MATH 334 - DIFFERENTIAL
EQUATIONS - Cauchy-Euler equations
Question Bank - Set 2
Liberty University
Question 1
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
Step 1: Assume the solution is 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′′ 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 xr.
xr(r2−4r+ 4) = 0
Step 6: Set the equation equal to zero for the characteristic equation.
r2−4r+ 4 = 0
Step 7: Solve the characteristic equation to find the roots r1and r2.
r= 2
Step 8: Form the general solution.
y=c1xr1+c2xr2
y=c1x2+c2x2
y= (c1+c2)x2
Therefore, the general solution to the Cauchy-Euler equation is y=c1x2+
c2x2= (c1+c2)x2.
Question 2
Question
Solve the Cauchy-Euler differential equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xm.
Step 2: Calculate the first and second derivatives of ywith respect to x.
y′=mxm−1and y′′ =m(m−1)xm−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation and simplify.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m2−m−3m+ 4 = m(m−4) = 0
Step 4: Solve the characteristic equation m(m−4) = 0 to find the roots.
m= 0 or m= 4
Step 5: Form the general solution using the roots found in Step 4.
y=c1x0+c2x4=c1+c2x4
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1+c2x4, where c1and c2are arbitrary constants.
Question 3
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
2
Solution
Step 1: Assume a solution of the form y=xrand differentiate to find y′and
y′′.
Let y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ back into the differential 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
Step 4: Solve for r.
r= 2
Step 5: Form the general solution using the roots r1= 2.
y=c1xr1+c2xr1ln(x)
y=c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln(x).
Question 4
Question
Find the general solution to the Cauchy-Euler equation:
x2y′′ −xy′+y= 0
3
Solution
Step 1: Assume a solution of the form y=xrand substitute it into the differ-
ential equation to find r.
Step 2: Calculate the first and second derivatives of y:
y′=rxr−1
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) + xr= 0
Step 4: Simplify the equation:
r(r−1)xr−rxr+xr= 0
Step 5: Combine like terms:
r(r−1) −r+ 1 = 0
Step 6: Solve for r:
r2−r−r+ 1 = 0
r2−2r+ 1 = 0
(r−1)2= 0
r= 1
Step 7: The solution has repeated root r= 1, so the general solution is of
the form:
y=c1x+c2xln(x)
where c1and c2are arbitrary constants.
Question 5
Question
Find the general solution to the Cauchy-Euler equation (x2−3x)y′′−xy′+y= 0.
Solution
Step 1: Assume a solution of the form y=xm.
Step 2: Find the first and second derivatives of y=xm.
y=xm⇒y′=mxm−1⇒y′′ =m(m−1)xm−2
4
Step 3: Substitute y=xm,y′, and y′′ into the Cauchy-Euler equation.
(x2−3x)(m(m−1)xm−2)−x(mxm−1) + xm= 0
Step 4: Simplify the equation.
(m(m−1)xm)−3mxm−(mxm) + xm= 0
m(m−1)xm−3mxm−mxm+xm= 0
m(m−1)xm−4mxm+xm= 0
m(m−1) −4m+ 1 = 0
Step 5: Solve for m.
m2−m−4m+ 1 = 0
m2−5m+ 1 = 0
m=5±√52−4∗1∗1
2∗1
m=5±√21
2
Step 6: Determine the roots of the characteristic equation.
m1=5 + √21
2and m2=5−√21
2
Step 7: Form the general solution.
y(x) = c1xm1+c2xm2
y(x) = c1x5+√21
2+c2x5−√21
2
Question 6
Question
Solve the Cauchy-Euler equation: x2y′′ + 3xy′+y= 0.
Solution
Step 1: Write the auxiliary equation by assuming that y=xr. The general
form of a Cauchy-Euler equation is ax2y′′ +bxy′+cy = 0. Substitute y=xr
into the given equation and solve for r:
x2r(r−1) + 3xr +xr= 0
Step 2: Solve the auxiliary equation. Simplify the equation to get:
r2−r+ 3r+ 1 = 0
5
r2+ 2r+ 1 = 0
(r+ 1)2= 0
r=−1
Step 3: Write down the general solution using the roots obtained from the
auxiliary equation. The general solution is given by:
y=c1x−1+c2x−1ln(x)
Therefore, the solution to the Cauchy-Euler equation x2y′′ + 3xy′+y= 0 is:
y=c1x−1+c2x−1ln(x)
Question 7
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xr.
Solution
Step 1: Write the differential equation in terms of rby making the substitution
y=xr.
x2(r)(r−1)xr−2−3x(r)xr−1+ 4xr= 0
Step 2: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Factor out xr.
xr(r2−4r+ 4) = 0
Step 4: Set the expression inside the parentheses equal to zero.
r2−4r+ 4 = (r−2)2= 0
Step 5: Find the roots of the characteristic equation.
r= 2
Step 6: Write the general solution using the roots found in step 5.
y=c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln(x).
6
Question 8
Question
Solve the Cauchy-Euler differential equation: x2y′′ −3xy′+ 5y= 0.
Solution
Step 1: Write the characteristic equation using the substitution y=xr. The
characteristic equation for a Cauchy-Euler equation is given by ar(r−1)+br+c=
0, where a,b, and care the coefficients of y′′,y′, and y, respectively.
For our equation x2y′′ −3xy′+ 5y= 0, the characteristic equation becomes
r(r−1) −3r+ 5 = r2−4r+ 5 = 0.
Step 2: Solve the characteristic equation for r. Using the quadratic formula,
r=−(−4)±√(−4)2−4(1)(5)
2(1) =4±√16−20
2=4±√−4
2= 2 ±2i.
Therefore, the solutions for rare r= 2 + 2iand r= 2 −2i.
Step 3: Write the general solution in terms of x. Since the roots are complex,
the general solution will be of the form y=xr=x2+2iand y=xr=x2−2i.
Hence, the general solution is y(x) = c1x2cos(2 ln(x)) + c2x2sin(2 ln(x)).
So, the solution to the Cauchy-Euler differential equation x2y′′−3xy′+5y= 0
is y(x) = c1x2cos(2 ln(x)) + c2x2sin(2 ln(x)).
Question 9
Question
Solve the Cauchy-Euler differential equation: x2y′′ −6xy′+ 12y= 0.
Solution
Step 1: Find the characteristic equation by substituting y=xrinto the differ-
ential equation.
x2r(r−1)xr−2−6xrxr−1+ 12xr= 0
Step 2: Simplify the equation.
r(r−1) −6r+ 12 = 0
Step 3: Solve for rby factoring or using the quadratic formula.
r2−r−6r+ 12 = 0
r(r−1) −6(r−1) = 0
(r−1)(r−6) = 0
r= 1 or r= 6
7
Step 4: Write down the complementary solution.
yc=C1x+C2x6
Step 5: Find the particular solution using the method of variation of param-
eters. Let yp=v1(x)x+v2(x)x6.
Step 6: Calculate y′
pand y′′
p.
y′
p=v1+v′
1x+v2x6+ 6v2x5
y′′
p=v′
1+v′
1x+ 2v2x5+ 6v2x4
Step 7: Substitute yp,y′
p, and y′′
pinto the differential equation.
x2(y′′
p−6y′
p+ 12yp)−6x(y′
p−6yp) + 12yp= 0
Step 8: Solve for v′
1and v′
2.
v′
1=72x6
x2and v′
2= 0
Step 9: Integrate v′
1and v′
2to find v1and v2.
v1= 12x5+C3and v2=C4
Step 10: Write down the particular solution.
yp= (12x5+C3)x+C4x6
Step 11: Write down the general solution.
y=C1x+C2x6+ (12x5+C3)x+C4x6
Step 12: Simplify the general solution.
y=C1x+ 12x6+ 12x6+C3x+C4x6
Step 13: Combine like terms and rewrite the general solution.
y= (C1+C3)x+ (12 + C4)x6
Therefore, the general solution to the Cauchy-Euler equation is y= (C1+
C3)x+ (12 + C4)x6.
Question 10
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 8y= 0.
8
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−5x(rxr−1)+8xr= 0
Step 4: Simplify the equation.
r(r−1)xr−5rxr+ 8xr= 0
Step 5: Factor out xr.
xr(r(r−1) −5r+ 8) = 0
Step 6: Set the expression in the parentheses equal to zero and solve for r.
r(r−1) −5r+ 8 = 0
r2−r−5r+ 8 = 0
r2−6r+ 8 = 0
(r−4)(r−2) = 0
r1= 4 and r2= 2
Step 7: Write down the general solution using both roots.
y=c1x4+c2x2
where c1and c2are arbitrary constants.
Question 11
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y= 0 using the substitution
y=xm.
9
Solution
Step 1: Find the derivatives of ywith respect to x.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the expressions for y,y′, and y′′ into the differential
equation.
x2y′′ −xy′+y= 0
x2(m(m−1)xm−2)−x(mxm−1) + xm= 0
m(m−1)xm−mxm+xm= 0
Step 3: Simplify the equation.
m(m−1)xm−mxm+xm= 0
m(m−1) −m+ 1 = 0
m2−m−m+ 1 = 0
m2−2m+ 1 = 0
(m−1)2= 0
Step 4: Since (m−1)2= 0, we have a repeated root m= 1.
Step 5: Write down the general solution using the roots.
y=c1x+c2xln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′−xy′+y=
0 is y=c1x+c2xln(x), where c1and c2are arbitrary constants.
Question 12
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
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
10
Step 2: Substitute y=xm,y′, and y′′ into the differential equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
Step 3: Simplify the equation.
m(m−1)xm−3mxm+ 4xm= 0
m(m−1) −3m+ 4 = 0 (Divide by xm)
Step 4: Solve for m.
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
m= 2 (Repeated root)
Step 5: Form the solution using m= 2.The solution to the Cauchy-
Euler equation is y=c1x2+c2x2ln x, where c1and c2are arbitrary constants.
Question 13
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−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ 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 xr.
xr(r2−4r+ 4) = 0
Step 6: Solve for r.
r2−4r+ 4 = 0
11
(r−2)2= 0
r= 2
Step 7: General solution is of the form y=c1xr1+c2xr2.
y=c1x2+c2x2
y= (c1+c2)x2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2, or simply y=cx2, where cis an arbitrary constant.
Question 14
Question
Solve the Cauchy-Euler differential equation:
x2y′′ −3xy′+ 4y= 0
Solution
1. We first assume the solution has the form y=xr.
2. Differentiating y=xrwith respect to xgives us:
y′=rxr−1
3. Differentiating y′with respect to xagain gives us:
y′′ =r(r−1)xr−2
4. Substituting y=xr,y′, and y′′ into the differential equation, we get:
x2r(r−1)xr−2−3xrxr−1+ 4xr= 0
5. Simplifying the equation gives:
r(r−1) −3r+ 4 = 0
6. Solving this quadratic equation gives the values of r:
r2−4r+ 4 = (r−2)2= 0
r= 2
7. Since we have a repeated root, the general solution is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
12
Question 15
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−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the differential equation and
simplify.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
Step 4: Factor out xrfrom the equation.
xr(r2−4r+ 4) = 0
Step 5: Set the expression equal to zero and find the roots.
r2−4r+ 4 = 0
(r−2)2= 0
r= 2 (double root)
Step 6: Form the general solution using the roots found in Step 5.
y(x) = c1x2+c2x2ln x
where c1and c2are constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y(x) = c1x2+c2x2ln x.
Question 16
Question
Solve the Cauchy-Euler equation x2y′′−3xy′+9y= 0 by making the substitution
y(x) = xr.
13
Solution
Step 1: We substitute y(x) = xrinto the Cauchy-Euler equation x2y′′ −3xy′+
9y= 0:
x2(r(r−1)xr−2)−3x(rxr−1)+9xr= 0
Expanding and simplifying, we get:
r(r−1)xr−3rxr+ 9xr= 0
Step 2: Factor out xr:
xr(r2−4r+ 9) = 0
This gives us the auxiliary equation r2−4r+ 9 = 0.
Step 3: Solve the auxiliary equation r2−4r+ 9 = 0: The roots of this
quadratic equation can be found using the quadratic formula:
r=4±p(−4)2−4(1)(9)
2(1)
r=4±√16 −36
2
r=4±√−20
2
r=4±2i√5
2
r= 2 ±i√5
Step 4: The general solution is then given by:
y(x) = c1x2+i√5+c2x2−i√5
Step 5: Therefore, the solutions to the Cauchy-Euler equation x2y′′−3xy′+
9y= 0 are:
y(x) = c1x2+i√5+c2x2−i√5
Question 17
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
14
Solution
Step 1: Assume y=xmand find the first and second derivatives.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the expressions for y, y′, y′′ back into the differential equa-
tion.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
Step 3: Simplify and rewrite the equation in terms of m.
m(m−1)xm−3mxm+ 4xm= 0
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
m= 2
Step 4: Use the value of mto find the general solution. Since m= 2, the
general solution is
y=c1x2+c2x2ln x
where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is y=c1x2+c2x2ln x.
Question 18
Question
Solve the Cauchy-Euler equation x2y′′ +xy′−y= 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) + x(rxr−1)−xr= 0
15
Step 3: Simplify the equation.
r(r−1)xr+rxr−xr= 0
r2xr−rxr= 0
rxr(r−1) = 0
Step 4: Find the roots of the characteristic equation r(r−1) = 0.
r= 0,1
Step 5: Write down the general solution using the roots found. Since r= 0
and r= 1 are both roots, the general solution is
y=c1x0+c2x1
Step 6: Simplify the general solution.
y=c1+c2x
Therefore, the solution to the Cauchy-Euler equation x2y′′ +xy′−y= 0 is
y=c1+c2x.
Question 19
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 8y= 0
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Find
roots of the characteristic equation. Step 3: Form the general solution.
Step 1: Assume y=xr, then find y′and y′′:
y′=rxr−1, y′′ =r(r−1)xr−2
Substitute these into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+8xr= 0
Step 2: Simplify the equation:
r(r−1)xr−3rxr+ 8xr= 0
r2−r−3r+ 8 = 0
(r2−4r+ 8) = 0
16
Solve for rby using the quadratic formula:
r=4±p(−4)2−4(1)(8)
2
r=4±√16 −32
2
r=4±√−16
2
r=4±4i
2
r= 2 ±2i
Step 3: The general solution is:
y(x) = c1x2cos(2 ln(x)) + c2x2sin(2 ln(x))
where c1and c2are arbitrary constants.
Question 20
Question
Solve the Cauchy-Euler equation:
x2y′′ −4xy′+ 6y= 0
Solution
Step 1: Assume a solution of the form y=xr. Differentiate ytwice with respect
to x.
Step 2: Substitute y=xrand its derivatives into the Cauchy-Euler equation.
Step 3: Simplify the equation by dividing by xr.
Step 4: The characteristic equation is r(r−1) −4r+ 6 = 0.
Step 5: Solve the characteristic equation to find the roots r1and r2.
Step 6: Write down the general solution using the roots r1and r2.
Step 7: The general solution is y(x) = c1xr1+c2xr2, where c1and c2are
arbitrary constants.
Step 8: Check for linear dependence to ensure a nontrivial solution.
Step 9: The solution to the Cauchy-Euler equation is y(x) = c1x2+c2x3.
Question 21
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′+y=x.
17
Solution
Step 1: First, we assume a solution of the form y=xr, where ris a constant
to be determined.
Step 2: Compute the first and second derivatives of y=xr:
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y, y′,and y′′ back into the Cauchy-Euler equation:
x2·r(r−1)xr−2−x·rxr−1+xr=x
Step 4: Simplify the equation:
r(r−1)xr−rxr+xr=x
Step 5: Combine like terms:
(r2−r)xr=x
Step 6: Set the above expression to zero to find the characteristic equation:
r2−r= 0
Step 7: Solve the characteristic equation to find the roots:
r= 0,1
Step 8: Write down the general solution based on the roots:
y(x) = c1x0+c2x1=c1+c2x
Therefore, the general solution to the Cauchy-Euler equation x2y′′−xy′+y=
xis y(x) = c1+c2x, where c1and c2are arbitrary constants.
Question 22
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: We first assume a solution in the form of y=xr, and differentiate to
find y′and y′′.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
18
Step 2: Substitute the expressions for y,y′, and y′′ into the Cauchy-Euler
equation.
x2(r)(r−1)xr−2−5x(rxr−1)+6xr= 0
Step 3: Simplify the equation by combining like terms.
r(r−1)xr−5rxr+ 6xr= 0
Step 4: Factor out xrfrom the equation.
xr(r(r−1) −5r+ 6) = 0
Step 5: Since we assumed that xris not equal to zero, the term inside the
parentheses must be equal to zero.
r2−r−5r+ 6 = r2−6r+ 6 = 0
Step 6: Solve the quadratic equation for rto get the roots.
r=6±p(−6)2−4(1)(6)
2(1) =6±√36 −24
2=6±2
2
Step 7: The roots are r= 4 and r= 2. This gives the general solutions as
y=c1x4+c2x2
where c1and c2are constants.
Question 23
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 9y= 0
Solution
Step 1: Assume a solution of the form y=xmand find the characteristic
equation by substituting into the differential equation:
x2m(m−1)xm−2−5xmxm−1+ 9xm= 0
⇒m(m−1) −5m+ 9 = 0
⇒m2−6m+ 9 = 0
⇒(m−3)2= 0
⇒m= 3 (since we have a repeated root)
19
Step 2: Therefore, the solution to the differential equation is of the form:
y=c1x3+c2x3ln(x)
where c1and c2are arbitrary constants.
So, the general solution to the Cauchy-Euler equation is:
y=c1x3+c2x3ln(x)
Question 24
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: We first assume the solution has the form y=xr. Then we find the
derivatives needed for substitution into the equation.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives back into the Cauchy-Euler equation.
x2r(r−1)xr−2−5xrxr−1+ 6xr= 0
r(r−1)xr−5rxr+ 6xr= 0
r2−r−5r+ 6 = 0
r2−6r+ 6 = 0
r2−3r−2r+ 6 = 0
r(r−3) −2(r−3) = 0
(r−3)(r−2) = 0
Step 3: We now have the roots r= 3 and r= 2. So the general solution is
given by y(x) = c1x3+c2x2where c1and c2are constants.
Question 25
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 2y=x3.
20
Solution
Step 1: Start by assuming the solution has the form y=xr.
Step 2: Calculate the first and second derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute yand its derivatives back into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−3x(rxr−1)+2xr=x3
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 2xr=x3
Step 5: Divide both sides by xrto get a simple equation.
r(r−1) −3r+ 2 = 1 ⇒r2−4r+ 2 = 1
Step 6: Rearrange and solve for rusing the quadratic formula.
r=4±p(−4)2−4(1)(2)
2
r=4±√16 −8
2=4±2√2
2= 2 ±√2
Step 7: Therefore, the general solution to the Cauchy-Euler equation is
y=c1x2+√2+c2x2−√2
where c1and c2are arbitrary constants.
Question 26
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 9y=x3.
Solution
To solve the Cauchy-Euler equation, we’ll first assume a solution of the form
y=xrand then find the values of rthat satisfy the equation.
Step 1: Find the derivatives of y
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
21
Step 2: Substitute yand its derivatives into the differential equa-
tion Substitute y,y′, and y′′ into the differential equation x2y′′−5xy′+9y=x3:
x2[r(r−1)xr−2]−5x[rxr−1]+9xr=x3
Step 3: Simplify the equation Simplify the left side of the equation:
r(r−1)xr−5rxr+ 9xr=x3
r2xr−rxr−5rxr+ 9xr=x3
r2−6r+ 9 = 1
Step 4: Solve for r
r2−6r+ 9 = 1
r2−6r+ 8 = 0
(r−4)(r−2) = 0
Thus, r= 4 or r= 2.
Step 5: Write the general solution The general solution is a linear
combination of the solutions corresponding to r= 4 and r= 2:
y(x) = c1x4+c2x2
Therefore, the solution to the Cauchy-Euler equation x2y′′ −5xy′+ 9y=x3
is y(x) = c1x4+c2x2, where c1and c2are constants.
Question 27
Question
Consider the Cauchy-Euler equation (x2−3x)y′′ −4xy′+ 6y= 0. Find the
general solution to this differential equation.
Solution
Step 1: Assume a solution in the form of y=xr. Let y=xr, then y′=rxr−1
and y′′ =r(r−1)xr−2.
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation:
(x2−3x)(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 3: Simplify the equation:
r(r−1)xr−3rxr−4rxr+ 6xr= 0
Step 4: Combine like terms:
(r2−4r+ 6)xr= 0
22
Step 5: Since xris never zero, the equation becomes:
r2−4r+ 6 = 0
Step 6: Solve the quadratic equation r2−4r+ 6 = 0 to find the roots using
the quadratic formula:
r=4±p(−4)2−4(1)(6)
2(1) =4±√−8
2=4±2i√2
2= 2 ±i√2
Step 7: The general solution to the Cauchy-Euler equation is:
y=c1x2cos√2 ln(x)+c2x2sin√2 ln(x)
where c1and c2are arbitrary constants.
Question 28
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
We will solve the given Cauchy-Euler equation by assuming a solution of the
form y=xrand finding the values of rfor which this form satisfies the equation.
Step 1: Find the derivatives of y=xr.
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
given equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
Step 4: Factor out xrfrom the equation.
xr(r2−4r+ 4) = 0
Step 5: Set the expression in the parentheses equal to zero to find possible
values for r.
r2−4r+ 4 = 0
23
(r−2)2= 0
r= 2
Step 6: Write down the general solution using the value of r. Since r= 2
is a repeated root of the characteristic equation, the general solution is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 29
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ −4xy′+ 6y=x.
Solution
Step 1: We first assume a solution of the form y=xm. Substituting y=xm
into the differential equation, we get
(x2−4)x2m(m−1)xm−2−4xmxm−1+ 6xm=x
Step 2: Simplifying the equation, we obtain
(x2−4)m(m−1)xm−4mxm+ 6xm=x
Step 3: Rewrite the equation in terms of xm:
m(m−1)(x2−4)xm−4mxm+ 6xm=x
Step 4: Simplify further to get the characteristic equation:
m2x2−4m2−4mx + 6 = 0
Step 5: Solve the characteristic equation to find the roots m1and m2.
m2−4m−6=0
(m−6)(m+ 1) = 0
m1= 6, m2=−1
Step 6: Therefore, the two linearly independent solutions are y1=x6and
y2=x−1.
Step 7: The general solution is given by y=c1x6+c2x−1, where c1and c2
are constants.
Step 8: Finally, to find the particular solution, we substitute back into the
original equation
(x2−4)x2y′′ −4xy′+ 6y=x
24
[(x2−4)x2(6)(6x5)−4x(6x5) + 6(x6)] −4x(6x6) + 6(x6) = x
Step 9: Simplifying the equation gives
(x2−4)(36x7)−24x6+ 6x6−24x6+ 6x6=x
Step 10: This simplifies to
36x9−24x6+ 6x6−24x6+ 6x6=x
Step 11: Further simplifying, we find the particular solution to be y=1
36 x3.
Question 30
Question
Solve the Cauchy-Euler differential equation: x2y′′ −4xy′+ 6y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xrand find the derivatives. Step
2: Substitute y(x) and its derivatives into the differential equation. Step 3:
Solve the resulting equation for the characteristic equation. Step 4: Find the
roots of the characteristic equation and determine the general solution. Step 5:
Use the initial conditions to find the particular solution.
Step 1: Assume y(x) = xr, then find y′(x) and y′′(x):
y′(x) = rxr−1, y′′(x) = r(r−1)xr−2
Step 2: Substitute y(x) and its derivatives back into the differential equa-
tion:
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 3: Simplify the equation:
r(r−1)xr−4rxr+ 6xr= 0
r2−r−4r+ 6 = 0
r2−5r+ 6 = 0
Step 4: Solve the characteristic equation:
r2−5r+ 6 = (r−2)(r−3) = 0
r1= 2, r2= 3
The general solution is given by:
y(x) = c1xr1+c2xr2=c1x2+c2x3
Step 5: Use initial conditions to find the particular solution if needed.
25
Question 31
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0.
Solution
To solve this Cauchy-Euler equation, we’ll assume a solution of the form y=xr
and find the values of rthat satisfy the differential equation.
Step 1: Find the derivatives Let y=xr. Then, calculate the first and
second derivatives of ywith respect to x:
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute into the differential equation Substitute y=xr,
y′=rxr−1, and y′′ =r(r−1)xr−2back into the differential equation:
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 3: Simplify the equation Simplify the equation by combining like
terms:
r(r−1)xr−4rxr+ 6xr= 0
Step 4: Factor out xrFactor out xrfrom the equation:
xr(r(r−1) −4r+ 6) = 0
Step 5: Solve for rSet the expression inside the parentheses to zero:
r(r−1) −4r+ 6 = 0
r2−r−4r+ 6 = 0
r2−5r+ 6 = 0
r2−2r−3r+ 6 = 0
r(r−2) −3(r−2) = 0
(r−2)(r−3) = 0
Thus, r= 2 or r= 3.
Step 6: Form the general solutions Since we have two distinct roots,
the general solution to the Cauchy-Euler equation is:
y=c1x2+c2x3
where c1and c2are arbitrary constants.
26
Question 32
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+y= 0.
Solution
Step 1: Let’s first assume a solution of the form y=xr.
Step 2: Compute the first and second derivatives of 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)−3x(rxr−1) + xr= 0
Step 4: Simplify the equation:
r(r−1)xr−3rxr+xr= 0
Step 5: Factor out xrfrom the equation:
xr(r(r−1) −3r+ 1) = 0
Step 6: Solve for r:
r2−r−3r+ 1 = 0
r2−4r+ 1 = 0
r=4±√16 −4
2
r=4±√12
2
r=4±2√3
2
r= 2 ±√3
Step 7: Write down the general solution by considering both roots:
y=c1x2+√3+c2x2−√3
where c1and c2are arbitrary constants.
Question 33
Question
Solve the Cauchy-Euler equation x2y′′ +xy′−2y= 0.
27
Solution
Step 1: Write the Cauchy-Euler equation in standard form by dividing by x2:
(y′′ +1
xy′−2
x2y)=0
Step 2: Assume a solution of the form y=xr. Find the derivatives of ywith
respect to x:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute the expressions for y,y′, and y′′ into the Cauchy-Euler
equation:
r(r−1)xr−2+1
xrxr−1−2
x2xr= 0
Step 4: Simplify the equation by multiplying through by x2to get rid of the
denominators:
r(r−1)xr+rxr−2xr= 0
r2xr−xr= 0
xr(r2−1) = 0
Step 5: Set xr= 0 to find the first solution:
xr= 0 =⇒r= 0
Step 6: Set r2−1 = 0 to find the second solution:
r2−1=0
r2= 1
r=±1
Step 7: Write down the general solution based on the roots r= 0,1,−1:
y(x) = c1x0+c2x1+c3x−1
y(x) = c1+c2x+c3
x
Question 34
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 13y= 0.
28
Solution
Step 1: Assume a solution of the form y=xr. Then find y′and y′′. Step 2:
Substitute y,y′, and y′′ into the given differential equation. Step 3: Simplify the
equation and solve for rby characteristic equation r2−6r+13 = 0. Step 4: Find
the roots of the characteristic equation to get r1= 3 + 2iand r2= 3 −2i. Step
5: Form the general solution using the roots r1and r2:y=c1x3cos(2 ln(x)) +
c2x3sin(2 ln(x)).
Question 35
Question
Find the general solution to the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
To solve a Cauchy-Euler equation, we make the substitution y=xmwhere m
is a constant to be determined. We then find the characteristic equation by
substituting y=xminto the given differential equation.
Step 1: Find the derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y=xmand its derivatives into the differential
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.
m2−m−3m+ 4 = m2−4m+ 4 = 0
(m−2)2= 0
m= 2
Step 4: Form the general solution. Since we found that m= 2, the
solution corresponding to this value of mis y=x2. Therefore, the general
solution to the given Cauchy-Euler equation is:
y=c1x2+c2x2ln(x),where c1, c2are arbitrary constants
29
Step 8: Form the general solution.
y=c1xr1+c2xr2
y=c1x2+c2x2
y= (c1+c2)x2
Therefore, the general solution to the Cauchy-Euler equation is y=c1x2+
c2x2= (c1+c2)x2.
Question 2
Question
Solve the Cauchy-Euler differential equation: x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xm.
Step 2: Calculate the first and second derivatives of ywith respect to x.
y′=mxm−1and y′′ =m(m−1)xm−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation and simplify.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
m(m−1)xm−3mxm+ 4xm= 0
m2−m−3m+ 4 = m(m−4) = 0
Step 4: Solve the characteristic equation m(m−4) = 0 to find the roots.
m= 0 or m= 4
Step 5: Form the general solution using the roots found in Step 4.
y=c1x0+c2x4=c1+c2x4
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1+c2x4, where c1and c2are arbitrary constants.
Question 3
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
2
Solution
Step 1: Assume a solution of the form y=xrand differentiate to find y′and
y′′.
Let y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ back into the differential 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
Step 4: Solve for r.
r= 2
Step 5: Form the general solution using the roots r1= 2.
y=c1xr1+c2xr1ln(x)
y=c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln(x).
Question 4
Question
Find the general solution to the Cauchy-Euler equation:
x2y′′ −xy′+y= 0
3
Solution
Step 1: Assume a solution of the form y=xrand substitute it into the differ-
ential equation to find r.
Step 2: Calculate the first and second derivatives of y:
y′=rxr−1
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) + xr= 0
Step 4: Simplify the equation:
r(r−1)xr−rxr+xr= 0
Step 5: Combine like terms:
r(r−1) −r+ 1 = 0
Step 6: Solve for r:
r2−r−r+ 1 = 0
r2−2r+ 1 = 0
(r−1)2= 0
r= 1
Step 7: The solution has repeated root r= 1, so the general solution is of
the form:
y=c1x+c2xln(x)
where c1and c2are arbitrary constants.
Question 5
Question
Find the general solution to the Cauchy-Euler equation (x2−3x)y′′−xy′+y= 0.
Solution
Step 1: Assume a solution of the form y=xm.
Step 2: Find the first and second derivatives of y=xm.
y=xm⇒y′=mxm−1⇒y′′ =m(m−1)xm−2
4
Step 3: Substitute y=xm,y′, and y′′ into the Cauchy-Euler equation.
(x2−3x)(m(m−1)xm−2)−x(mxm−1) + xm= 0
Step 4: Simplify the equation.
(m(m−1)xm)−3mxm−(mxm) + xm= 0
m(m−1)xm−3mxm−mxm+xm= 0
m(m−1)xm−4mxm+xm= 0
m(m−1) −4m+ 1 = 0
Step 5: Solve for m.
m2−m−4m+ 1 = 0
m2−5m+ 1 = 0
m=5±√52−4∗1∗1
2∗1
m=5±√21
2
Step 6: Determine the roots of the characteristic equation.
m1=5 + √21
2and m2=5−√21
2
Step 7: Form the general solution.
y(x) = c1xm1+c2xm2
y(x) = c1x5+√21
2+c2x5−√21
2
Question 6
Question
Solve the Cauchy-Euler equation: x2y′′ + 3xy′+y= 0.
Solution
Step 1: Write the auxiliary equation by assuming that y=xr. The general
form of a Cauchy-Euler equation is ax2y′′ +bxy′+cy = 0. Substitute y=xr
into the given equation and solve for r:
x2r(r−1) + 3xr +xr= 0
Step 2: Solve the auxiliary equation. Simplify the equation to get:
r2−r+ 3r+ 1 = 0
5
r2+ 2r+ 1 = 0
(r+ 1)2= 0
r=−1
Step 3: Write down the general solution using the roots obtained from the
auxiliary equation. The general solution is given by:
y=c1x−1+c2x−1ln(x)
Therefore, the solution to the Cauchy-Euler equation x2y′′ + 3xy′+y= 0 is:
y=c1x−1+c2x−1ln(x)
Question 7
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the substitution
y=xr.
Solution
Step 1: Write the differential equation in terms of rby making the substitution
y=xr.
x2(r)(r−1)xr−2−3x(r)xr−1+ 4xr= 0
Step 2: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Factor out xr.
xr(r2−4r+ 4) = 0
Step 4: Set the expression inside the parentheses equal to zero.
r2−4r+ 4 = (r−2)2= 0
Step 5: Find the roots of the characteristic equation.
r= 2
Step 6: Write the general solution using the roots found in step 5.
y=c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln(x).
6
Question 8
Question
Solve the Cauchy-Euler differential equation: x2y′′ −3xy′+ 5y= 0.
Solution
Step 1: Write the characteristic equation using the substitution y=xr. The
characteristic equation for a Cauchy-Euler equation is given by ar(r−1)+br+c=
0, where a,b, and care the coefficients of y′′,y′, and y, respectively.
For our equation x2y′′ −3xy′+ 5y= 0, the characteristic equation becomes
r(r−1) −3r+ 5 = r2−4r+ 5 = 0.
Step 2: Solve the characteristic equation for r. Using the quadratic formula,
r=−(−4)±√(−4)2−4(1)(5)
2(1) =4±√16−20
2=4±√−4
2= 2 ±2i.
Therefore, the solutions for rare r= 2 + 2iand r= 2 −2i.
Step 3: Write the general solution in terms of x. Since the roots are complex,
the general solution will be of the form y=xr=x2+2iand y=xr=x2−2i.
Hence, the general solution is y(x) = c1x2cos(2 ln(x)) + c2x2sin(2 ln(x)).
So, the solution to the Cauchy-Euler differential equation x2y′′−3xy′+5y= 0
is y(x) = c1x2cos(2 ln(x)) + c2x2sin(2 ln(x)).
Question 9
Question
Solve the Cauchy-Euler differential equation: x2y′′ −6xy′+ 12y= 0.
Solution
Step 1: Find the characteristic equation by substituting y=xrinto the differ-
ential equation.
x2r(r−1)xr−2−6xrxr−1+ 12xr= 0
Step 2: Simplify the equation.
r(r−1) −6r+ 12 = 0
Step 3: Solve for rby factoring or using the quadratic formula.
r2−r−6r+ 12 = 0
r(r−1) −6(r−1) = 0
(r−1)(r−6) = 0
r= 1 or r= 6
7
Step 4: Write down the complementary solution.
yc=C1x+C2x6
Step 5: Find the particular solution using the method of variation of param-
eters. Let yp=v1(x)x+v2(x)x6.
Step 6: Calculate y′
pand y′′
p.
y′
p=v1+v′
1x+v2x6+ 6v2x5
y′′
p=v′
1+v′
1x+ 2v2x5+ 6v2x4
Step 7: Substitute yp,y′
p, and y′′
pinto the differential equation.
x2(y′′
p−6y′
p+ 12yp)−6x(y′
p−6yp) + 12yp= 0
Step 8: Solve for v′
1and v′
2.
v′
1=72x6
x2and v′
2= 0
Step 9: Integrate v′
1and v′
2to find v1and v2.
v1= 12x5+C3and v2=C4
Step 10: Write down the particular solution.
yp= (12x5+C3)x+C4x6
Step 11: Write down the general solution.
y=C1x+C2x6+ (12x5+C3)x+C4x6
Step 12: Simplify the general solution.
y=C1x+ 12x6+ 12x6+C3x+C4x6
Step 13: Combine like terms and rewrite the general solution.
y= (C1+C3)x+ (12 + C4)x6
Therefore, the general solution to the Cauchy-Euler equation is y= (C1+
C3)x+ (12 + C4)x6.
Question 10
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 8y= 0.
8
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−5x(rxr−1)+8xr= 0
Step 4: Simplify the equation.
r(r−1)xr−5rxr+ 8xr= 0
Step 5: Factor out xr.
xr(r(r−1) −5r+ 8) = 0
Step 6: Set the expression in the parentheses equal to zero and solve for r.
r(r−1) −5r+ 8 = 0
r2−r−5r+ 8 = 0
r2−6r+ 8 = 0
(r−4)(r−2) = 0
r1= 4 and r2= 2
Step 7: Write down the general solution using both roots.
y=c1x4+c2x2
where c1and c2are arbitrary constants.
Question 11
Question
Solve the Cauchy-Euler equation x2y′′ −xy′+y= 0 using the substitution
y=xm.
9
Solution
Step 1: Find the derivatives of ywith respect to x.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the expressions for y,y′, and y′′ into the differential
equation.
x2y′′ −xy′+y= 0
x2(m(m−1)xm−2)−x(mxm−1) + xm= 0
m(m−1)xm−mxm+xm= 0
Step 3: Simplify the equation.
m(m−1)xm−mxm+xm= 0
m(m−1) −m+ 1 = 0
m2−m−m+ 1 = 0
m2−2m+ 1 = 0
(m−1)2= 0
Step 4: Since (m−1)2= 0, we have a repeated root m= 1.
Step 5: Write down the general solution using the roots.
y=c1x+c2xln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′−xy′+y=
0 is y=c1x+c2xln(x), where c1and c2are arbitrary constants.
Question 12
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
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
10
Step 2: Substitute y=xm,y′, and y′′ into the differential equation.
x2y′′ −3xy′+ 4y= 0
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
Step 3: Simplify the equation.
m(m−1)xm−3mxm+ 4xm= 0
m(m−1) −3m+ 4 = 0 (Divide by xm)
Step 4: Solve for m.
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
m= 2 (Repeated root)
Step 5: Form the solution using m= 2.The solution to the Cauchy-
Euler equation is y=c1x2+c2x2ln x, where c1and c2are arbitrary constants.
Question 13
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−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ 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 xr.
xr(r2−4r+ 4) = 0
Step 6: Solve for r.
r2−4r+ 4 = 0
11
(r−2)2= 0
r= 2
Step 7: General solution is of the form y=c1xr1+c2xr2.
y=c1x2+c2x2
y= (c1+c2)x2
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2, or simply y=cx2, where cis an arbitrary constant.
Question 14
Question
Solve the Cauchy-Euler differential equation:
x2y′′ −3xy′+ 4y= 0
Solution
1. We first assume the solution has the form y=xr.
2. Differentiating y=xrwith respect to xgives us:
y′=rxr−1
3. Differentiating y′with respect to xagain gives us:
y′′ =r(r−1)xr−2
4. Substituting y=xr,y′, and y′′ into the differential equation, we get:
x2r(r−1)xr−2−3xrxr−1+ 4xr= 0
5. Simplifying the equation gives:
r(r−1) −3r+ 4 = 0
6. Solving this quadratic equation gives the values of r:
r2−4r+ 4 = (r−2)2= 0
r= 2
7. Since we have a repeated root, the general solution is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
12
Question 15
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−1and y′′ =r(r−1)xr−2
Step 3: Substitute y,y′, and y′′ back into the differential equation and
simplify.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
Step 4: Factor out xrfrom the equation.
xr(r2−4r+ 4) = 0
Step 5: Set the expression equal to zero and find the roots.
r2−4r+ 4 = 0
(r−2)2= 0
r= 2 (double root)
Step 6: Form the general solution using the roots found in Step 5.
y(x) = c1x2+c2x2ln x
where c1and c2are constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y(x) = c1x2+c2x2ln x.
Question 16
Question
Solve the Cauchy-Euler equation x2y′′−3xy′+9y= 0 by making the substitution
y(x) = xr.
13
Solution
Step 1: We substitute y(x) = xrinto the Cauchy-Euler equation x2y′′ −3xy′+
9y= 0:
x2(r(r−1)xr−2)−3x(rxr−1)+9xr= 0
Expanding and simplifying, we get:
r(r−1)xr−3rxr+ 9xr= 0
Step 2: Factor out xr:
xr(r2−4r+ 9) = 0
This gives us the auxiliary equation r2−4r+ 9 = 0.
Step 3: Solve the auxiliary equation r2−4r+ 9 = 0: The roots of this
quadratic equation can be found using the quadratic formula:
r=4±p(−4)2−4(1)(9)
2(1)
r=4±√16 −36
2
r=4±√−20
2
r=4±2i√5
2
r= 2 ±i√5
Step 4: The general solution is then given by:
y(x) = c1x2+i√5+c2x2−i√5
Step 5: Therefore, the solutions to the Cauchy-Euler equation x2y′′−3xy′+
9y= 0 are:
y(x) = c1x2+i√5+c2x2−i√5
Question 17
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
14
Solution
Step 1: Assume y=xmand find the first and second derivatives.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the expressions for y, y′, y′′ back into the differential equa-
tion.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
Step 3: Simplify and rewrite the equation in terms of m.
m(m−1)xm−3mxm+ 4xm= 0
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
m= 2
Step 4: Use the value of mto find the general solution. Since m= 2, the
general solution is
y=c1x2+c2x2ln x
where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is y=c1x2+c2x2ln x.
Question 18
Question
Solve the Cauchy-Euler equation x2y′′ +xy′−y= 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) + x(rxr−1)−xr= 0
15
Step 3: Simplify the equation.
r(r−1)xr+rxr−xr= 0
r2xr−rxr= 0
rxr(r−1) = 0
Step 4: Find the roots of the characteristic equation r(r−1) = 0.
r= 0,1
Step 5: Write down the general solution using the roots found. Since r= 0
and r= 1 are both roots, the general solution is
y=c1x0+c2x1
Step 6: Simplify the general solution.
y=c1+c2x
Therefore, the solution to the Cauchy-Euler equation x2y′′ +xy′−y= 0 is
y=c1+c2x.
Question 19
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 8y= 0
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Find
roots of the characteristic equation. Step 3: Form the general solution.
Step 1: Assume y=xr, then find y′and y′′:
y′=rxr−1, y′′ =r(r−1)xr−2
Substitute these into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−3x(rxr−1)+8xr= 0
Step 2: Simplify the equation:
r(r−1)xr−3rxr+ 8xr= 0
r2−r−3r+ 8 = 0
(r2−4r+ 8) = 0
16
Solve for rby using the quadratic formula:
r=4±p(−4)2−4(1)(8)
2
r=4±√16 −32
2
r=4±√−16
2
r=4±4i
2
r= 2 ±2i
Step 3: The general solution is:
y(x) = c1x2cos(2 ln(x)) + c2x2sin(2 ln(x))
where c1and c2are arbitrary constants.
Question 20
Question
Solve the Cauchy-Euler equation:
x2y′′ −4xy′+ 6y= 0
Solution
Step 1: Assume a solution of the form y=xr. Differentiate ytwice with respect
to x.
Step 2: Substitute y=xrand its derivatives into the Cauchy-Euler equation.
Step 3: Simplify the equation by dividing by xr.
Step 4: The characteristic equation is r(r−1) −4r+ 6 = 0.
Step 5: Solve the characteristic equation to find the roots r1and r2.
Step 6: Write down the general solution using the roots r1and r2.
Step 7: The general solution is y(x) = c1xr1+c2xr2, where c1and c2are
arbitrary constants.
Step 8: Check for linear dependence to ensure a nontrivial solution.
Step 9: The solution to the Cauchy-Euler equation is y(x) = c1x2+c2x3.
Question 21
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′+y=x.
17
Solution
Step 1: First, we assume a solution of the form y=xr, where ris a constant
to be determined.
Step 2: Compute the first and second derivatives of y=xr:
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute y, y′,and y′′ back into the Cauchy-Euler equation:
x2·r(r−1)xr−2−x·rxr−1+xr=x
Step 4: Simplify the equation:
r(r−1)xr−rxr+xr=x
Step 5: Combine like terms:
(r2−r)xr=x
Step 6: Set the above expression to zero to find the characteristic equation:
r2−r= 0
Step 7: Solve the characteristic equation to find the roots:
r= 0,1
Step 8: Write down the general solution based on the roots:
y(x) = c1x0+c2x1=c1+c2x
Therefore, the general solution to the Cauchy-Euler equation x2y′′−xy′+y=
xis y(x) = c1+c2x, where c1and c2are arbitrary constants.
Question 22
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: We first assume a solution in the form of y=xr, and differentiate to
find y′and y′′.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
18
Step 2: Substitute the expressions for y,y′, and y′′ into the Cauchy-Euler
equation.
x2(r)(r−1)xr−2−5x(rxr−1)+6xr= 0
Step 3: Simplify the equation by combining like terms.
r(r−1)xr−5rxr+ 6xr= 0
Step 4: Factor out xrfrom the equation.
xr(r(r−1) −5r+ 6) = 0
Step 5: Since we assumed that xris not equal to zero, the term inside the
parentheses must be equal to zero.
r2−r−5r+ 6 = r2−6r+ 6 = 0
Step 6: Solve the quadratic equation for rto get the roots.
r=6±p(−6)2−4(1)(6)
2(1) =6±√36 −24
2=6±2
2
Step 7: The roots are r= 4 and r= 2. This gives the general solutions as
y=c1x4+c2x2
where c1and c2are constants.
Question 23
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 9y= 0
Solution
Step 1: Assume a solution of the form y=xmand find the characteristic
equation by substituting into the differential equation:
x2m(m−1)xm−2−5xmxm−1+ 9xm= 0
⇒m(m−1) −5m+ 9 = 0
⇒m2−6m+ 9 = 0
⇒(m−3)2= 0
⇒m= 3 (since we have a repeated root)
19
Step 2: Therefore, the solution to the differential equation is of the form:
y=c1x3+c2x3ln(x)
where c1and c2are arbitrary constants.
So, the general solution to the Cauchy-Euler equation is:
y=c1x3+c2x3ln(x)
Question 24
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: We first assume the solution has the form y=xr. Then we find the
derivatives needed for substitution into the equation.
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives back into the Cauchy-Euler equation.
x2r(r−1)xr−2−5xrxr−1+ 6xr= 0
r(r−1)xr−5rxr+ 6xr= 0
r2−r−5r+ 6 = 0
r2−6r+ 6 = 0
r2−3r−2r+ 6 = 0
r(r−3) −2(r−3) = 0
(r−3)(r−2) = 0
Step 3: We now have the roots r= 3 and r= 2. So the general solution is
given by y(x) = c1x3+c2x2where c1and c2are constants.
Question 25
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 2y=x3.
20
Solution
Step 1: Start by assuming the solution has the form y=xr.
Step 2: Calculate the first and second derivatives of y.
y′=rxr−1and y′′ =r(r−1)xr−2
Step 3: Substitute yand its derivatives back into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−3x(rxr−1)+2xr=x3
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 2xr=x3
Step 5: Divide both sides by xrto get a simple equation.
r(r−1) −3r+ 2 = 1 ⇒r2−4r+ 2 = 1
Step 6: Rearrange and solve for rusing the quadratic formula.
r=4±p(−4)2−4(1)(2)
2
r=4±√16 −8
2=4±2√2
2= 2 ±√2
Step 7: Therefore, the general solution to the Cauchy-Euler equation is
y=c1x2+√2+c2x2−√2
where c1and c2are arbitrary constants.
Question 26
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 9y=x3.
Solution
To solve the Cauchy-Euler equation, we’ll first assume a solution of the form
y=xrand then find the values of rthat satisfy the equation.
Step 1: Find the derivatives of y
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
21
Step 2: Substitute yand its derivatives into the differential equa-
tion Substitute y,y′, and y′′ into the differential equation x2y′′−5xy′+9y=x3:
x2[r(r−1)xr−2]−5x[rxr−1]+9xr=x3
Step 3: Simplify the equation Simplify the left side of the equation:
r(r−1)xr−5rxr+ 9xr=x3
r2xr−rxr−5rxr+ 9xr=x3
r2−6r+ 9 = 1
Step 4: Solve for r
r2−6r+ 9 = 1
r2−6r+ 8 = 0
(r−4)(r−2) = 0
Thus, r= 4 or r= 2.
Step 5: Write the general solution The general solution is a linear
combination of the solutions corresponding to r= 4 and r= 2:
y(x) = c1x4+c2x2
Therefore, the solution to the Cauchy-Euler equation x2y′′ −5xy′+ 9y=x3
is y(x) = c1x4+c2x2, where c1and c2are constants.
Question 27
Question
Consider the Cauchy-Euler equation (x2−3x)y′′ −4xy′+ 6y= 0. Find the
general solution to this differential equation.
Solution
Step 1: Assume a solution in the form of y=xr. Let y=xr, then y′=rxr−1
and y′′ =r(r−1)xr−2.
Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler equation:
(x2−3x)(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 3: Simplify the equation:
r(r−1)xr−3rxr−4rxr+ 6xr= 0
Step 4: Combine like terms:
(r2−4r+ 6)xr= 0
22
Step 5: Since xris never zero, the equation becomes:
r2−4r+ 6 = 0
Step 6: Solve the quadratic equation r2−4r+ 6 = 0 to find the roots using
the quadratic formula:
r=4±p(−4)2−4(1)(6)
2(1) =4±√−8
2=4±2i√2
2= 2 ±i√2
Step 7: The general solution to the Cauchy-Euler equation is:
y=c1x2cos√2 ln(x)+c2x2sin√2 ln(x)
where c1and c2are arbitrary constants.
Question 28
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
We will solve the given Cauchy-Euler equation by assuming a solution of the
form y=xrand finding the values of rfor which this form satisfies the equation.
Step 1: Find the derivatives of y=xr.
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
given equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
Step 3: Simplify the equation.
r(r−1)xr−3rxr+ 4xr= 0
Step 4: Factor out xrfrom the equation.
xr(r2−4r+ 4) = 0
Step 5: Set the expression in the parentheses equal to zero to find possible
values for r.
r2−4r+ 4 = 0
23
(r−2)2= 0
r= 2
Step 6: Write down the general solution using the value of r. Since r= 2
is a repeated root of the characteristic equation, the general solution is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 29
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ −4xy′+ 6y=x.
Solution
Step 1: We first assume a solution of the form y=xm. Substituting y=xm
into the differential equation, we get
(x2−4)x2m(m−1)xm−2−4xmxm−1+ 6xm=x
Step 2: Simplifying the equation, we obtain
(x2−4)m(m−1)xm−4mxm+ 6xm=x
Step 3: Rewrite the equation in terms of xm:
m(m−1)(x2−4)xm−4mxm+ 6xm=x
Step 4: Simplify further to get the characteristic equation:
m2x2−4m2−4mx + 6 = 0
Step 5: Solve the characteristic equation to find the roots m1and m2.
m2−4m−6=0
(m−6)(m+ 1) = 0
m1= 6, m2=−1
Step 6: Therefore, the two linearly independent solutions are y1=x6and
y2=x−1.
Step 7: The general solution is given by y=c1x6+c2x−1, where c1and c2
are constants.
Step 8: Finally, to find the particular solution, we substitute back into the
original equation
(x2−4)x2y′′ −4xy′+ 6y=x
24
[(x2−4)x2(6)(6x5)−4x(6x5) + 6(x6)] −4x(6x6) + 6(x6) = x
Step 9: Simplifying the equation gives
(x2−4)(36x7)−24x6+ 6x6−24x6+ 6x6=x
Step 10: This simplifies to
36x9−24x6+ 6x6−24x6+ 6x6=x
Step 11: Further simplifying, we find the particular solution to be y=1
36 x3.
Question 30
Question
Solve the Cauchy-Euler differential equation: x2y′′ −4xy′+ 6y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xrand find the derivatives. Step
2: Substitute y(x) and its derivatives into the differential equation. Step 3:
Solve the resulting equation for the characteristic equation. Step 4: Find the
roots of the characteristic equation and determine the general solution. Step 5:
Use the initial conditions to find the particular solution.
Step 1: Assume y(x) = xr, then find y′(x) and y′′(x):
y′(x) = rxr−1, y′′(x) = r(r−1)xr−2
Step 2: Substitute y(x) and its derivatives back into the differential equa-
tion:
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 3: Simplify the equation:
r(r−1)xr−4rxr+ 6xr= 0
r2−r−4r+ 6 = 0
r2−5r+ 6 = 0
Step 4: Solve the characteristic equation:
r2−5r+ 6 = (r−2)(r−3) = 0
r1= 2, r2= 3
The general solution is given by:
y(x) = c1xr1+c2xr2=c1x2+c2x3
Step 5: Use initial conditions to find the particular solution if needed.
25
Question 31
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0.
Solution
To solve this Cauchy-Euler equation, we’ll assume a solution of the form y=xr
and find the values of rthat satisfy the differential equation.
Step 1: Find the derivatives Let y=xr. Then, calculate the first and
second derivatives of ywith respect to x:
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute into the differential equation Substitute y=xr,
y′=rxr−1, and y′′ =r(r−1)xr−2back into the differential equation:
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 3: Simplify the equation Simplify the equation by combining like
terms:
r(r−1)xr−4rxr+ 6xr= 0
Step 4: Factor out xrFactor out xrfrom the equation:
xr(r(r−1) −4r+ 6) = 0
Step 5: Solve for rSet the expression inside the parentheses to zero:
r(r−1) −4r+ 6 = 0
r2−r−4r+ 6 = 0
r2−5r+ 6 = 0
r2−2r−3r+ 6 = 0
r(r−2) −3(r−2) = 0
(r−2)(r−3) = 0
Thus, r= 2 or r= 3.
Step 6: Form the general solutions Since we have two distinct roots,
the general solution to the Cauchy-Euler equation is:
y=c1x2+c2x3
where c1and c2are arbitrary constants.
26
Question 32
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+y= 0.
Solution
Step 1: Let’s first assume a solution of the form y=xr.
Step 2: Compute the first and second derivatives of 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)−3x(rxr−1) + xr= 0
Step 4: Simplify the equation:
r(r−1)xr−3rxr+xr= 0
Step 5: Factor out xrfrom the equation:
xr(r(r−1) −3r+ 1) = 0
Step 6: Solve for r:
r2−r−3r+ 1 = 0
r2−4r+ 1 = 0
r=4±√16 −4
2
r=4±√12
2
r=4±2√3
2
r= 2 ±√3
Step 7: Write down the general solution by considering both roots:
y=c1x2+√3+c2x2−√3
where c1and c2are arbitrary constants.
Question 33
Question
Solve the Cauchy-Euler equation x2y′′ +xy′−2y= 0.
27
Solution
Step 1: Write the Cauchy-Euler equation in standard form by dividing by x2:
(y′′ +1
xy′−2
x2y)=0
Step 2: Assume a solution of the form y=xr. Find the derivatives of ywith
respect to x:
y′=rxr−1
y′′ =r(r−1)xr−2
Step 3: Substitute the expressions for y,y′, and y′′ into the Cauchy-Euler
equation:
r(r−1)xr−2+1
xrxr−1−2
x2xr= 0
Step 4: Simplify the equation by multiplying through by x2to get rid of the
denominators:
r(r−1)xr+rxr−2xr= 0
r2xr−xr= 0
xr(r2−1) = 0
Step 5: Set xr= 0 to find the first solution:
xr= 0 =⇒r= 0
Step 6: Set r2−1 = 0 to find the second solution:
r2−1=0
r2= 1
r=±1
Step 7: Write down the general solution based on the roots r= 0,1,−1:
y(x) = c1x0+c2x1+c3x−1
y(x) = c1+c2x+c3
x
Question 34
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 13y= 0.
28
Solution
Step 1: Assume a solution of the form y=xr. Then find y′and y′′. Step 2:
Substitute y,y′, and y′′ into the given differential equation. Step 3: Simplify the
equation and solve for rby characteristic equation r2−6r+13 = 0. Step 4: Find
the roots of the characteristic equation to get r1= 3 + 2iand r2= 3 −2i. Step
5: Form the general solution using the roots r1and r2:y=c1x3cos(2 ln(x)) +
c2x3sin(2 ln(x)).
Question 35
Question
Find the general solution to the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
To solve a Cauchy-Euler equation, we make the substitution y=xmwhere m
is a constant to be determined. We then find the characteristic equation by
substituting y=xminto the given differential equation.
Step 1: Find the derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y=xmand its derivatives into the differential
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.
m2−m−3m+ 4 = m2−4m+ 4 = 0
(m−2)2= 0
m= 2
Step 4: Form the general solution. Since we found that m= 2, the
solution corresponding to this value of mis y=x2. Therefore, the general
solution to the given Cauchy-Euler equation is:
y=c1x2+c2x2ln(x),where c1, c2are arbitrary constants
29
