= 0
r(r−1) −5r+ 8 = 0
r2−6r+ 8 = 0
(r−2)(r−4) = 0
Step 2: Find the roots of the characteristic equation to determine the form
of the general solution. The roots are r1= 2 and r2= 4.
Step 3: Write down the general solution to the Cauchy-Euler equation using
the roots obtained in Step 2.
y(x) = c1x2+c2x4
Thus, the general solution to the given Cauchy-Euler equation x2y′′ −5xy′+
8y= 0 is y(x) = c1x2+c2x4, where c1and c2are arbitrary constants.
Question 27
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y= 0
Solution
Step 1: Assume the solution is of the form y=xr.
Step 2: Find the derivatives of y.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −5xy′+ 6y=x2[r(r−1)xr−2]−5x[rxr−1]+6xr
=r(r−1)xr−5rxr+ 6xr
Step 4: Simplify the expression.
r(r−1)xr−5rxr+ 6xr= 0
Step 5: Factor out xr.
xr[r(r−1) −5r+ 6] = 0
21
Step 6: We get the characteristic equation from r(r−1) −5r+ 6 = 0.
r2−r−5r+ 6 = r2−6r+ 6 = 0
Step 7: Solve the characteristic equation:
For r1and r2such that r1×r2= 6 and r1+r2= 6:
r1= 2 and r2= 3
Step 8: The solutions are then y1=x2and y2=x3.
Step 9: Therefore, the general solution to the Cauchy-Euler equation is:
y=c1x2+c2x3
where c1and c2are arbitrary constants.
Question 28
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: We first express the equation in standard form by dividing through by
x2:
y′′ −3
xy′+4
x2y= 0
Step 2: Now, we assume a solution of the form y=xmand differentiate to
find y′and y′′.
y′=mxm−1
y′′ =m(m−1)xm−2
Step 3: Substitute the expressions for y,y′, and y′′ into the differential
equation:
m(m−1)xm−2−3
xmxm−1+4
x2xm= 0
Step 4: Simplify the equation by multiplying through by x2:
m(m−1)xm−3mxm+ 4xm= 0
Step 5: Combine the terms involving xm:
(m2−3m+ 4)xm= 0
Step 6: The characteristic equation is m2−3m+ 4 = 0. Solve for musing
the quadratic formula:
m=3±p(−3)2−4∗1∗4
2=3±√−7
2=3±i√7
2
22
Step 7: Since the roots are complex, the general solution is given by:
y=c1x3+i√7
2+c2x3−i√7
2
Question 29
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xmand then
solve for the characteristic equation.
Step 1: Find the first and second derivatives of y.
Let y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
Step 3: Simplify the equation.
m(m−1) −3m+ 4 = 0
Step 4: Solve the characteristic equation m2−4m+ 4 = 0.
m2−4m+ 4 = 0
(m−2)2= 0
m= 2 (double root)
Step 5: Find the general solution using the root m= 2.
y(x) = c1x2+c2x2ln x
where c1and c2are arbitrary constants.
Therefore, the general solution to the Cauchy-Euler equation is y(x) = c1x2+
c2x2ln x.
23
Question 30
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
We will solve the given Cauchy-Euler equation using the substitution y=xm.
Step 1: Find the first and second derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the derivatives of yinto the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m(m−1) −3m+ 4 = 0 (dividing by xm)
m2−4m+ 4 = 0
(m−2)2= 0
m= 2 (repeated root)
Step 3: Write the general solution using the repeated root m= 2.
y=c1xm1+c2xm2
=c1x2+c2x2
= (c1+c2)x2
Step 4: Therefore, the general solution to the Cauchy-Euler equation x2y′′−
3xy′+ 4y= 0 is y= (c1+c2)x2, where c1and c2are arbitrary constants.
Question 31
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xr.
24
Solution
Step 1: Find the derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′, and y′′ into the Cauchy-Euler equation.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
r2xr−rxr−3rxr+ 4xr= 0
(r2−r−3r+ 4)xr= 0
(r2−4r+ 4)xr= 0
(r−2)2xr= 0
Step 3: Set the characteristic equation (r−2)2= 0 equal to zero and solve
for r.
(r−2)2= 0
r−2=0
r= 2
Step 4: Write down the general solution using the roots of the auxiliary
equation. Since we have a double root at r= 2, the general solution is given by
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 32
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
To solve a Cauchy-Euler equation, we make the substitution y=xrfor some
real number r. Then we differentiate ywith respect to xto find y′and y′′, and
substitute them into the differential equation.
25
Step 1: Find y′and y′′. Let y=xr. Then:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
r(r−1) −3r+ 4 = 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2
Step 3: Write the general solution using the found value of r. Since r= 2,
one of the solutions is y1=x2. The general solution is then:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 33
Question
Solve the Cauchy-Euler equation: x2y′′ + 3xy′+ 4y= 0.
Solution
Step 1: We first assume a solution of the form y=xrand find the derivatives.
Step 2: Compute the first and second derivatives of y. Step 3: Substitute the
derivatives of yinto the differential equation x2y′′ + 3xy′+ 4y= 0. Step 4:
Simplify the equation obtained by substituting y, y′,and y′′. Step 5: Divide
the entire equation by xrto simplify further. Step 6: Set the resulting equation
equal to zero. Step 7: Solve the characteristic equation to find the roots. Step
8: Write down the general solution using the roots found in the previous step.
Step 9: Write down the final solution by multiplying the general solution by xr.
Question 34
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y=x3.
26
Solution
Let’s first assume that the solution has the form y=xr. We can then find the
characteristic equation by substituting yand its derivatives into the Cauchy-
Euler equation.
Step 1: Find the derivatives Let y=xr. Then,
y′=rxr−1and y′′ =r(r−1)xr−2.
Step 2: Substitute into the equation Substitute y,y′, and y′′ into the
equation x2y′′ −xy′+y=x3:
x2(r(r−1)xr−2)−x(rxr−1) + xr=x3.
Step 3: Simplify and solve for rSimplify the equation:
r(r−1)xr−rxr+xr=x3.
r(r−1)xr−rxr+xr=x3.
Factor out xr:
xr(r(r−1) −r+ 1) = x3.
Solve for r:
r2−r−r+ 1 = 0
r2−2r+ 1 = 0
(r−1)2= 0
r= 1.
Step 4: General solution With r= 1, we get a single root. Therefore,
the general solution is given by
y=c1x1+c2x1ln(x),
where c1and c2are arbitrary constants.
Thus, the solution to the Cauchy-Euler equation x2y′′ −xy′+y=x3is
y=c1x+c2xln(x).
Question 35
Question
Solve the Cauchy-Euler differential equation: x2y′′ +xy′−4y= 0.
27
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Differ-
entiate y=xrtwice to find y′and y′′. Step 3: Substitute y,y′, and y′′ into
the differential equation. Step 4: Simplify the equation and solve for r. Step 5:
Write down the general solution in terms of r. Step 6: Find the specific solution
by considering the cases for r.
Step 1: Write the characteristic equation by assuming y=xr.
The characteristic equation is given by ar(r−1)xr−2+bxr−1−4xr= 0.
Step 2: Differentiate y=xrtwice to find y′and y′′.
y′=rxr−1and y′′ =r(r−1)xr−2.
Step 3: Substitute y,y′, and y′′ into the differential equation.
x2(r)(r−1)xr−2+x(rxr−1)−4xr= 0.
Step 4: Simplify the equation and solve for r.
r(r−1) + r−4 = 0. r2−r+r−4 = 0. r2−4 = 0. r2= 4. r=±2.
Step 5: Write down the general solution in terms of r.
The general solution is y(x) = c1x2+c2x−2.
Step 6: Find the specific solution by considering the cases for r.
When r= 2, y(x) = c1x2. When r=−2, y(x) = c2x−2.
Therefore, the solutions to the Cauchy-Euler differential equation are y(x) =
c1x2+c2x−2.
28
