MATH 334 - DIFFERENTIAL
EQUATIONS - Cauchy-Euler equations
Question Bank - Set 1
Liberty University
Question 1
Question
Solve the Cauchy-Euler equation: x2y′′ −4xy′+ 6y= 0.
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xmwhere
mis a constant to be determined.
Step 1: Find the first and second derivatives of y
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute into the differential equation
x2y′′ −4xy′+ 6y= 0
x2[m(m−1)xm−2]−4x[mxm−1]+6xm= 0
m(m−1)xm−4mxm+ 6xm= 0
Step 3: Combine like terms
m(m−1)xm−4mxm+ 6xm= 0
m(m−1) −4m+ 6 = 0
m2−m−4m+ 6 = 0
m2−5m+ 6 = 0
(m−2)(m−3) = 0
Step 4: Find the values of m
m= 2 or m= 3
Step 5: Write the general solution The general solution is
y=c1x2+c2x3
where c1and c2are constants.
Question 2
Question
Solve the Cauchy-Euler equation (x2−1)y′′−3xy′+4y= 0 using the substitution
y=xm.
Solution
Step 1: Find the first and second derivatives of y=xm. The first derivative of
yis y′=mxm−1and the second derivative is y′′ =m(m−1)xm−2.
Step 2: Substitute y=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the
Cauchy-Euler equation. We get (x2−1)(m(m−1)xm−2)−3x(mxm−1)+4xm= 0.
Step 3: Simplify the equation. Expanding and combining like terms gives
us (m2−m)xm−3mxm+ 4xm= 0.
Step 4: Combine like terms. We have (m2−4m+ 4)xm= 0.
Step 5: Equate the coefficient of xmto zero. Setting (m2−4m+ 4) = 0, we
find the characteristic equation m2−4m+ 4 = 0.
Step 6: Solve the characteristic equation. Factoring the characteristic equa-
tion gives us (m−2)2= 0, so m= 2 (with multiplicity 2).
Step 7: Write down the general solution. Since m= 2 has multiplicity 2,
the general solution is y(x) = c1x2+c2x2ln(x), where c1and c2are arbitrary
constants.
Question 3
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y=x2ln x
2
Solution
Step 1: Assume a solution of the form y=xr. Step 2: Differentiate yto find y′
and y′′. Step 3: Substitute y, y′, and y′′ into the Cauchy-Euler equation. Step
4: Solve for the characteristic equation. Step 5: Determine the roots of the
characteristic equation and use them to form the general solution. Step 6: Find
the particular solution using the method of variation of parameters. Step 7:
Combine the general solution and particular solution to get the final solution.
Question 4
Question
Solve the Cauchy-Euler equation
x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume the solution has the form y(x) = xrand differentiate to find 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 to get
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0.
Step 3: Simplify the equation by combining like terms.
r(r−1) −3r+ 4 = 0.
Step 4: Solve the quadratic equation for r.
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2.
Step 5: Form the general solution using the value of r.
y(x) = c1x2+c2x2ln(x),
where c1and c2are constants.
3
Question 5
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives of yinto 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
xr(r2−6r+ 6) = 0
Step 5: Find the roots of the characteristic equation r2−6r+ 6 = 0.
r=6±p(−6)2−4(1)(6)
2(1)
=6±√36 −24
2
=6±√12
2
= 3 ±√3
Step 6: Write down the general solution.
y=c1x3+√3+c2x3−√3
where c1and c2are arbitrary constants.
Question 6
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
4
Solution
Step 1: Write the characteristic equation by substituting y=xminto the
Cauchy-Euler equation.
x2m(m−1)xm−2−5xmxm−1+ 6xm= 0
m(m−1) −5m+ 6 = 0
m2−6m+ 6 = 0
Step 2: Solve the characteristic equation to find the roots m1and m2.
m=6±p(−6)2−4∗1∗6
2=6±√12
2= 3 ±√3
Step 3: Write down the general solution using the roots m1and m2.
y(x) = c1x3+√3+c2x3−√3
So the general solution to the Cauchy-Euler equation is y(x) = c1x3+√3+
c2x3−√3.
Question 7
Question
Find the general solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Guess a solution of the form y=xr. Substitute this into the differential
equation to find the characteristic equation:
r(r−1)xr−3rxr+ 4xr= 0
Step 2: Simplify the equation:
r2xr−rxr−3rxr+ 4xr= 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 3: The characteristic equation has a repeated root r= 2. Therefore,
we have the solutions y1=x2and y2=x2ln(x).
Step 4: The general solution to the Cauchy-Euler equation is given by:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
5
Question 8
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−4x(rxr−1)+6xr=xr(r(r−1) −4r+ 6) = 0
Step 3: Simplify the expression.
r(r−1) −4r+ 6 = r2−r−4r+ 6 = r2−5r+ 6 = (r−2)(r−3) = 0
Step 4: Solve for the roots of the characteristic equation.
r= 2 or r= 3
Step 5: Write down the general solution.
y=c1x2+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y=c1x2+c2x3, where c1and c2are arbitrary constants.
Question 9
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 16y= 0
Solution
Step 1: Assume a solution of the form y(x) = xr. Then, compute the first and
second derivatives of y(x). Step 2: Substitute y(x), y′(x), and y′′(x) into the
differential equation. Step 3: Simplify the equation by factoring out xr. Step 4:
Solve for the values of r. Step 5: Form the general solution based on the values
of rfound in step 4.
Step 1: Assume y(x) = xr. Then, compute the first and second derivatives
of y(x).
6
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2
Step 2: Substitute y(x), y′(x), and y′′(x) into the differential equation.
x2(r(r−1)xr−2)−5x(rxr−1) + 16xr= 0
Step 3: Simplify the equation by factoring out xr.
r(r−1) −5r+ 16 = 0
Step 4: Solve for the values of r.
r2−r−5r+ 16 = r2−6r+ 16 = (r−2)(r−8) = 0
So, r= 2 or r= 8.
Step 5: Form the general solution based on the values of rfound in step 4.
The general solution is:
y(x) = c1x2+c2x8
where c1and c2are arbitrary constants.
Question 10
Question
Solve the Cauchy-Euler equation (x2−x−6)y′′−3xy′−4y= 0 using the method
of undetermined coefficients.
Solution
1. First, we assume a solution of the form y(x) = xr. We can rewrite the
given differential equation as:
(x2−x−6)r(r−1)xr−2−3xrxr−1−4xr= 0
2. Simplifying the above expression:
(r2−7r+ 12)xr−3rxr−4xr= 0
r2−10r+ 12 = 0
(r−2)(r−6) = 0
The roots are r1= 2 and r2= 6. So, the general solution is given by:
y(x) = c1x2+c2x6
7
Question 11
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr. Substituting this into the
equation, we get:
x2(r)(r−1)xr−2−3x(r)xr−1+ 4xr= 0
Step 2: Simplify the equation:
r(r−1) −3r+ 4 = 0
Step 3: Rearrange the terms to get a quadratic equation in terms of r:
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 4: Since we have a repeated root, the general solution is:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 12
Question
Solve the Cauchy-Euler equation: x3y′′′ −2x2y′′ +xy′−y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Find the first three derivatives of y(x):
y′(x) = rxr−1, y′′(x) = r(r−1)xr−2, y′′′(x) = r(r−1)(r−2)xr−3
Step 3: Substitute y(x) and its derivatives back into the original equation and
simplify:
x3(r(r−1)(r−2)xr−3)−2x2(r(r−1)xr−2) + x(rxr−1)−xr= 0
8
Step 4: Simplify the equation:
r(r−1)(r−2)xr−2r(r−1)xr+rxr−xr= 0
Step 5: Combine like terms:
r3xr−3r2xr+rxr−xr= 0
r3−2r2+r−1 = r(r2−2r+ 1) = r(r−1)2= 0
Step 6: Solve for r:r= 0,1 (with multiplicity 2)
Step 7: Write down the general solutions for r:
y(x) = c1x0+c2x+c3xln(x)
where c1, c2, c3are arbitrary constants.
Question 13
Question
Solve the Cauchy-Euler equation (x2−4)y′′ −xy′+y= 0.
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Dif-
ferentiate y=xrto find y′and y′′. Step 3: Substitute y,y′, and y′′ into the
Cauchy-Euler equation. Step 4: Divide the equation by x2to simplify it. Step
5: Solve the resulting auxiliary equation to find the roots. Step 6: Write down
the general solution using the roots found.
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0
Solution
Step 1: Let’s assume a solution of the form y(x) = xr. We will substitute this
into the differential equation to find the possible values of r.
Step 2: Calculate the first and second derivatives of y(x):
y′=rxr−1and y′′ =r(r−1)xr−2
9
Step 3: Now substitute y,y′, and y′′ into the differential equation:
x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
Step 4: Simplify the expression by combining like terms:
r(r−1)xr+ 3rxr−4xr= 0
Step 5: Factor out xr:
xr(r2−r+ 3r−4) = 0
Step 6: Simplify the expression further:
xr(r2+ 2r−4) = 0
Step 7: Set the expression equal to zero and solve for r:
r2+ 2r−4=0
Step 8: Use the quadratic formula to find the roots of the equation:
r=−2±p22−4(1)(−4)
2(1) =−2±√4 + 16
2=−2±√20
2=−1±√5
Step 9: The general solution is given by y(x) = c1x(−1+√5) +c2x(−1−√5).
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x2ln(x).
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) and its derivatives into the differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x2ln(x)
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr=x2ln(x)
Step 5: Combine like terms.
r2−r−3r+ 4 = ln(x)
10
Step 6: Simplify the equation.
r2−4r+ 4 = ln(x)
Step 7: Solve the characteristic equation r2−4r+ 4 = 0 to find the roots of
r.
(r−2)2= 0 =⇒r= 2
Step 8: The solution to the characteristic equation gives us a repeated root
r= 2.
Step 9: Therefore, the general solution to the Cauchy-Euler equation is given
by
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Step 10: Now we find the particular solution to the original differential
equation by substituting y(x) = c1x2+c2x2ln(x) into the equation x2y′′ −
3xy′+ 4y=x2ln(x).
Step 11: After substitution and simplification, we equate coefficients of like
terms to obtain c1=−2 and c2= 1.
Step 12: Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y=x2ln(x) is
y(x) = −2x2+x2ln(x)
Question 16
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y, y′, y′′ back into the Cauchy-Euler 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.
r2−r+ 4 = 0
11
Step 6: Solve the quadratic equation r2−r+ 4 = 0 for r.
r=1±p1−4(1)(4)
2=1±√−15
2=1±i√15
2
Step 7: Write down the solutions.
y1=x1+i√15
2and y2=x1−i√15
2
Question 17
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Begin by assuming a solution of the form y=xrand substitute it
into the differential equation. Then find the characteristic equation by solving
x2r(r−1) −3xr + 4 = 0.
Step 2: Simplify the characteristic equation to r2−4r+ 4 = 0.
Step 3: Factor the characteristic equation to (r−2)2= 0.
Step 4: There is a repeated root at r= 2. Therefore, the general solution
takes the form y=C1x2+C2x2ln(x), where C1and C2are constants.
Step 5: Thus, the general solution to the Cauchy-Euler equation x2y′′ −
3xy′+ 4y= 0 is y=C1x2+C2x2ln(x).
Question 18
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x4.
Solution
Step 1: Write the characteristic equation by substituting y=xrinto the homo-
geneous form of the equation (xny′′ +pxn−1y′+qxny= 0).
r(r−1)xr−2−3rxr−1+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2(multiplicity 2)
So, the auxiliary equation has the repeated root r= 2.
12
Step 2: Form the complementary function yc(x) = c1x2+c2x2ln xwhere c1
and c2are constants to be determined.
Step 3: Find a particular solution yp(x) using the method of undetermined
coefficients. Since x4is a polynomial of degree 4, we guess yp(x) = Ax4+Bx3+
Cx2+Dx +E.
x2(4A)−3x(4Ax3+ 3Bx2+ 2Cx +D) + 4(Ax4+Bx3+Cx2+Dx +E) = x4
4Ax2−12Ax2−9Bx2−8Cx −3Dx + 4Ax4+ 4Bx3+ 4Cx2+ 4Dx + 4E=x4
4Ax4+ (4B−12A)x3+ (−9B+ 4C)x2+ (−8C−3D)x+ 4A=x4
By comparing coefficients, we get:
4A= 1,4B−12A= 0,−9B+ 4C= 0,−8C−3D= 0,4A= 0
Solving the system of equations, we find A=1
4, B =3
16 , C =9
64 , D =
−27
128 , E = 0.
Therefore, the particular solution is yp(x) = 1
4x4+3
16 x3+9
64 x2−27
128 x.
Step 4: The general solution is given by y(x) = yc(x) + yp(x). Thus, the
solution to the Cauchy-Euler equation is
y(x) = c1x2+c2x2ln x+1
4x4+3
16x3+9
64x2−27
128x
Question 19
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ −3
xy′+4
x2y= 0.
Step 2: Assume a solution of the form y=xr. Substitute yand its deriva-
tives into the equation and solve for r:
r(r−1)xr−2−3
xrxr−1+4
x2xr= 0,
r2−r−3r+ 4 = 0,
r2−4r+ 4 = 0,
(r−2)2= 0.
13
Step 3: The characteristic equation has a repeated root r= 2. Therefore,
the general solution is given by:
y(x) = c1x2+c2x2ln(x),
where c1and c2are arbitrary constants.
Question 20
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xm.
Solution
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=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the
given differential equation.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
Step 3: Find the roots of the characteristic equation.
m= 2
Step 4: Write down the general solution based on the roots found.
y=c1x2+c2x2ln(x)
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 3y= 0 using the substitution
y=xm.
14
Solution
Step 1: Let’s begin by making the substitution y=xminto the differential
equation x2y′′ −3xy′+ 3y= 0, and find the derivatives of y.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Substitute y,y′, and y′′ back into the differential equation:
x2(m(m−1)xm−2)−3x(mxm−1)+3xm= 0
Simplify the equation:
m(m−1)xm−3mxm+ 3xm= 0
m(m−1) −3m+ 3 = 0
Step 2: Solve the resulting characteristic equation m(m−1) −3m+ 3 = 0:
m2−m−3m+ 3 = 0
m2−4m+ 3 = 0
(m−3)(m−1) = 0
So, we have m= 1 or m= 3.
Step 3: Form the general solution:
y=c1x1+c2x3
y=c1x+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
3y= 0 is y=c1x+c2x3.
Question 22
Question
Solve the Cauchy-Euler equation:
x2y′′ +xy′−y= 0
15
Solution
Step 1: Write the equation in standard form by dividing through by x2:
y′′ +1
xy′−1
x2y= 0
Step 2: Guess a solution of the form y=xr. Substitute this into the
equation:
r(r−1)xr−2+1
xrxr−1−1
x2xr= 0
Step 3: Simplify the equation:
r2−r+r−1 = 0
r2−1 = 0
(r−1)(r+ 1) = 0
Step 4: Find the roots of the characteristic equation:
r1= 1 and r2=−1
Step 5: Write down the general solution:
y=c1x1+c2x−1
y=c1x+c2
x
Question 23
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the differential equation.
x2(r)(r−1)xr−2−3x(rxr−1)+4xr= 0
16
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 5: Solve for r.
r= 2
Step 6: Form the general solution. Since there is a repeated root at r= 2,
the general solution is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are constants.
Question 24
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 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.
x2(r(r−1)xr−2)−5x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−5rxr+ 4xr= 0
r2−r−5r+ 4 = 0
r2−6r+ 4 = 0
Step 4: Solve the quadratic equation for r.
r=6±p62−4(1)(4)
2=6±√32
2=6±4√2
2
17
Step 5: Find the roots of the equation.
r1= 3 + 2√2, r2= 3 −2√2
Step 6: Write the general solution using the roots r1and r2.
y=c1x3+2√2+c2x3−2√2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
4y= 0 is y=c1x3+2√2+c2x3−2√2.
Question 25
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y=x3
Solution
Step 1: We first identify the characteristic equation by substituting y=xrinto
the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x3
Step 2: Simplifying the expression gives:
r(r−1)xr−3rxr+ 4xr=x3
Step 3: Combining like terms yields the characteristic equation:
r2−4r+ 4 = 0
Step 4: The characteristic equation can be factored as (r−2)2= 0, so we
have a repeated root r= 2.
Step 5: The solution to the Cauchy-Euler equation is therefore of the form:
y(x) = c1x2+c2x2ln(x)
Step 6: To find a particular solution, we use the method of undetermined
coefficients. Assume a particular solution of the form:
yp=Ax3
Step 7: Substituting yp=Ax3back into the original differential equation,
we get:
(Ax3)′′ −3x(Ax3)′+ 4Ax3=x3
18
Step 8: Simplifying the expression gives:
6Ax + 4Ax3=x3
Step 9: Equating coefficients gives 4A= 1, so A=1
4.
Step 10. The general solution to the Cauchy-Euler equation is:
y(x) = c1x2+c2x2ln(x) + 1
4x3
Question 26
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: Write down the characteristic equation. In the case of a Cauchy-Euler
equation, the characteristic equation is obtained by substituting y=xrinto the
differential equation.
r(r−1)xr−5rxr+ 6xr= 0
Step 2: Simplify the characteristic equation.
r2−r−5r+ 6 = 0
r2−6r+ 6 = 0
Step 3: Solve for the roots of the characteristic equation using the quadratic
formula.
r=6±√62−4·1·6
2=6±√12
2=6±2√3
2
r= 3 ±√3
Step 4: Write down the general solution to the Cauchy-Euler equation.
y(x) = c1x3+√3+c2x3−√3
Therefore, the general solution to the given Cauchy-Euler equation is y(x) =
c1x3+√3+c2x3−√3, where c1and c2are arbitrary constants.
Question 27
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 8y= 0
19
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Dif-
ferentiate ytwice and substitute into the differential equation. Step 3: Solve
the characteristic equation to find the roots. Step 4: Write down the general
solution.
Step 1: Assume y=xrand differentiate.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −5xy′+ 8y= 0
x2(r(r−1)xr−2)−5x(rxr−1)+8xr= 0
r(r−1)xr−5rxr+ 8xr= 0
Step 3: Solve the characteristic equation r(r−1) −5r+ 8 = 0.
r2−r−5r+ 8 = 0
r2−6r+ 8 = 0
(r−2)(r−4) = 0
So, the roots are r= 2 and r= 4.
Step 4: Write down the general solution.
The general solution of the Cauchy-Euler equation is:
y(x) = c1x2+c2x4
where c1and c2are constants.
Question 28
Question
Solve the Cauchy-Euler equation x2y′′ +xy′−6y= 0 using the substitution
y=xr.
20
Solution
Step 1: Find the derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives into the Cauchy-Euler equation.
x2(r(r−1)xr−2) + x(rxr−1)−6xr= 0
Step 3: Simplify the equation.
r(r−1)xr+rxr−6xr= 0
Step 4: Combine like terms.
r2xr−r2xr−6xr= 0
Step 5: Simplify further.
−6xr= 0
Step 6: Divide by xr.
−6=0
Step 7: Since −6= 0, there are no solutions to this differential equation
using the substitution y=xr. Thus, we conclude that y=xrdoes not solve
the Cauchy-Euler equation x2y′′ +xy′−6y= 0.
Question 29
Question
Solve the Cauchy-Euler equation x2y′′ + 5xy′−6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the differential equation.
x2(r(r−1)xr−2)+5x(rxr−1)−6xr= 0
r(r−1)xr+ 5rxr−6xr= 0
r2−r+ 5r−6=0
r2+ 4r−6=0
(r+ 3)(r−2) = 0
21
Step 3: Solve for the roots of the characteristic equation: r=−3 and r= 2.
Step 4: Write down the general solution using the roots r1and r2.
y=c1x−3+c2x2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ + 5xy′−
6y= 0 is y=c1x−3+c2x2, where c1and c2are arbitrary constants.
Question 30
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y= 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) back into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr= 0
Step 4: Simplify the equation by combining like terms:
r(r−1)xr−5rxr+ 6xr= 0
r2−r−5r+ 6 = 0
Step 5: Solve the quadratic equation r2−6r+ 6 = 0 to find the roots:
r=6±p(−6)2−4∗1∗6
2∗1=6±√36 −24
2=6±√12
2
r=6±2√3
2= 3 ±√3
Step 6: Write the general solution using the roots r1= 3 + √3 and r2= 3 −√3:
y(x) = c1x3+√3+c2x3−√3
where c1and c2are constants.
Question 31
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 7y= 0.
22
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Calculate 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 differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+7xr= 0
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 7xr= 0
r(r−1) −3r+ 7 = 0
Step 5: Solve the quadratic equation.
r2−r−3r+ 7 = 0
r2−4r+ 7 = 0
r=4±p(−4)2−4(1)(7)
2(1)
r=4±√16 −28
2
r=4±√−12
2
r=4±2i√3
2
r= 2 ±i√3
Step 6: Write the general solution.
y(x) = c1x2+√3i+c2x2−√3i
y(x) = c1x2cos√3 ln(x)+c2x2sin√3 ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
7y= 0 is y(x) = c1x2cos√3 ln(x)+c2x2sin√3 ln(x).
Question 32
Question
Solve the Cauchy-Euler differential equation: x2y′′ + 5xy′+ 7y= 0.
23
Solution
Step 1: Write the equation in standard form by finding the characteristic equa-
tion. Assume y=xr.
x2r(r−1)xr−2+ 5xrxr−1+ 7xr= 0
r(r−1) + 5r+ 7 = 0
r2+ 4r+ 7 = 0
Step 2: Solve the characteristic equation for rusing the quadratic formula.
r=−4±p42−4(1)(7)
2(1)
r=−4±√16 −28
2
r=−4±√−12
2
r=−4±2i√3
2
r=−2±i√3
Step 3: Form the general solution using the roots of the characteristic equa-
tion.
y(x) = c1x−2cos√3 ln x+c2x−2sin√3 ln x
where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler differential equation x2y′′ +
5xy′+ 7y= 0 is y(x) = c1x−2cos√3 ln x+c2x−2sin√3 ln x.
Question 33
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm.
Solution
To solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm, we need to find the values of mthat satisfy the equation.
Step 1: Find the derivatives of y=xm. Start by finding the first and
second derivatives of y:
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
24
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation. Substitute
y=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the Cauchy-Euler equation:
x2(m(m−1)xm−2)−5x(mxm−1)+6xm= 0
Step 3: Simplify the equation. Simplify the equation from Step 2:
m(m−1)xm−5mxm+ 6xm= 0
Step 4: Factor out xm. Factor out xmfrom the equation:
xm(m(m−1) −5m+ 6) = 0
Step 5: Set the expression equal to zero. Set the expression equal to zero
to find the values of m:
m(m−1) −5m+ 6 = 0
m2−m−5m+ 6 = 0
m2−6m+ 6 = 0
Step 6: Solve for m. Solve the quadratic equation m2−6m+ 6 = 0 to find
the values of m. The solutions are:
m= 3 ±√3
Step 7: Write the general solution. The general solution to the Cauchy-
Euler equation is given by:
y(x) = c1x3+√3+c2x3−√3
where c1and c2are constants determined by initial conditions.
Question 34
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand differentiate twice to find
y′and y′′.
y(x) = xr
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
25
Step 2: Substitute y,y′, and y′′ back into the Cauchy-Euler equation.
x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
Step 3: Simplify the equation and express it in terms of r.
r(r−1)xr+ 3rxr−4xr= 0
r2−r+ 3r−4=0
r2+ 2r−4=0
(r+ 2)(r−2) = 0
Step 4: Find the roots of the characteristic equation to determine the solu-
tions.
The roots are r=−2 and r= 2.
Step 5: Form the general solution using the roots.
y(x) = c1x−2+c2x2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ + 3xy′−
4y= 0 is y(x) = c1x−2+c2x2, where c1and c2are arbitrary constants.
Question 35
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand compute the derivatives.
Step 2: Substitute y(x) and its derivatives into the Cauchy-Euler equation.
Step 3: Simplify the resulting equation by regrouping terms.
Step 4: Solve the resulting characteristic equation for r.
Step 5: Write down the general solution of the Cauchy-Euler equation.
Step 1: Assume y(x) = xr, then find the first and second derivatives:
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
Step 2: Substitute y(x) and its derivatives into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
26
(m−2)(m−3) = 0
Step 4: Find the values of m
m= 2 or m= 3
Step 5: Write the general solution The general solution is
y=c1x2+c2x3
where c1and c2are constants.
Question 2
Question
Solve the Cauchy-Euler equation (x2−1)y′′−3xy′+4y= 0 using the substitution
y=xm.
Solution
Step 1: Find the first and second derivatives of y=xm. The first derivative of
yis y′=mxm−1and the second derivative is y′′ =m(m−1)xm−2.
Step 2: Substitute y=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the
Cauchy-Euler equation. We get (x2−1)(m(m−1)xm−2)−3x(mxm−1)+4xm= 0.
Step 3: Simplify the equation. Expanding and combining like terms gives
us (m2−m)xm−3mxm+ 4xm= 0.
Step 4: Combine like terms. We have (m2−4m+ 4)xm= 0.
Step 5: Equate the coefficient of xmto zero. Setting (m2−4m+ 4) = 0, we
find the characteristic equation m2−4m+ 4 = 0.
Step 6: Solve the characteristic equation. Factoring the characteristic equa-
tion gives us (m−2)2= 0, so m= 2 (with multiplicity 2).
Step 7: Write down the general solution. Since m= 2 has multiplicity 2,
the general solution is y(x) = c1x2+c2x2ln(x), where c1and c2are arbitrary
constants.
Question 3
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y=x2ln x
2
Solution
Step 1: Assume a solution of the form y=xr. Step 2: Differentiate yto find y′
and y′′. Step 3: Substitute y, y′, and y′′ into the Cauchy-Euler equation. Step
4: Solve for the characteristic equation. Step 5: Determine the roots of the
characteristic equation and use them to form the general solution. Step 6: Find
the particular solution using the method of variation of parameters. Step 7:
Combine the general solution and particular solution to get the final solution.
Question 4
Question
Solve the Cauchy-Euler equation
x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume the solution has the form y(x) = xrand differentiate to find 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 to get
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0.
Step 3: Simplify the equation by combining like terms.
r(r−1) −3r+ 4 = 0.
Step 4: Solve the quadratic equation for r.
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2.
Step 5: Form the general solution using the value of r.
y(x) = c1x2+c2x2ln(x),
where c1and c2are constants.
3
Question 5
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives of yinto 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
xr(r2−6r+ 6) = 0
Step 5: Find the roots of the characteristic equation r2−6r+ 6 = 0.
r=6±p(−6)2−4(1)(6)
2(1)
=6±√36 −24
2
=6±√12
2
= 3 ±√3
Step 6: Write down the general solution.
y=c1x3+√3+c2x3−√3
where c1and c2are arbitrary constants.
Question 6
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
4
Solution
Step 1: Write the characteristic equation by substituting y=xminto the
Cauchy-Euler equation.
x2m(m−1)xm−2−5xmxm−1+ 6xm= 0
m(m−1) −5m+ 6 = 0
m2−6m+ 6 = 0
Step 2: Solve the characteristic equation to find the roots m1and m2.
m=6±p(−6)2−4∗1∗6
2=6±√12
2= 3 ±√3
Step 3: Write down the general solution using the roots m1and m2.
y(x) = c1x3+√3+c2x3−√3
So the general solution to the Cauchy-Euler equation is y(x) = c1x3+√3+
c2x3−√3.
Question 7
Question
Find the general solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Guess a solution of the form y=xr. Substitute this into the differential
equation to find the characteristic equation:
r(r−1)xr−3rxr+ 4xr= 0
Step 2: Simplify the equation:
r2xr−rxr−3rxr+ 4xr= 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 3: The characteristic equation has a repeated root r= 2. Therefore,
we have the solutions y1=x2and y2=x2ln(x).
Step 4: The general solution to the Cauchy-Euler equation is given by:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
5
Question 8
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−4x(rxr−1)+6xr=xr(r(r−1) −4r+ 6) = 0
Step 3: Simplify the expression.
r(r−1) −4r+ 6 = r2−r−4r+ 6 = r2−5r+ 6 = (r−2)(r−3) = 0
Step 4: Solve for the roots of the characteristic equation.
r= 2 or r= 3
Step 5: Write down the general solution.
y=c1x2+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y=c1x2+c2x3, where c1and c2are arbitrary constants.
Question 9
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 16y= 0
Solution
Step 1: Assume a solution of the form y(x) = xr. Then, compute the first and
second derivatives of y(x). Step 2: Substitute y(x), y′(x), and y′′(x) into the
differential equation. Step 3: Simplify the equation by factoring out xr. Step 4:
Solve for the values of r. Step 5: Form the general solution based on the values
of rfound in step 4.
Step 1: Assume y(x) = xr. Then, compute the first and second derivatives
of y(x).
6
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2
Step 2: Substitute y(x), y′(x), and y′′(x) into the differential equation.
x2(r(r−1)xr−2)−5x(rxr−1) + 16xr= 0
Step 3: Simplify the equation by factoring out xr.
r(r−1) −5r+ 16 = 0
Step 4: Solve for the values of r.
r2−r−5r+ 16 = r2−6r+ 16 = (r−2)(r−8) = 0
So, r= 2 or r= 8.
Step 5: Form the general solution based on the values of rfound in step 4.
The general solution is:
y(x) = c1x2+c2x8
where c1and c2are arbitrary constants.
Question 10
Question
Solve the Cauchy-Euler equation (x2−x−6)y′′−3xy′−4y= 0 using the method
of undetermined coefficients.
Solution
1. First, we assume a solution of the form y(x) = xr. We can rewrite the
given differential equation as:
(x2−x−6)r(r−1)xr−2−3xrxr−1−4xr= 0
2. Simplifying the above expression:
(r2−7r+ 12)xr−3rxr−4xr= 0
r2−10r+ 12 = 0
(r−2)(r−6) = 0
The roots are r1= 2 and r2= 6. So, the general solution is given by:
y(x) = c1x2+c2x6
7
Question 11
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr. Substituting this into the
equation, we get:
x2(r)(r−1)xr−2−3x(r)xr−1+ 4xr= 0
Step 2: Simplify the equation:
r(r−1) −3r+ 4 = 0
Step 3: Rearrange the terms to get a quadratic equation in terms of r:
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 4: Since we have a repeated root, the general solution is:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 12
Question
Solve the Cauchy-Euler equation: x3y′′′ −2x2y′′ +xy′−y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Find the first three derivatives of y(x):
y′(x) = rxr−1, y′′(x) = r(r−1)xr−2, y′′′(x) = r(r−1)(r−2)xr−3
Step 3: Substitute y(x) and its derivatives back into the original equation and
simplify:
x3(r(r−1)(r−2)xr−3)−2x2(r(r−1)xr−2) + x(rxr−1)−xr= 0
8
Step 4: Simplify the equation:
r(r−1)(r−2)xr−2r(r−1)xr+rxr−xr= 0
Step 5: Combine like terms:
r3xr−3r2xr+rxr−xr= 0
r3−2r2+r−1 = r(r2−2r+ 1) = r(r−1)2= 0
Step 6: Solve for r:r= 0,1 (with multiplicity 2)
Step 7: Write down the general solutions for r:
y(x) = c1x0+c2x+c3xln(x)
where c1, c2, c3are arbitrary constants.
Question 13
Question
Solve the Cauchy-Euler equation (x2−4)y′′ −xy′+y= 0.
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Dif-
ferentiate y=xrto find y′and y′′. Step 3: Substitute y,y′, and y′′ into the
Cauchy-Euler equation. Step 4: Divide the equation by x2to simplify it. Step
5: Solve the resulting auxiliary equation to find the roots. Step 6: Write down
the general solution using the roots found.
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0
Solution
Step 1: Let’s assume a solution of the form y(x) = xr. We will substitute this
into the differential equation to find the possible values of r.
Step 2: Calculate the first and second derivatives of y(x):
y′=rxr−1and y′′ =r(r−1)xr−2
9
Step 3: Now substitute y,y′, and y′′ into the differential equation:
x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
Step 4: Simplify the expression by combining like terms:
r(r−1)xr+ 3rxr−4xr= 0
Step 5: Factor out xr:
xr(r2−r+ 3r−4) = 0
Step 6: Simplify the expression further:
xr(r2+ 2r−4) = 0
Step 7: Set the expression equal to zero and solve for r:
r2+ 2r−4=0
Step 8: Use the quadratic formula to find the roots of the equation:
r=−2±p22−4(1)(−4)
2(1) =−2±√4 + 16
2=−2±√20
2=−1±√5
Step 9: The general solution is given by y(x) = c1x(−1+√5) +c2x(−1−√5).
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x2ln(x).
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) and its derivatives into the differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x2ln(x)
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr=x2ln(x)
Step 5: Combine like terms.
r2−r−3r+ 4 = ln(x)
10
Step 6: Simplify the equation.
r2−4r+ 4 = ln(x)
Step 7: Solve the characteristic equation r2−4r+ 4 = 0 to find the roots of
r.
(r−2)2= 0 =⇒r= 2
Step 8: The solution to the characteristic equation gives us a repeated root
r= 2.
Step 9: Therefore, the general solution to the Cauchy-Euler equation is given
by
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Step 10: Now we find the particular solution to the original differential
equation by substituting y(x) = c1x2+c2x2ln(x) into the equation x2y′′ −
3xy′+ 4y=x2ln(x).
Step 11: After substitution and simplification, we equate coefficients of like
terms to obtain c1=−2 and c2= 1.
Step 12: Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y=x2ln(x) is
y(x) = −2x2+x2ln(x)
Question 16
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y, y′, y′′ back into the Cauchy-Euler 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.
r2−r+ 4 = 0
11
Step 6: Solve the quadratic equation r2−r+ 4 = 0 for r.
r=1±p1−4(1)(4)
2=1±√−15
2=1±i√15
2
Step 7: Write down the solutions.
y1=x1+i√15
2and y2=x1−i√15
2
Question 17
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Begin by assuming a solution of the form y=xrand substitute it
into the differential equation. Then find the characteristic equation by solving
x2r(r−1) −3xr + 4 = 0.
Step 2: Simplify the characteristic equation to r2−4r+ 4 = 0.
Step 3: Factor the characteristic equation to (r−2)2= 0.
Step 4: There is a repeated root at r= 2. Therefore, the general solution
takes the form y=C1x2+C2x2ln(x), where C1and C2are constants.
Step 5: Thus, the general solution to the Cauchy-Euler equation x2y′′ −
3xy′+ 4y= 0 is y=C1x2+C2x2ln(x).
Question 18
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x4.
Solution
Step 1: Write the characteristic equation by substituting y=xrinto the homo-
geneous form of the equation (xny′′ +pxn−1y′+qxny= 0).
r(r−1)xr−2−3rxr−1+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2(multiplicity 2)
So, the auxiliary equation has the repeated root r= 2.
12
Step 2: Form the complementary function yc(x) = c1x2+c2x2ln xwhere c1
and c2are constants to be determined.
Step 3: Find a particular solution yp(x) using the method of undetermined
coefficients. Since x4is a polynomial of degree 4, we guess yp(x) = Ax4+Bx3+
Cx2+Dx +E.
x2(4A)−3x(4Ax3+ 3Bx2+ 2Cx +D) + 4(Ax4+Bx3+Cx2+Dx +E) = x4
4Ax2−12Ax2−9Bx2−8Cx −3Dx + 4Ax4+ 4Bx3+ 4Cx2+ 4Dx + 4E=x4
4Ax4+ (4B−12A)x3+ (−9B+ 4C)x2+ (−8C−3D)x+ 4A=x4
By comparing coefficients, we get:
4A= 1,4B−12A= 0,−9B+ 4C= 0,−8C−3D= 0,4A= 0
Solving the system of equations, we find A=1
4, B =3
16 , C =9
64 , D =
−27
128 , E = 0.
Therefore, the particular solution is yp(x) = 1
4x4+3
16 x3+9
64 x2−27
128 x.
Step 4: The general solution is given by y(x) = yc(x) + yp(x). Thus, the
solution to the Cauchy-Euler equation is
y(x) = c1x2+c2x2ln x+1
4x4+3
16x3+9
64x2−27
128x
Question 19
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ −3
xy′+4
x2y= 0.
Step 2: Assume a solution of the form y=xr. Substitute yand its deriva-
tives into the equation and solve for r:
r(r−1)xr−2−3
xrxr−1+4
x2xr= 0,
r2−r−3r+ 4 = 0,
r2−4r+ 4 = 0,
(r−2)2= 0.
13
Step 3: The characteristic equation has a repeated root r= 2. Therefore,
the general solution is given by:
y(x) = c1x2+c2x2ln(x),
where c1and c2are arbitrary constants.
Question 20
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xm.
Solution
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=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the
given differential equation.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
Step 3: Find the roots of the characteristic equation.
m= 2
Step 4: Write down the general solution based on the roots found.
y=c1x2+c2x2ln(x)
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 3y= 0 using the substitution
y=xm.
14
Solution
Step 1: Let’s begin by making the substitution y=xminto the differential
equation x2y′′ −3xy′+ 3y= 0, and find the derivatives of y.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Substitute y,y′, and y′′ back into the differential equation:
x2(m(m−1)xm−2)−3x(mxm−1)+3xm= 0
Simplify the equation:
m(m−1)xm−3mxm+ 3xm= 0
m(m−1) −3m+ 3 = 0
Step 2: Solve the resulting characteristic equation m(m−1) −3m+ 3 = 0:
m2−m−3m+ 3 = 0
m2−4m+ 3 = 0
(m−3)(m−1) = 0
So, we have m= 1 or m= 3.
Step 3: Form the general solution:
y=c1x1+c2x3
y=c1x+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
3y= 0 is y=c1x+c2x3.
Question 22
Question
Solve the Cauchy-Euler equation:
x2y′′ +xy′−y= 0
15
Solution
Step 1: Write the equation in standard form by dividing through by x2:
y′′ +1
xy′−1
x2y= 0
Step 2: Guess a solution of the form y=xr. Substitute this into the
equation:
r(r−1)xr−2+1
xrxr−1−1
x2xr= 0
Step 3: Simplify the equation:
r2−r+r−1 = 0
r2−1 = 0
(r−1)(r+ 1) = 0
Step 4: Find the roots of the characteristic equation:
r1= 1 and r2=−1
Step 5: Write down the general solution:
y=c1x1+c2x−1
y=c1x+c2
x
Question 23
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the differential equation.
x2(r)(r−1)xr−2−3x(rxr−1)+4xr= 0
16
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 5: Solve for r.
r= 2
Step 6: Form the general solution. Since there is a repeated root at r= 2,
the general solution is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are constants.
Question 24
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 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.
x2(r(r−1)xr−2)−5x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−5rxr+ 4xr= 0
r2−r−5r+ 4 = 0
r2−6r+ 4 = 0
Step 4: Solve the quadratic equation for r.
r=6±p62−4(1)(4)
2=6±√32
2=6±4√2
2
17
Step 5: Find the roots of the equation.
r1= 3 + 2√2, r2= 3 −2√2
Step 6: Write the general solution using the roots r1and r2.
y=c1x3+2√2+c2x3−2√2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
4y= 0 is y=c1x3+2√2+c2x3−2√2.
Question 25
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y=x3
Solution
Step 1: We first identify the characteristic equation by substituting y=xrinto
the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x3
Step 2: Simplifying the expression gives:
r(r−1)xr−3rxr+ 4xr=x3
Step 3: Combining like terms yields the characteristic equation:
r2−4r+ 4 = 0
Step 4: The characteristic equation can be factored as (r−2)2= 0, so we
have a repeated root r= 2.
Step 5: The solution to the Cauchy-Euler equation is therefore of the form:
y(x) = c1x2+c2x2ln(x)
Step 6: To find a particular solution, we use the method of undetermined
coefficients. Assume a particular solution of the form:
yp=Ax3
Step 7: Substituting yp=Ax3back into the original differential equation,
we get:
(Ax3)′′ −3x(Ax3)′+ 4Ax3=x3
18
Step 8: Simplifying the expression gives:
6Ax + 4Ax3=x3
Step 9: Equating coefficients gives 4A= 1, so A=1
4.
Step 10. The general solution to the Cauchy-Euler equation is:
y(x) = c1x2+c2x2ln(x) + 1
4x3
Question 26
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: Write down the characteristic equation. In the case of a Cauchy-Euler
equation, the characteristic equation is obtained by substituting y=xrinto the
differential equation.
r(r−1)xr−5rxr+ 6xr= 0
Step 2: Simplify the characteristic equation.
r2−r−5r+ 6 = 0
r2−6r+ 6 = 0
Step 3: Solve for the roots of the characteristic equation using the quadratic
formula.
r=6±√62−4·1·6
2=6±√12
2=6±2√3
2
r= 3 ±√3
Step 4: Write down the general solution to the Cauchy-Euler equation.
y(x) = c1x3+√3+c2x3−√3
Therefore, the general solution to the given Cauchy-Euler equation is y(x) =
c1x3+√3+c2x3−√3, where c1and c2are arbitrary constants.
Question 27
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 8y= 0
19
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Dif-
ferentiate ytwice and substitute into the differential equation. Step 3: Solve
the characteristic equation to find the roots. Step 4: Write down the general
solution.
Step 1: Assume y=xrand differentiate.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −5xy′+ 8y= 0
x2(r(r−1)xr−2)−5x(rxr−1)+8xr= 0
r(r−1)xr−5rxr+ 8xr= 0
Step 3: Solve the characteristic equation r(r−1) −5r+ 8 = 0.
r2−r−5r+ 8 = 0
r2−6r+ 8 = 0
(r−2)(r−4) = 0
So, the roots are r= 2 and r= 4.
Step 4: Write down the general solution.
The general solution of the Cauchy-Euler equation is:
y(x) = c1x2+c2x4
where c1and c2are constants.
Question 28
Question
Solve the Cauchy-Euler equation x2y′′ +xy′−6y= 0 using the substitution
y=xr.
20
Solution
Step 1: Find the derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives into the Cauchy-Euler equation.
x2(r(r−1)xr−2) + x(rxr−1)−6xr= 0
Step 3: Simplify the equation.
r(r−1)xr+rxr−6xr= 0
Step 4: Combine like terms.
r2xr−r2xr−6xr= 0
Step 5: Simplify further.
−6xr= 0
Step 6: Divide by xr.
−6=0
Step 7: Since −6= 0, there are no solutions to this differential equation
using the substitution y=xr. Thus, we conclude that y=xrdoes not solve
the Cauchy-Euler equation x2y′′ +xy′−6y= 0.
Question 29
Question
Solve the Cauchy-Euler equation x2y′′ + 5xy′−6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the differential equation.
x2(r(r−1)xr−2)+5x(rxr−1)−6xr= 0
r(r−1)xr+ 5rxr−6xr= 0
r2−r+ 5r−6=0
r2+ 4r−6=0
(r+ 3)(r−2) = 0
21
Step 3: Solve for the roots of the characteristic equation: r=−3 and r= 2.
Step 4: Write down the general solution using the roots r1and r2.
y=c1x−3+c2x2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ + 5xy′−
6y= 0 is y=c1x−3+c2x2, where c1and c2are arbitrary constants.
Question 30
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y= 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) back into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr= 0
Step 4: Simplify the equation by combining like terms:
r(r−1)xr−5rxr+ 6xr= 0
r2−r−5r+ 6 = 0
Step 5: Solve the quadratic equation r2−6r+ 6 = 0 to find the roots:
r=6±p(−6)2−4∗1∗6
2∗1=6±√36 −24
2=6±√12
2
r=6±2√3
2= 3 ±√3
Step 6: Write the general solution using the roots r1= 3 + √3 and r2= 3 −√3:
y(x) = c1x3+√3+c2x3−√3
where c1and c2are constants.
Question 31
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 7y= 0.
22
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Calculate 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 differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+7xr= 0
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 7xr= 0
r(r−1) −3r+ 7 = 0
Step 5: Solve the quadratic equation.
r2−r−3r+ 7 = 0
r2−4r+ 7 = 0
r=4±p(−4)2−4(1)(7)
2(1)
r=4±√16 −28
2
r=4±√−12
2
r=4±2i√3
2
r= 2 ±i√3
Step 6: Write the general solution.
y(x) = c1x2+√3i+c2x2−√3i
y(x) = c1x2cos√3 ln(x)+c2x2sin√3 ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
7y= 0 is y(x) = c1x2cos√3 ln(x)+c2x2sin√3 ln(x).
Question 32
Question
Solve the Cauchy-Euler differential equation: x2y′′ + 5xy′+ 7y= 0.
23
Solution
Step 1: Write the equation in standard form by finding the characteristic equa-
tion. Assume y=xr.
x2r(r−1)xr−2+ 5xrxr−1+ 7xr= 0
r(r−1) + 5r+ 7 = 0
r2+ 4r+ 7 = 0
Step 2: Solve the characteristic equation for rusing the quadratic formula.
r=−4±p42−4(1)(7)
2(1)
r=−4±√16 −28
2
r=−4±√−12
2
r=−4±2i√3
2
r=−2±i√3
Step 3: Form the general solution using the roots of the characteristic equa-
tion.
y(x) = c1x−2cos√3 ln x+c2x−2sin√3 ln x
where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler differential equation x2y′′ +
5xy′+ 7y= 0 is y(x) = c1x−2cos√3 ln x+c2x−2sin√3 ln x.
Question 33
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm.
Solution
To solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm, we need to find the values of mthat satisfy the equation.
Step 1: Find the derivatives of y=xm. Start by finding the first and
second derivatives of y:
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
24
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation. Substitute
y=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the Cauchy-Euler equation:
x2(m(m−1)xm−2)−5x(mxm−1)+6xm= 0
Step 3: Simplify the equation. Simplify the equation from Step 2:
m(m−1)xm−5mxm+ 6xm= 0
Step 4: Factor out xm. Factor out xmfrom the equation:
xm(m(m−1) −5m+ 6) = 0
Step 5: Set the expression equal to zero. Set the expression equal to zero
to find the values of m:
m(m−1) −5m+ 6 = 0
m2−m−5m+ 6 = 0
m2−6m+ 6 = 0
Step 6: Solve for m. Solve the quadratic equation m2−6m+ 6 = 0 to find
the values of m. The solutions are:
m= 3 ±√3
Step 7: Write the general solution. The general solution to the Cauchy-
Euler equation is given by:
y(x) = c1x3+√3+c2x3−√3
where c1and c2are constants determined by initial conditions.
Question 34
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand differentiate twice to find
y′and y′′.
y(x) = xr
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
25
Step 2: Substitute y,y′, and y′′ back into the Cauchy-Euler equation.
x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
Step 3: Simplify the equation and express it in terms of r.
r(r−1)xr+ 3rxr−4xr= 0
r2−r+ 3r−4=0
r2+ 2r−4=0
(r+ 2)(r−2) = 0
Step 4: Find the roots of the characteristic equation to determine the solu-
tions.
The roots are r=−2 and r= 2.
Step 5: Form the general solution using the roots.
y(x) = c1x−2+c2x2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ + 3xy′−
4y= 0 is y(x) = c1x−2+c2x2, where c1and c2are arbitrary constants.
Question 35
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand compute the derivatives.
Step 2: Substitute y(x) and its derivatives into the Cauchy-Euler equation.
Step 3: Simplify the resulting equation by regrouping terms.
Step 4: Solve the resulting characteristic equation for r.
Step 5: Write down the general solution of the Cauchy-Euler equation.
Step 1: Assume y(x) = xr, then find the first and second derivatives:
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
Step 2: Substitute y(x) and its derivatives into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
26
(m−2)(m−3) = 0
Step 4: Find the values of m
m= 2 or m= 3
Step 5: Write the general solution The general solution is
y=c1x2+c2x3
where c1and c2are constants.
Question 2
Question
Solve the Cauchy-Euler equation (x2−1)y′′−3xy′+4y= 0 using the substitution
y=xm.
Solution
Step 1: Find the first and second derivatives of y=xm. The first derivative of
yis y′=mxm−1and the second derivative is y′′ =m(m−1)xm−2.
Step 2: Substitute y=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the
Cauchy-Euler equation. We get (x2−1)(m(m−1)xm−2)−3x(mxm−1)+4xm= 0.
Step 3: Simplify the equation. Expanding and combining like terms gives
us (m2−m)xm−3mxm+ 4xm= 0.
Step 4: Combine like terms. We have (m2−4m+ 4)xm= 0.
Step 5: Equate the coefficient of xmto zero. Setting (m2−4m+ 4) = 0, we
find the characteristic equation m2−4m+ 4 = 0.
Step 6: Solve the characteristic equation. Factoring the characteristic equa-
tion gives us (m−2)2= 0, so m= 2 (with multiplicity 2).
Step 7: Write down the general solution. Since m= 2 has multiplicity 2,
the general solution is y(x) = c1x2+c2x2ln(x), where c1and c2are arbitrary
constants.
Question 3
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y=x2ln x
2
Solution
Step 1: Assume a solution of the form y=xr. Step 2: Differentiate yto find y′
and y′′. Step 3: Substitute y, y′, and y′′ into the Cauchy-Euler equation. Step
4: Solve for the characteristic equation. Step 5: Determine the roots of the
characteristic equation and use them to form the general solution. Step 6: Find
the particular solution using the method of variation of parameters. Step 7:
Combine the general solution and particular solution to get the final solution.
Question 4
Question
Solve the Cauchy-Euler equation
x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume the solution has the form y(x) = xrand differentiate to find 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 to get
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0.
Step 3: Simplify the equation by combining like terms.
r(r−1) −3r+ 4 = 0.
Step 4: Solve the quadratic equation for r.
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2.
Step 5: Form the general solution using the value of r.
y(x) = c1x2+c2x2ln(x),
where c1and c2are constants.
3
Question 5
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives of yinto 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
xr(r2−6r+ 6) = 0
Step 5: Find the roots of the characteristic equation r2−6r+ 6 = 0.
r=6±p(−6)2−4(1)(6)
2(1)
=6±√36 −24
2
=6±√12
2
= 3 ±√3
Step 6: Write down the general solution.
y=c1x3+√3+c2x3−√3
where c1and c2are arbitrary constants.
Question 6
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
4
Solution
Step 1: Write the characteristic equation by substituting y=xminto the
Cauchy-Euler equation.
x2m(m−1)xm−2−5xmxm−1+ 6xm= 0
m(m−1) −5m+ 6 = 0
m2−6m+ 6 = 0
Step 2: Solve the characteristic equation to find the roots m1and m2.
m=6±p(−6)2−4∗1∗6
2=6±√12
2= 3 ±√3
Step 3: Write down the general solution using the roots m1and m2.
y(x) = c1x3+√3+c2x3−√3
So the general solution to the Cauchy-Euler equation is y(x) = c1x3+√3+
c2x3−√3.
Question 7
Question
Find the general solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Guess a solution of the form y=xr. Substitute this into the differential
equation to find the characteristic equation:
r(r−1)xr−3rxr+ 4xr= 0
Step 2: Simplify the equation:
r2xr−rxr−3rxr+ 4xr= 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 3: The characteristic equation has a repeated root r= 2. Therefore,
we have the solutions y1=x2and y2=x2ln(x).
Step 4: The general solution to the Cauchy-Euler equation is given by:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
5
Question 8
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−4x(rxr−1)+6xr=xr(r(r−1) −4r+ 6) = 0
Step 3: Simplify the expression.
r(r−1) −4r+ 6 = r2−r−4r+ 6 = r2−5r+ 6 = (r−2)(r−3) = 0
Step 4: Solve for the roots of the characteristic equation.
r= 2 or r= 3
Step 5: Write down the general solution.
y=c1x2+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y=c1x2+c2x3, where c1and c2are arbitrary constants.
Question 9
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 16y= 0
Solution
Step 1: Assume a solution of the form y(x) = xr. Then, compute the first and
second derivatives of y(x). Step 2: Substitute y(x), y′(x), and y′′(x) into the
differential equation. Step 3: Simplify the equation by factoring out xr. Step 4:
Solve for the values of r. Step 5: Form the general solution based on the values
of rfound in step 4.
Step 1: Assume y(x) = xr. Then, compute the first and second derivatives
of y(x).
6
y′(x) = rxr−1and y′′(x) = r(r−1)xr−2
Step 2: Substitute y(x), y′(x), and y′′(x) into the differential equation.
x2(r(r−1)xr−2)−5x(rxr−1) + 16xr= 0
Step 3: Simplify the equation by factoring out xr.
r(r−1) −5r+ 16 = 0
Step 4: Solve for the values of r.
r2−r−5r+ 16 = r2−6r+ 16 = (r−2)(r−8) = 0
So, r= 2 or r= 8.
Step 5: Form the general solution based on the values of rfound in step 4.
The general solution is:
y(x) = c1x2+c2x8
where c1and c2are arbitrary constants.
Question 10
Question
Solve the Cauchy-Euler equation (x2−x−6)y′′−3xy′−4y= 0 using the method
of undetermined coefficients.
Solution
1. First, we assume a solution of the form y(x) = xr. We can rewrite the
given differential equation as:
(x2−x−6)r(r−1)xr−2−3xrxr−1−4xr= 0
2. Simplifying the above expression:
(r2−7r+ 12)xr−3rxr−4xr= 0
r2−10r+ 12 = 0
(r−2)(r−6) = 0
The roots are r1= 2 and r2= 6. So, the general solution is given by:
y(x) = c1x2+c2x6
7
Question 11
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr. Substituting this into the
equation, we get:
x2(r)(r−1)xr−2−3x(r)xr−1+ 4xr= 0
Step 2: Simplify the equation:
r(r−1) −3r+ 4 = 0
Step 3: Rearrange the terms to get a quadratic equation in terms of r:
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 4: Since we have a repeated root, the general solution is:
y=c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 12
Question
Solve the Cauchy-Euler equation: x3y′′′ −2x2y′′ +xy′−y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Find the first three derivatives of y(x):
y′(x) = rxr−1, y′′(x) = r(r−1)xr−2, y′′′(x) = r(r−1)(r−2)xr−3
Step 3: Substitute y(x) and its derivatives back into the original equation and
simplify:
x3(r(r−1)(r−2)xr−3)−2x2(r(r−1)xr−2) + x(rxr−1)−xr= 0
8
Step 4: Simplify the equation:
r(r−1)(r−2)xr−2r(r−1)xr+rxr−xr= 0
Step 5: Combine like terms:
r3xr−3r2xr+rxr−xr= 0
r3−2r2+r−1 = r(r2−2r+ 1) = r(r−1)2= 0
Step 6: Solve for r:r= 0,1 (with multiplicity 2)
Step 7: Write down the general solutions for r:
y(x) = c1x0+c2x+c3xln(x)
where c1, c2, c3are arbitrary constants.
Question 13
Question
Solve the Cauchy-Euler equation (x2−4)y′′ −xy′+y= 0.
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Dif-
ferentiate y=xrto find y′and y′′. Step 3: Substitute y,y′, and y′′ into the
Cauchy-Euler equation. Step 4: Divide the equation by x2to simplify it. Step
5: Solve the resulting auxiliary equation to find the roots. Step 6: Write down
the general solution using the roots found.
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0
Solution
Step 1: Let’s assume a solution of the form y(x) = xr. We will substitute this
into the differential equation to find the possible values of r.
Step 2: Calculate the first and second derivatives of y(x):
y′=rxr−1and y′′ =r(r−1)xr−2
9
Step 3: Now substitute y,y′, and y′′ into the differential equation:
x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
Step 4: Simplify the expression by combining like terms:
r(r−1)xr+ 3rxr−4xr= 0
Step 5: Factor out xr:
xr(r2−r+ 3r−4) = 0
Step 6: Simplify the expression further:
xr(r2+ 2r−4) = 0
Step 7: Set the expression equal to zero and solve for r:
r2+ 2r−4=0
Step 8: Use the quadratic formula to find the roots of the equation:
r=−2±p22−4(1)(−4)
2(1) =−2±√4 + 16
2=−2±√20
2=−1±√5
Step 9: The general solution is given by y(x) = c1x(−1+√5) +c2x(−1−√5).
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x2ln(x).
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) and its derivatives into the differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x2ln(x)
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr=x2ln(x)
Step 5: Combine like terms.
r2−r−3r+ 4 = ln(x)
10
Step 6: Simplify the equation.
r2−4r+ 4 = ln(x)
Step 7: Solve the characteristic equation r2−4r+ 4 = 0 to find the roots of
r.
(r−2)2= 0 =⇒r= 2
Step 8: The solution to the characteristic equation gives us a repeated root
r= 2.
Step 9: Therefore, the general solution to the Cauchy-Euler equation is given
by
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Step 10: Now we find the particular solution to the original differential
equation by substituting y(x) = c1x2+c2x2ln(x) into the equation x2y′′ −
3xy′+ 4y=x2ln(x).
Step 11: After substitution and simplification, we equate coefficients of like
terms to obtain c1=−2 and c2= 1.
Step 12: Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y=x2ln(x) is
y(x) = −2x2+x2ln(x)
Question 16
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y, y′, y′′ back into the Cauchy-Euler 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.
r2−r+ 4 = 0
11
Step 6: Solve the quadratic equation r2−r+ 4 = 0 for r.
r=1±p1−4(1)(4)
2=1±√−15
2=1±i√15
2
Step 7: Write down the solutions.
y1=x1+i√15
2and y2=x1−i√15
2
Question 17
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Begin by assuming a solution of the form y=xrand substitute it
into the differential equation. Then find the characteristic equation by solving
x2r(r−1) −3xr + 4 = 0.
Step 2: Simplify the characteristic equation to r2−4r+ 4 = 0.
Step 3: Factor the characteristic equation to (r−2)2= 0.
Step 4: There is a repeated root at r= 2. Therefore, the general solution
takes the form y=C1x2+C2x2ln(x), where C1and C2are constants.
Step 5: Thus, the general solution to the Cauchy-Euler equation x2y′′ −
3xy′+ 4y= 0 is y=C1x2+C2x2ln(x).
Question 18
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x4.
Solution
Step 1: Write the characteristic equation by substituting y=xrinto the homo-
geneous form of the equation (xny′′ +pxn−1y′+qxny= 0).
r(r−1)xr−2−3rxr−1+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2(multiplicity 2)
So, the auxiliary equation has the repeated root r= 2.
12
Step 2: Form the complementary function yc(x) = c1x2+c2x2ln xwhere c1
and c2are constants to be determined.
Step 3: Find a particular solution yp(x) using the method of undetermined
coefficients. Since x4is a polynomial of degree 4, we guess yp(x) = Ax4+Bx3+
Cx2+Dx +E.
x2(4A)−3x(4Ax3+ 3Bx2+ 2Cx +D) + 4(Ax4+Bx3+Cx2+Dx +E) = x4
4Ax2−12Ax2−9Bx2−8Cx −3Dx + 4Ax4+ 4Bx3+ 4Cx2+ 4Dx + 4E=x4
4Ax4+ (4B−12A)x3+ (−9B+ 4C)x2+ (−8C−3D)x+ 4A=x4
By comparing coefficients, we get:
4A= 1,4B−12A= 0,−9B+ 4C= 0,−8C−3D= 0,4A= 0
Solving the system of equations, we find A=1
4, B =3
16 , C =9
64 , D =
−27
128 , E = 0.
Therefore, the particular solution is yp(x) = 1
4x4+3
16 x3+9
64 x2−27
128 x.
Step 4: The general solution is given by y(x) = yc(x) + yp(x). Thus, the
solution to the Cauchy-Euler equation is
y(x) = c1x2+c2x2ln x+1
4x4+3
16x3+9
64x2−27
128x
Question 19
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Write the equation in standard form by dividing by x2:
y′′ −3
xy′+4
x2y= 0.
Step 2: Assume a solution of the form y=xr. Substitute yand its deriva-
tives into the equation and solve for r:
r(r−1)xr−2−3
xrxr−1+4
x2xr= 0,
r2−r−3r+ 4 = 0,
r2−4r+ 4 = 0,
(r−2)2= 0.
13
Step 3: The characteristic equation has a repeated root r= 2. Therefore,
the general solution is given by:
y(x) = c1x2+c2x2ln(x),
where c1and c2are arbitrary constants.
Question 20
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xm.
Solution
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=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the
given differential equation.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
Step 3: Find the roots of the characteristic equation.
m= 2
Step 4: Write down the general solution based on the roots found.
y=c1x2+c2x2ln(x)
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 3y= 0 using the substitution
y=xm.
14
Solution
Step 1: Let’s begin by making the substitution y=xminto the differential
equation x2y′′ −3xy′+ 3y= 0, and find the derivatives of y.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Substitute y,y′, and y′′ back into the differential equation:
x2(m(m−1)xm−2)−3x(mxm−1)+3xm= 0
Simplify the equation:
m(m−1)xm−3mxm+ 3xm= 0
m(m−1) −3m+ 3 = 0
Step 2: Solve the resulting characteristic equation m(m−1) −3m+ 3 = 0:
m2−m−3m+ 3 = 0
m2−4m+ 3 = 0
(m−3)(m−1) = 0
So, we have m= 1 or m= 3.
Step 3: Form the general solution:
y=c1x1+c2x3
y=c1x+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
3y= 0 is y=c1x+c2x3.
Question 22
Question
Solve the Cauchy-Euler equation:
x2y′′ +xy′−y= 0
15
Solution
Step 1: Write the equation in standard form by dividing through by x2:
y′′ +1
xy′−1
x2y= 0
Step 2: Guess a solution of the form y=xr. Substitute this into the
equation:
r(r−1)xr−2+1
xrxr−1−1
x2xr= 0
Step 3: Simplify the equation:
r2−r+r−1 = 0
r2−1 = 0
(r−1)(r+ 1) = 0
Step 4: Find the roots of the characteristic equation:
r1= 1 and r2=−1
Step 5: Write down the general solution:
y=c1x1+c2x−1
y=c1x+c2
x
Question 23
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the differential equation.
x2(r)(r−1)xr−2−3x(rxr−1)+4xr= 0
16
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 5: Solve for r.
r= 2
Step 6: Form the general solution. Since there is a repeated root at r= 2,
the general solution is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are constants.
Question 24
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 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.
x2(r(r−1)xr−2)−5x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−5rxr+ 4xr= 0
r2−r−5r+ 4 = 0
r2−6r+ 4 = 0
Step 4: Solve the quadratic equation for r.
r=6±p62−4(1)(4)
2=6±√32
2=6±4√2
2
17
Step 5: Find the roots of the equation.
r1= 3 + 2√2, r2= 3 −2√2
Step 6: Write the general solution using the roots r1and r2.
y=c1x3+2√2+c2x3−2√2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
4y= 0 is y=c1x3+2√2+c2x3−2√2.
Question 25
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y=x3
Solution
Step 1: We first identify the characteristic equation by substituting y=xrinto
the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x3
Step 2: Simplifying the expression gives:
r(r−1)xr−3rxr+ 4xr=x3
Step 3: Combining like terms yields the characteristic equation:
r2−4r+ 4 = 0
Step 4: The characteristic equation can be factored as (r−2)2= 0, so we
have a repeated root r= 2.
Step 5: The solution to the Cauchy-Euler equation is therefore of the form:
y(x) = c1x2+c2x2ln(x)
Step 6: To find a particular solution, we use the method of undetermined
coefficients. Assume a particular solution of the form:
yp=Ax3
Step 7: Substituting yp=Ax3back into the original differential equation,
we get:
(Ax3)′′ −3x(Ax3)′+ 4Ax3=x3
18
Step 8: Simplifying the expression gives:
6Ax + 4Ax3=x3
Step 9: Equating coefficients gives 4A= 1, so A=1
4.
Step 10. The general solution to the Cauchy-Euler equation is:
y(x) = c1x2+c2x2ln(x) + 1
4x3
Question 26
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: Write down the characteristic equation. In the case of a Cauchy-Euler
equation, the characteristic equation is obtained by substituting y=xrinto the
differential equation.
r(r−1)xr−5rxr+ 6xr= 0
Step 2: Simplify the characteristic equation.
r2−r−5r+ 6 = 0
r2−6r+ 6 = 0
Step 3: Solve for the roots of the characteristic equation using the quadratic
formula.
r=6±√62−4·1·6
2=6±√12
2=6±2√3
2
r= 3 ±√3
Step 4: Write down the general solution to the Cauchy-Euler equation.
y(x) = c1x3+√3+c2x3−√3
Therefore, the general solution to the given Cauchy-Euler equation is y(x) =
c1x3+√3+c2x3−√3, where c1and c2are arbitrary constants.
Question 27
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 8y= 0
19
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Dif-
ferentiate ytwice and substitute into the differential equation. Step 3: Solve
the characteristic equation to find the roots. Step 4: Write down the general
solution.
Step 1: Assume y=xrand differentiate.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −5xy′+ 8y= 0
x2(r(r−1)xr−2)−5x(rxr−1)+8xr= 0
r(r−1)xr−5rxr+ 8xr= 0
Step 3: Solve the characteristic equation r(r−1) −5r+ 8 = 0.
r2−r−5r+ 8 = 0
r2−6r+ 8 = 0
(r−2)(r−4) = 0
So, the roots are r= 2 and r= 4.
Step 4: Write down the general solution.
The general solution of the Cauchy-Euler equation is:
y(x) = c1x2+c2x4
where c1and c2are constants.
Question 28
Question
Solve the Cauchy-Euler equation x2y′′ +xy′−6y= 0 using the substitution
y=xr.
20
Solution
Step 1: Find the derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute the derivatives into the Cauchy-Euler equation.
x2(r(r−1)xr−2) + x(rxr−1)−6xr= 0
Step 3: Simplify the equation.
r(r−1)xr+rxr−6xr= 0
Step 4: Combine like terms.
r2xr−r2xr−6xr= 0
Step 5: Simplify further.
−6xr= 0
Step 6: Divide by xr.
−6=0
Step 7: Since −6= 0, there are no solutions to this differential equation
using the substitution y=xr. Thus, we conclude that y=xrdoes not solve
the Cauchy-Euler equation x2y′′ +xy′−6y= 0.
Question 29
Question
Solve the Cauchy-Euler equation x2y′′ + 5xy′−6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the derivatives of y=xr.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the differential equation.
x2(r(r−1)xr−2)+5x(rxr−1)−6xr= 0
r(r−1)xr+ 5rxr−6xr= 0
r2−r+ 5r−6=0
r2+ 4r−6=0
(r+ 3)(r−2) = 0
21
Step 3: Solve for the roots of the characteristic equation: r=−3 and r= 2.
Step 4: Write down the general solution using the roots r1and r2.
y=c1x−3+c2x2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ + 5xy′−
6y= 0 is y=c1x−3+c2x2, where c1and c2are arbitrary constants.
Question 30
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y= 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) back into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr= 0
Step 4: Simplify the equation by combining like terms:
r(r−1)xr−5rxr+ 6xr= 0
r2−r−5r+ 6 = 0
Step 5: Solve the quadratic equation r2−6r+ 6 = 0 to find the roots:
r=6±p(−6)2−4∗1∗6
2∗1=6±√36 −24
2=6±√12
2
r=6±2√3
2= 3 ±√3
Step 6: Write the general solution using the roots r1= 3 + √3 and r2= 3 −√3:
y(x) = c1x3+√3+c2x3−√3
where c1and c2are constants.
Question 31
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 7y= 0.
22
Solution
Step 1: Assume a solution of the form y(x) = xr.
Step 2: Calculate 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 differential equation.
x2(r(r−1)xr−2)−3x(rxr−1)+7xr= 0
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 7xr= 0
r(r−1) −3r+ 7 = 0
Step 5: Solve the quadratic equation.
r2−r−3r+ 7 = 0
r2−4r+ 7 = 0
r=4±p(−4)2−4(1)(7)
2(1)
r=4±√16 −28
2
r=4±√−12
2
r=4±2i√3
2
r= 2 ±i√3
Step 6: Write the general solution.
y(x) = c1x2+√3i+c2x2−√3i
y(x) = c1x2cos√3 ln(x)+c2x2sin√3 ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
7y= 0 is y(x) = c1x2cos√3 ln(x)+c2x2sin√3 ln(x).
Question 32
Question
Solve the Cauchy-Euler differential equation: x2y′′ + 5xy′+ 7y= 0.
23
Solution
Step 1: Write the equation in standard form by finding the characteristic equa-
tion. Assume y=xr.
x2r(r−1)xr−2+ 5xrxr−1+ 7xr= 0
r(r−1) + 5r+ 7 = 0
r2+ 4r+ 7 = 0
Step 2: Solve the characteristic equation for rusing the quadratic formula.
r=−4±p42−4(1)(7)
2(1)
r=−4±√16 −28
2
r=−4±√−12
2
r=−4±2i√3
2
r=−2±i√3
Step 3: Form the general solution using the roots of the characteristic equa-
tion.
y(x) = c1x−2cos√3 ln x+c2x−2sin√3 ln x
where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler differential equation x2y′′ +
5xy′+ 7y= 0 is y(x) = c1x−2cos√3 ln x+c2x−2sin√3 ln x.
Question 33
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm.
Solution
To solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm, we need to find the values of mthat satisfy the equation.
Step 1: Find the derivatives of y=xm. Start by finding the first and
second derivatives of y:
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
24
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation. Substitute
y=xm,y′=mxm−1, and y′′ =m(m−1)xm−2into the Cauchy-Euler equation:
x2(m(m−1)xm−2)−5x(mxm−1)+6xm= 0
Step 3: Simplify the equation. Simplify the equation from Step 2:
m(m−1)xm−5mxm+ 6xm= 0
Step 4: Factor out xm. Factor out xmfrom the equation:
xm(m(m−1) −5m+ 6) = 0
Step 5: Set the expression equal to zero. Set the expression equal to zero
to find the values of m:
m(m−1) −5m+ 6 = 0
m2−m−5m+ 6 = 0
m2−6m+ 6 = 0
Step 6: Solve for m. Solve the quadratic equation m2−6m+ 6 = 0 to find
the values of m. The solutions are:
m= 3 ±√3
Step 7: Write the general solution. The general solution to the Cauchy-
Euler equation is given by:
y(x) = c1x3+√3+c2x3−√3
where c1and c2are constants determined by initial conditions.
Question 34
Question
Solve the Cauchy-Euler equation:
x2y′′ + 3xy′−4y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand differentiate twice to find
y′and y′′.
y(x) = xr
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
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Step 2: Substitute y,y′, and y′′ back into the Cauchy-Euler equation.
x2(r(r−1)xr−2)+3x(rxr−1)−4xr= 0
Step 3: Simplify the equation and express it in terms of r.
r(r−1)xr+ 3rxr−4xr= 0
r2−r+ 3r−4=0
r2+ 2r−4=0
(r+ 2)(r−2) = 0
Step 4: Find the roots of the characteristic equation to determine the solu-
tions.
The roots are r=−2 and r= 2.
Step 5: Form the general solution using the roots.
y(x) = c1x−2+c2x2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ + 3xy′−
4y= 0 is y(x) = c1x−2+c2x2, where c1and c2are arbitrary constants.
Question 35
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume a solution of the form y(x) = xrand compute the derivatives.
Step 2: Substitute y(x) and its derivatives into the Cauchy-Euler equation.
Step 3: Simplify the resulting equation by regrouping terms.
Step 4: Solve the resulting characteristic equation for r.
Step 5: Write down the general solution of the Cauchy-Euler equation.
Step 1: Assume y(x) = xr, then find the first and second derivatives:
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
Step 2: Substitute y(x) and its derivatives into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
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Step 3: Regroup the terms:
r(r−1)xr−3rxr+ 4xr= 0
Step 4: Solve for rby combining like terms:
r(r−1−3 + 4)xr= 0
r2= 0
r= 0
Step 5: The general solution is given by:
y(x) = c1x0+c2x0ln(x)
y(x) = c1+c2ln(x)
where c1and c2are arbitrary constants.
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