MATH 334 - DIFFERENTIAL
EQUATIONS - Cauchy-Euler equations
Question Bank - Set 5
Liberty University
Question 1
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm.
Solution
Step 1: We start by differentiating the substitution y=xmto find y′and y′′.
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
Cauchy-Euler equation.
x2(m(m−1)xm−2)−5x(mxm−1)+6xm= 0
m(m−1)xm−5mxm+ 6xm= 0
Step 3: Simplify the equation by combining like terms.
m(m−1) −5m+ 6 = 0
m2−m−5m+ 6 = 0
m2−6m+ 6 = 0
Step 4: Solve the quadratic equation m2−6m+6 = 0 to find the roots using
the quadratic formula.
m=−(−6) ±p(−6)2−4(1)(6)
2(1)
m=6±√36 −24
2
m=6±√12
2
m=6±2√3
2
m= 3 ±√3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
6y= 0 is y=c1x3+√3+c2x3−√3, where c1and c2are arbitrary constants.
Question 2
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Substitute y=xrinto the differential equation x2y′′ −4xy′+ 6y= 0.
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 2: Simplify the equation by combining like terms.
r(r−1)xr−4rxr+ 6xr= 0
r2xr−rxr−4rxr+ 6xr= 0
Step 3: Factor out xr.
xr(r2−4r+ 6) = 0
Step 4: Set the expression equal to zero.
r2−4r+ 6 = 0
Step 5: Solve for rusing the quadratic formula.
r=4±p(−4)2−4(1)(6)
2(1)
2
r=4±√16 −24
2
r=4±√−8
2
r=4±2i√2
2
r= 2 ±i√2
Step 6: Write down the general solution.
y(x) = c1x2+√2i+c2x2−√2i
Thus, the solution to the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 is
y(x) = c1x2+√2i+c2x2−√2i.
Question 3
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y= 0
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Rewrite
the equation with the exponentials from the characteristic equation. Step 3:
Solve for the roots to determine the form of the general solution. Step 4: Write
the general solution to the Cauchy-Euler equation.
Step 1: Assume y=xrand find the first and second derivatives:
y′=rxr−1, y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ back into the equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr= 0
Step 3: Simplify the equation:
r(r−1)xr−5rxr+ 6xr= 0
r2−r−5r+ 6 = 0
r2−6r+ 6 = 0
Solve the quadratic equation to find the roots:
r=6±p(−6)2−4(1)(6)
2·1= 3 ±√3
Step 4: Write the general solution using the roots:
y(x) = c1x3+√3+c2x3−√3
3
Question 4
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the first and second derivatives of y=xr.
y=xr
y′=rxr−1=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′=rxr−1, and y′′ =r(r−1)xr−2into the
Cauchy-Euler equation.
x2y′′ −5xy′+ 6y= 0
x2[r(r−1)xr−2]−5x[rxr−1]+6xr= 0
r(r−1)xr−5rxr+ 6xr= 0
r(r−1) −5r+ 6 = 0 (divide by xr)
r2−6r+ 6 = 0
Step 3: Solve the characteristic equation.
r2−6r+ 6 = 0
(r−3)2−3 = 0
(r−3 + √3)(r−3−√3) = 0
r= 3 ±√3
Step 4: Write the general solution. The general solution is given by y=
c1x3+√3+c2x3−√3, where c1and c2are arbitrary constants.
Question 5
Question
Solve the Cauchy-Euler equation: x2y′′ −4xy′+ 6y= 0.
Solution
Step 1: Assume the solution has the form y(x) = xr.
4
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 Cauchy-Euler equation:
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 4: Simplify the equation by collecting like terms and dividing by xr:
r(r−1) −4r+ 6 = 0
Step 5: Simplify the equation further to obtain the characteristic equation:
r2−5r+ 6 = 0
Step 6: Solve the characteristic equation to find the roots r1and r2:
r1= 2 and r2= 3
Step 7: Write down the general solution using the roots r1and r2:
y(x) = c1x2+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y(x) = c1x2+c2x3, where c1and c2are arbitrary constants.
Question 6
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ −2xy′+ 2y= 0.
Solution
Step 1: We first rewrite the given differential equation in standard form by
dividing every term by (x2−4)x2:
(x2−4)y′′ −2
xy′+2
x2y= 0.
Step 2: We make a substitution y=xmto solve the equation. Then,
calculate y′and y′′:
y′=mxm−1, y′′ =m(m−1)xm−2.
Step 3: Substitute y,y′, and y′′ into the equation to get:
(x2−4)m(m−1)xm−2−2
xmxm−1+2
x2xm= 0.
5
Step 4: Simplify the equation to obtain the characteristic equation:
m(m−1)(x2−4) −2mx + 2 = 0.
Step 5: Factor the characteristic equation and solve for m. We find roots
m= 2 and m=−1.
Step 6: The general solution is given by:
y(x) = c1x2+c2x−1,where c1, c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler equation (x2−4)x2y′′ −2xy′+
2y= 0 is y(x) = c1x2+c2x−1.
Question 7
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x3.
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
dy
dx =rxr−1
d2y
dx2=r(r−1)xr−2
Step 3: Substitute y,dy
dx , and d2y
dx2into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x3
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr=x3
Step 5: Combine like terms.
r2xr−rxr−3rxr+ 4xr=x3
Step 6: Simplify further.
(r2−4r+ 4)xr=x3
Step 7: Divide by xr.
r2−4r+ 4 = 1
6
Step 8: Rearrange the equation to obtain a quadratic equation.
r2−4r+ 3 = 0
Step 9: Solve the quadratic equation to find the roots r1= 1 and r2= 3.
Step 10: Form the general solution.
y=Ax1+Bx3
Step 11: Find the particular solution.
yp=Cx3
Step 12: Substitute ypback into the Cauchy-Euler equation and solve for C.
x2(3C)−3x(3Cx)+4Cx3=x3
3Cx2−9Cx2+ 4Cx3=x3
−6Cx2+ 4Cx3=x3
−6C+ 4C= 1
−2C= 1
C=−1
2
Step 13: The particular solution is yp=−1
2x3.
Step 14: The general solution is y=Ax +Bx3−1
2x3=Ax +1
2x3where A
is a constant.
Question 8
Question
Solve the Cauchy-Euler equation (x2+1)y′′−3xy′+3y= 0 using the substitution
x=et.
Solution
Step 1: Find the derivatives of y. Let x=et, then we have y(t) = y(x(t)) =
y(et). Using the chain rule, we can find the derivatives of ywith respect to tin
terms of x:
dy
dt =dy
dx
dx
dt =dy
dx et
d2y
dt2=d
dt dy
dt =d
dt dy
dx et=d2y
dx2e2t+dy
dx et
7
Step 2: Substitute derivatives into the differential equation. Substitute the
derivatives of yfrom Step 1 into the Cauchy-Euler equation (x2+ 1)y′′ −3xy′+
3y= 0:
(e2t+ 1) d2y
dx2e2t+dy
dx et−3etdy
dx + 3y= 0
Step 3: Simplify the equation. Rewrite the equation in terms of y,dy
dx , and
d2y
dx2:
(e2t+ 1) d2y
dx2e2t+ (e2t+ 1) dy
dx et−3etdy
dx + 3y= 0
Step 4: Rearrange the equation. Simplify the terms:
e2td2y
dx2+e3tdy
dx −3etdy
dx + 3y= 0
e2td2y
dx2−2etdy
dx + 3y= 0
Step 5: Solve the Cauchy-Euler equation. The resulting equation is a
Cauchy-Euler equation with the substitution x=et. Let z=y(x) = y(et),
then we can rewrite the equation as:
e2tz′′ −2etz′+ 3z= 0
This is a second order linear differential equation with constant coefficients,
which can be solved using standard methods.
Question 9
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the method of
undetermined coefficients.
Solution
1. Let’s first find the characteristic equation of the Cauchy-Euler equation
x2y′′ −5xy′+ 6y= 0 by assuming y=xm.
2. Step 1: Substitute y=xminto the equation to get m(m−1)xm−5mxm+
6xm= 0.
3. Step 2: Simplify the equation to get m2−6m+ 6 = 0.
4. Step 3: Solve the quadratic equation m2−6m+ 6 = 0 to find the roots
m1= 3 and m2= 2.
5. Therefore, the solution to the homogeneous Cauchy-Euler equation is
yh(x) = c1x3+c2x2.
8
6. Step 4: Assume a particular solution of the form yp=Ax2+Bx and
substitute it back into the original equation.
7. Step 5: Calculate y′
pand y′′
pand substitute into the differential equation
to get 2A−5(Ax +B) + 6(Ax2+Bx) = 0.
8. Step 6: Simplify the equation and solve for A and B to find A= 1 and
B= 2.
9. So, the particular solution is yp(x) = x2+ 2x.
10. Step 7: The general solution to the Cauchy-Euler equation is y(x) =
yh(x) + yp(x) = c1x3+c2x2+x2+ 2x.
Question 10
Question
Solve the Cauchy-Euler differential equation:
x2y′′ −5xy′+ 8y= 0
Solution
Step 1: Let’s assume a solution of the form y=xrand find the derivatives.
y′=rxr−1and 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)+8xr= 0
Step 3: Simplify the equation.
r(r−1)xr−5rxr+ 8xr= 0
Step 4: Factor out xr.
xr(r2−6r+ 8) = 0
Step 5: Set r2−6r+ 8 = 0 to find the characteristic equation.
r2−6r+ 8 = (r−2)(r−4) = 0
r1= 2 and r2= 4
Step 6: Form the general solution.
y(x) = c1xr1+c2xr2=c1x2+c2x4
where c1and c2are arbitrary constants.
9
Question 11
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.
Step 3: Substitute y,y′, and y′′ into the differential equation.
Step 4: Simplify the equation using the assumption that y=xr.
Step 5: Solve the resulting equation for rto obtain the characteristic equa-
tion.
Step 6: Solve the characteristic equation to find the values of r.
Step 7: Form the general solution using the roots of the characteristic equa-
tion.
Step 8: If roots are complex, use Euler’s formula to rewrite the general
solution in terms of sine and cosine functions. Otherwise, proceed to the next
step.
Step 9: Substitute the values of rinto the general solution to obtain the
final solution.
Step 10: Simplify the final solution if necessary.
Therefore, the solution to the Cauchy-Euler equation x2y′′ + 3xy′−4y= 0
is y(x) = c1x+c2x−4.
Question 12
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y=x3
Solution
Step 1: We first assume the solution has the form y=xrfor some constant r.
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′′ back into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr=x3
r(r−1)xr−5rxr+ 6xr=x3
Step 4: Combine like terms and simplify:
r2xr−rxr−5rxr+ 6xr=x3
10
(r2−6r+ 6)xr=x3
Step 5: Equate the coefficients of like powers of xon both sides:
r2−6r+ 6 = 0
Step 6: Solve the quadratic equation r2−6r+ 6 = 0 to find the roots r1and
r2using the quadratic formula:
r=6±√62−4∗1∗6
2
r=6±√12
2
r= 3 ±√3
Step 7: Write down the general solution to the Cauchy-Euler equation:
y=c1x3+√3+c2x3−√3
where c1and c2are arbitrary constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
6y=x3is y=c1x3+√3+c2x3−√3.
Question 13
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 3y=x2ln x.
Solution
Step 1: Assume a solution of the form y=xr. Then, find the first and sec-
ond derivatives of y. Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler
equation. Step 3: Simplify the equation by collecting like terms and setting the
non-homogeneous term equal to zero. Step 4: Solve the resulting homogeneous
equation x2y′′−3xy′+3y= 0 by substituting y=xrand finding the characteris-
tic equation. Step 5: Find the roots of the characteristic equation and determine
the general solution of the homogeneous equation. Step 6: Use the method of
undetermined coefficients to find a particular solution to the non-homogeneous
equation. Step 7: Combine the general solution of the homogeneous equation
with the particular solution to obtain the complete solution.
11
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
To solve the Cauchy-Euler equation, we will assume a solution of the form
y=xr. We will then substitute this into the given differential equation and
find the values of rthat satisfy it.
Step 1: Substitute y=xrinto the differential equation. We have:
x2y′′ −3xy′+ 4y= 0
Substitute y=xr:
x2r(r−1)xr−2−3xrxr−1+ 4xr= 0
Step 2: Simplify the equation. Simplify the equation by combining like
terms:
r(r−1)xr−3rxr+ 4xr= 0
r2xr−rxr−3rxr+ 4xr= 0
r2−4r+ 1 = 0
Step 3: Find the roots of the characteristic equation. Solve the
characteristic equation r2−4r+ 1 = 0 to find the roots:
r=4±p42−4(1)(1)
2(1)
r=4±√16 −4
2
r=4±√12
2
r=4±2√3
2
r= 2 ±√3
Step 4: Write the general solution. Since the roots are distinct, the
general solution is:
y(x) = c1x2+√3+c2x2−√3
where c1and c2are arbitrary constants.
12
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x3.
Solution
Step 1: Assume a solution of the form y=xm.
Step 2: Find the first and second derivatives of y.
y′=mxm−1and y′′ =m(m−1)xm−2.
Step 3: Substitute yand its derivatives back into the Cauchy-Euler equation.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm=x3.
Step 4: Simplify the equation.
m(m−1)xm−3mxm+ 4xm=x3.
Step 5: Combine like terms.
m2−4m+ 4 = 1.
Step 6: Solve the quadratic equation m2−4m+ 4 = 1.
m2−4m+ 3 = 0.
(m−1)(m−3) = 0.
m= 1 or m= 3.
Step 7: Form the general solution.
y=c1x+c2x3,
where c1and c2are arbitrary constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y=x3is y=c1x+c2x3.
Question 16
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y=x4.
13
Solution
Step 1: Let’s first find the characteristic equation by substituting y=xminto
the differential equation. Then, we have
y′=mxm−1and y′′ =m(m−1)xm−2.
Substitute these expressions into the Cauchy-Euler equation to get
x2(m(m−1)xm−2)−5x(mxm−1)+6xm=x4.
Step 2: Simplify the equation to get
m(m−1)xm−5mxm+ 6xm=x4.
Step 3: Combine the terms to obtain
m2−m−5m+ 6 = 1.
Step 4: Simplify the equation to get
m2−6m+ 6 = 0.
Step 5: Solve the quadratic equation to find the roots:
m1= 3 + √3 and m2= 3 −√3.
Step 6: The general solution to the Cauchy-Euler equation is given by
y(x) = c1xm1+c2xm2,
where c1and c2are constants.
Step 7: Therefore, the solution to the Cauchy-Euler equation x2y′′ −5xy′+
6y=x4is
y(x) = c1x3+√3+c2x3−√3.
Question 17
Question
Find the general solution of the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xmand find
the roots of the characteristic equation.
Step 1: Find the characteristic equation The characteristic equation
corresponding to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 is given by:
m(m−1) −3m+ 4 = 0
14
Simplifying, we get:
m2−4m+ 4 = 0
(m−2)2= 0
The characteristic equation has a repeated root m= 2.
Step 2: Find the solution Since m= 2 is a repeated root, the general
solution of the Cauchy-Euler equation is given by:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Therefore, the general solution of the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is:
y(x) = c1x2+c2x2ln(x)
Question 18
Question
Solve the Cauchy-Euler equation: (x2−3x)y′′ + (4x−6)y′−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′′ into the Cauchy-Euler equation:
(x2−3x)(r(r−1)xr−2) + (4x−6)(rxr−1)−4xr= 0
.
Step 4: Simplify the equation:
r(r−1)xr−3r(r−1)xr+ 4rxr−6rxr−4xr= 0
.
Step 5: Combine like terms:
r2xr−3r2xr+ 4rxr−6rxr−4xr= 0
.
Step 6: Factor out xr:
(r2−3r2+ 4r−6r−4)xr= 0
15
.
Step 7: Simplify the expression:
(r2−3r+ 4r−6r−4)xr= 0
.
Step 8: Combine like terms:
(r2+r−4)xr= 0
.
Step 9: Solve the characteristic equation r2+r−4 = 0 to find the roots r1
and r2.
Step 10: The roots of the characteristic equation are r1= 1 and r2=−4.
Step 11: The general solution is given by:
y=c1x1+c2x−4
, where c1and c2are constants.
Question 19
Question
Find the general solution to the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: Assume the solution has the form y=xm. Step 2: Take derivatives
of y. Step 3: Substitute derivatives back into the differential equation. Step
4: Solve the resulting characteristic equation. Step 5: Determine the general
solution.
Step 1: Assume the solution has the form y=xm.
Let y=xm.
Step 2: Take derivatives of y.
Then, y′=mxm−1and y′′ =m(m−1)xm−2.
Step 3: Substitute derivatives back into the differential equation.
The Cauchy-Euler equation becomes: x2(m(m−1)xm−2)−5x(mxm−1) +
6xm= 0
This simplifies to: m(m−1)xm−5mxm+ 6xm= 0
Step 4: Solve the resulting characteristic equation.
Dividing by xm, we get the characteristic equation: m(m−1) −5m+ 6 = 0
Solving this equation, we find m= 2 and m= 3.
Step 5: Determine the general solution.
Thus, the general solution to the Cauchy-Euler equation x2y′′−5xy′+6y= 0
is: y=c1x2+c2x3, where c1and c2are arbitrary constants.
16
Question 20
Question
Solve the Cauchy-Euler equation x2y′′−3xy′+7y= 0 by making the substitution
y=xm.
Solution
Step 1: Substitute y=xminto the Cauchy-Euler equation x2y′′−3xy′+7y= 0.
x2(m(m−1)xm−2)−3x(mxm−1)+7xm= 0
Step 2: Simplify the equation by grouping the terms with x.
m(m−1)xm−3mxm+ 7xm= 0
m(m−1) −3m+ 7 = 0
Step 3: Solve the equation m(m−1) −3m+ 7 = 0 for m.
m2−m−3m+ 7 = 0
m2−4m+ 7 = 0
Step 4: Use the quadratic formula to solve for m.
m=4±p(−4)2−4(1)(7)
2(1)
m=4±√16 −28
2
m=4±√−12
2
m=4±2i√3
2
m= 2 ±i√3
Step 5: Write down the general solution using the roots m= 2 + i√3 and
m= 2 −i√3.
y(x) = c1x2+i√3+c2x2−i√3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
7y= 0 is y(x) = c1x2+i√3+c2x2−i√3.
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 using the substitution
y=xm.
17
Solution
Step 1: Compute the first and second derivatives of y.
y=xm⇒y′=mxm−1and y′′ =m(m−1)xm−2
Step 2: Substitute y,y′, and y′′ back into the Cauchy-Euler equation.
x2y′′ −4xy′+ 6y= 0 ⇒x2(m(m−1)xm−2)−4x(mxm−1)+6xm= 0
Step 3: Simplify the equation.
m(m−1)xm−4mxm+ 6xm= 0
Step 4: Factor out xmand solve for m.
m(m−1) −4m+ 6 = 0 ⇒m2−m−4m+ 6 = 0
Step 5: Simplify and solve for m.
m2−5m+ 6 = 0 ⇒(m−2)(m−3) = 0 ⇒m= 2 or m= 3
Step 6: Write down the general form of the solution using both values of m.
y=c1x2+c2x3where c1and c2are arbitrary constants.
Therefore, the general solution of the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y=c1x2+c2x3.
Question 22
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the method of
undetermined coefficients.
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:
x2y′′ −3xy′+ 4y=x2[r(r−1)xr−2]−3x[rxr−1]+4xr= 0.
Step 4: Simplify the equation to obtain:
r(r−1)xr−3rxr+ 4xr=xr[r(r−1) −3r+ 4] = 0.
18
Step 5: Solve the characteristic equation r2−4r+ 4 = (r−2)2= 0 to find
the repeated root r1=r2= 2.
Step 6: The general solution to the Cauchy-Euler equation is given by y(x) =
c1xr1+c2xr2=c1x2+c2x2= (c1+c2)x2.
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is y(x) = c1x2+c2x2= (c1+c2)x2, where c1and c2are arbitrary constants.
Question 23
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xrand differentiate to find y′and
y′′.
Let y(x) = xr
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Factor out xrand simplify the equation.
xr(r(r−1) −3r+ 4) = 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 4: Find the roots of the characteristic equation.
r= 2
Step 5: Form the general solution using the roots found.
y(x) = c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y(x) = c1x2+c2x2ln(x).
19
Question 24
Question
Solve the Cauchy-Euler equation:
x2y′′ −xy′+y= 0
Solution
Step 1: Assume a solution of the form y=xmand find the derivatives. Substi-
tute into the differential equation to get an auxiliary equation.
Step 2: Differentiate y=xmtwice to find y′and y′′:
y′=mxm−1and y′′ =m(m−1)xm−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation to get:
x2(m(m−1)xm−2)−x(mxm−1) + xm= 0
Step 4: Simplify the equation to get the auxiliary equation:
m(m−1)xm−mxm+xm= 0
Step 5: Factor out xmfrom the equation:
xm(m(m−1) −m+ 1) = 0
Step 6: Solve the auxiliary equation m(m−1) −m+ 1 = 0 to find the roots.
Step 7: The roots of the auxiliary equation are m1= 1, m2= 1, which are
equal and real.
Step 8: Write down the general solution for equal and real roots:
y(x) = c1x1ln(x) + c2x1
where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler equation is y(x) = c1xln(x) +
c2x.
Question 25
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm.
20
Solution
Step 1: Find the derivatives of y=xm.
Let y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the derivatives into the Cauchy-Euler equation.
x2y′′ −5xy′+ 6y= 0
x2(m(m−1)xm−2)−5x(mxm−1)+6xm= 0
m(m−1)xm−5mxm+ 6xm= 0
m(m−1) −5m+ 6 = 0
Step 3: Solve for m.
m2−m−5m+ 6 = 0
m2−6m+ 6 = 0
(m−3)(m−2) = 0
This gives two possible solutions: m= 3 or m= 2.
Step 4: Form the general solution using the roots m1= 3 and m2= 2.
For m=3:y1=x3
For m=2:y2=x2
Therefore, the general solution is
y=c1x3+c2x2
where c1and c2are arbitrary constants.
Question 26
Question
Solve the Cauchy-Euler equation x2y′′−3xy′+3y= 0 by making the substitution
y(x) = xr.
Solution
Step 1: Write the Cauchy-Euler equation in terms of the new variable r. Let
y(x) = xr. Then, we can find y′and y′′ as follows:
y′=rxr−1and y′′ =r(r−1)xr−2
21
Substitute y(x) = xr,y′, and y′′ into the Cauchy-Euler equation to get:
x2(r(r−1)xr−2)−3x(rxr−1)+3xr= 0
Step 2: Simplify the equation. Simplify the equation to get: r(r−1)xr−
3rxr+ 3xr= 0
Step 3: Factor out xr. Factor out xrfrom the equation to get: xr(r(r−
1) −3r+ 3) = 0
Step 4: Solve for r. Set the expression inside the parentheses equal to zero:
r(r−1) −3r+ 3 = 0
Solve this quadratic equation to find the roots r1and r2.
Step 5: Write the general solution. The general solution to the Cauchy-
Euler equation is:
y(x) = c1xr1+c2xr2
where c1and c2are arbitrary constants.
Question 27
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
To solve this Cauchy-Euler equation, we will assume y=xmand find the values
of mthat satisfy the equation.
Step 1: Find the derivatives of yLet’s find the first and second deriva-
tives of y=xm:
y′=mxm−1and y′′ =m(m−1)xm−2
Step 2: Substitute yand its derivatives into the equation Substitute
y, y′,and y′′ into the Cauchy-Euler equation:
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
Simplify the equation:
m(m−1)xm−3mxm+ 4xm= 0
Step 3: Solve for mCombining like terms, we get:
m2−m−3m+ 4 = 0
Simplify further:
m2−4m+ 4 = 0
22
(m−2)2= 0
Step 4: Find the solutions for mThe solution to the quadratic equation
is:
m= 2
Step 5: Form the general solution Since m= 2, we have one solution
y1=x2.
Step 6: Find the second linearly independent solution To find the
second linearly independent solution, we use the method of reduction of order.
We can assume y2=vx2, where vis a function of x.
Differentiating y2:
y′
2=v′x2+ 2vx
y′′
2=v′′x2+ 4v′x+ 2v
Substitute y2, y′
2, and y′′
2back into the Cauchy-Euler equation:
x2(v′′x2+ 4v′x+ 2v)−3x(v′x2+ 2vx)+4vx2= 0
Simplify the equation and solve for v(x).
Step 7: Write the general solution The general solution to the Cauchy-
Euler equation x2y′′ −3xy′+ 4y= 0 is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 28
Question
Solve the Cauchy-Euler equation:
x2y′′ −9xy′+ 25y= 0.
Solution
Step 1: Write the characteristic equation associated with the Cauchy-Euler
equation by assuming y=xr. Step 2: Find the roots of the characteristic
equation. Step 3: Form the general solution using the roots found in step 2.
Step 1: Write the characteristic equation by assuming y=xr:
r(r−1)xr−9rxr+ 25xr= 0
Step 2: Find the roots of the characteristic equation: Simplify the equation
to get:
r2−10r+ 25 = 0
23
Solve this quadratic equation to find the roots, r1and r2:
r1=r2= 5
Step 3: Form the general solution: Since we have a repeated root (r1=
r2= 5), the general solution is:
y(x) = c1x5+c2x5ln(x)
where c1and c2are arbitrary constants.
Question 29
Question
Solve the Cauchy-Euler differential equation (x2−3x+ 2)y′′ −2xy′+ 2y= 0
using the method of undetermined coefficients.
Solution
To solve the given Cauchy-Euler equation (x2−3x+ 2)y′′ −2xy′+ 2y= 0, we
first assume a solution of the form y=xr.
Step 1: Find the derivatives Calculate the first and second derivatives
of y:
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute into the differential equation Substitute y,y′, and
y′′ into the differential equation:
(x2−3x+ 2)(r(r−1)xr−2)−2x(rxr−1)+2xr= 0
Step 3: Simplify the equation Simplify the equation by combining like
terms:
r(r−1)xr−3(rxr)+2xr−2rxr+ 2xr= 0
r(r−1)xr−3rxr+ 2xr−2rxr+ 2xr= 0
r(r−1)xr−3rxr+ 2xr−2rxr+ 2xr= 0
r2xr−5rxr+ 2xr= 0
Step 4: Solve for rSet the expression equal to zero and solve for r:
r2−5r+ 2 = 0
Using the quadratic formula, we find the roots to be r1= 2 and r2=1
2.
Step 5: General solution The general solution will be in the form of
y=c1xr1+c2xr2, where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler differential equation (x2−3x+
2)y′′ −2xy′+ 2y= 0 is y=c1x2+c2x1/2.
24
Question 30
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr, where ris a constant to be
determined. Then we have:
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the original differential equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Simplify the equation:
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 4: Since we have a repeated root r= 2, we have the solution:
y=c1x2+c2x2ln(x)
where c1and c2are constants.
Question 31
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Write the characteristic equation by assuming y=xr.
Characteristic equation: x2r(r−1) −3xr + 4 = 0
r(r−1) −3r+ 4 = 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2 (double root)
25
Step 2: Write the general solution using the double root. Since r= 2 is a
double root, the general solution is given by:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Step 3: Find the specific solution by substituting the general solution back
into the differential equation.
x2y′′ −3xy′+ 4y=x2(2c2x+c2)−3x(2c1x+ 2c2) + 4(c1x2+c2x2ln(x))
= 2c2x3+c2x2−6c1x2−6c2x+ 4c1x2+ 4c2x2ln(x)
= (2c2−6c1+ 4c1)x2+ (c2−6c2)x+ 4c2x2ln(x)
Step 4: Equate the coefficients to zero. From the last step, we have:
(2c2−6c1+ 4c1= 0
c2−6c2= 0
Solving these equations gives c1=−1
2and c2= 0.
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is:
y(x) = −1
2x2
Question 32
Question
Solve the Cauchy-Euler equation x2y′′ + 3xy′+y= 0 using the substitution
y=xr.
Solution
Step 1: Compute the first and second derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the Cauchy-Euler equation.
x2(r(r−1)xr−2)+3x(rxr−1) + xr= 0
Step 3: Simplify the equation by combining like terms.
r(r−1)xr+ 3rxr+xr= 0
26
Step 4: Factor out xrfrom the equation.
xr(r2−r+ 3r+ 1) = 0
Step 5: Simplify the equation further.
xr(r2+ 2r+ 1) = 0
Step 6: Find the roots of the characteristic equation r2+ 2r+ 1 = 0.
r2+ 2r+ 1 = (r+ 1)2= 0
r=−1
Step 7: Write down the general solution using the roots r=−1.
y(x) = c1x−1+c2x−1ln(x)
Question 33
Question
Solve the Cauchy-Euler equation (x−1)y′′−2xy′+2y= 0 using the substitution
y=xm.
Solution
Step 1: We start by substituting y=xminto the differential equation.
(x−1)[m(m−1)xm−2]−2x[mxm−1]+2xm= 0
Step 2: Simplify the expression by expanding and collecting like terms.
m(m−1)xm−m(2x)xm−1+ 2xm= 0
Step 3: Divide through by xmto simplify the equation.
m(m−1) −2mx + 2 = 0
Step 4: Rewrite the equation in a standard form by rearranging the terms.
m2−m−2mx + 2 = 0
Step 5: This is a quadratic equation in m, so we can solve for mby using
the quadratic formula.
m=1±√1+8x
2
Step 6: There are two roots, m1and m2. Substituting these back into the
equation y=xm, we get the following two solution functions.
y1=x(1+√1+8x)/2and y2=x(1−√1+8x)/2
Step 7: Therefore, the general solution to the Cauchy-Euler equation (x−
1)y′′ −2xy′+ 2y= 0 is given by
y(x) = c1x(1+√1+8x)/2+c2x(1−√1+8x)/2
27
Question 34
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′+ 6y= 0.
Solution
Step 1: Write the Cauchy-Euler equation in standard form. Assume y=xr.
x2y′′ −xy′+ 6y= 0
x2(r)(r−1)xr−2−x(r)xr−1+ 6xr= 0
r(r−1)xr−rxr+ 6xr= 0
r(r−1) −r+ 6 = 0
r2−2r+ 6 = 0
(r−3)(r+ 2) = 0
r= 3,−2
Step 2: Write the general solution using r1= 3 and r2=−2.
y=c1x3+c2x−2
Step 3: Check linear independence by verifying that the Wronskian is
nonzero.
W=
x3x−2
3x2−2x−3
=x3(−2x−3)−x−2(3x2)
=−2−3x2= 0
Since W= 0, the functions are linearly independent.
Step 4: The general solution to the Cauchy-Euler equation is:
y=c1x3+c2x−2
Question 35
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
28
Solution
To solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0, we make the substi-
tution y=xmfor some real number m.
Step 1: Find the first and second derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y,y′, and y′′ back 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
m(m−1) −3m+ 4 = 0 (divide by xm)
Step 3: Solve the characteristic equation m(m−1) −3m+ 4 = 0.
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
m= 2
Step 4: Form the general solution with m= 2.
y=c1x2+c2x2ln x
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln x.
29
Step 4: Solve the quadratic equation m2−6m+6 = 0 to find the roots using
the quadratic formula.
m=−(−6) ±p(−6)2−4(1)(6)
2(1)
m=6±√36 −24
2
m=6±√12
2
m=6±2√3
2
m= 3 ±√3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
6y= 0 is y=c1x3+√3+c2x3−√3, where c1and c2are arbitrary constants.
Question 2
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Substitute y=xrinto the differential equation x2y′′ −4xy′+ 6y= 0.
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 2: Simplify the equation by combining like terms.
r(r−1)xr−4rxr+ 6xr= 0
r2xr−rxr−4rxr+ 6xr= 0
Step 3: Factor out xr.
xr(r2−4r+ 6) = 0
Step 4: Set the expression equal to zero.
r2−4r+ 6 = 0
Step 5: Solve for rusing the quadratic formula.
r=4±p(−4)2−4(1)(6)
2(1)
2
r=4±√16 −24
2
r=4±√−8
2
r=4±2i√2
2
r= 2 ±i√2
Step 6: Write down the general solution.
y(x) = c1x2+√2i+c2x2−√2i
Thus, the solution to the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 is
y(x) = c1x2+√2i+c2x2−√2i.
Question 3
Question
Solve the Cauchy-Euler equation:
x2y′′ −5xy′+ 6y= 0
Solution
Step 1: Write the characteristic equation by assuming y=xr. Step 2: Rewrite
the equation with the exponentials from the characteristic equation. Step 3:
Solve for the roots to determine the form of the general solution. Step 4: Write
the general solution to the Cauchy-Euler equation.
Step 1: Assume y=xrand find the first and second derivatives:
y′=rxr−1, y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ back into the equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr= 0
Step 3: Simplify the equation:
r(r−1)xr−5rxr+ 6xr= 0
r2−r−5r+ 6 = 0
r2−6r+ 6 = 0
Solve the quadratic equation to find the roots:
r=6±p(−6)2−4(1)(6)
2·1= 3 ±√3
Step 4: Write the general solution using the roots:
y(x) = c1x3+√3+c2x3−√3
3
Question 4
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xr.
Solution
Step 1: Find the first and second derivatives of y=xr.
y=xr
y′=rxr−1=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute y=xr,y′=rxr−1, and y′′ =r(r−1)xr−2into the
Cauchy-Euler equation.
x2y′′ −5xy′+ 6y= 0
x2[r(r−1)xr−2]−5x[rxr−1]+6xr= 0
r(r−1)xr−5rxr+ 6xr= 0
r(r−1) −5r+ 6 = 0 (divide by xr)
r2−6r+ 6 = 0
Step 3: Solve the characteristic equation.
r2−6r+ 6 = 0
(r−3)2−3 = 0
(r−3 + √3)(r−3−√3) = 0
r= 3 ±√3
Step 4: Write the general solution. The general solution is given by y=
c1x3+√3+c2x3−√3, where c1and c2are arbitrary constants.
Question 5
Question
Solve the Cauchy-Euler equation: x2y′′ −4xy′+ 6y= 0.
Solution
Step 1: Assume the solution has the form y(x) = xr.
4
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 Cauchy-Euler equation:
x2(r(r−1)xr−2)−4x(rxr−1)+6xr= 0
Step 4: Simplify the equation by collecting like terms and dividing by xr:
r(r−1) −4r+ 6 = 0
Step 5: Simplify the equation further to obtain the characteristic equation:
r2−5r+ 6 = 0
Step 6: Solve the characteristic equation to find the roots r1and r2:
r1= 2 and r2= 3
Step 7: Write down the general solution using the roots r1and r2:
y(x) = c1x2+c2x3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y(x) = c1x2+c2x3, where c1and c2are arbitrary constants.
Question 6
Question
Solve the Cauchy-Euler equation (x2−4)x2y′′ −2xy′+ 2y= 0.
Solution
Step 1: We first rewrite the given differential equation in standard form by
dividing every term by (x2−4)x2:
(x2−4)y′′ −2
xy′+2
x2y= 0.
Step 2: We make a substitution y=xmto solve the equation. Then,
calculate y′and y′′:
y′=mxm−1, y′′ =m(m−1)xm−2.
Step 3: Substitute y,y′, and y′′ into the equation to get:
(x2−4)m(m−1)xm−2−2
xmxm−1+2
x2xm= 0.
5
Step 4: Simplify the equation to obtain the characteristic equation:
m(m−1)(x2−4) −2mx + 2 = 0.
Step 5: Factor the characteristic equation and solve for m. We find roots
m= 2 and m=−1.
Step 6: The general solution is given by:
y(x) = c1x2+c2x−1,where c1, c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler equation (x2−4)x2y′′ −2xy′+
2y= 0 is y(x) = c1x2+c2x−1.
Question 7
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x3.
Solution
Step 1: Assume a solution of the form y=xr.
Step 2: Find the first and second derivatives of y.
dy
dx =rxr−1
d2y
dx2=r(r−1)xr−2
Step 3: Substitute y,dy
dx , and d2y
dx2into the Cauchy-Euler equation.
x2(r(r−1)xr−2)−3x(rxr−1)+4xr=x3
Step 4: Simplify the equation.
r(r−1)xr−3rxr+ 4xr=x3
Step 5: Combine like terms.
r2xr−rxr−3rxr+ 4xr=x3
Step 6: Simplify further.
(r2−4r+ 4)xr=x3
Step 7: Divide by xr.
r2−4r+ 4 = 1
6
Step 8: Rearrange the equation to obtain a quadratic equation.
r2−4r+ 3 = 0
Step 9: Solve the quadratic equation to find the roots r1= 1 and r2= 3.
Step 10: Form the general solution.
y=Ax1+Bx3
Step 11: Find the particular solution.
yp=Cx3
Step 12: Substitute ypback into the Cauchy-Euler equation and solve for C.
x2(3C)−3x(3Cx)+4Cx3=x3
3Cx2−9Cx2+ 4Cx3=x3
−6Cx2+ 4Cx3=x3
−6C+ 4C= 1
−2C= 1
C=−1
2
Step 13: The particular solution is yp=−1
2x3.
Step 14: The general solution is y=Ax +Bx3−1
2x3=Ax +1
2x3where A
is a constant.
Question 8
Question
Solve the Cauchy-Euler equation (x2+1)y′′−3xy′+3y= 0 using the substitution
x=et.
Solution
Step 1: Find the derivatives of y. Let x=et, then we have y(t) = y(x(t)) =
y(et). Using the chain rule, we can find the derivatives of ywith respect to tin
terms of x:
dy
dt =dy
dx
dx
dt =dy
dx et
d2y
dt2=d
dt dy
dt =d
dt dy
dx et=d2y
dx2e2t+dy
dx et
7
Step 2: Substitute derivatives into the differential equation. Substitute the
derivatives of yfrom Step 1 into the Cauchy-Euler equation (x2+ 1)y′′ −3xy′+
3y= 0:
(e2t+ 1) d2y
dx2e2t+dy
dx et−3etdy
dx + 3y= 0
Step 3: Simplify the equation. Rewrite the equation in terms of y,dy
dx , and
d2y
dx2:
(e2t+ 1) d2y
dx2e2t+ (e2t+ 1) dy
dx et−3etdy
dx + 3y= 0
Step 4: Rearrange the equation. Simplify the terms:
e2td2y
dx2+e3tdy
dx −3etdy
dx + 3y= 0
e2td2y
dx2−2etdy
dx + 3y= 0
Step 5: Solve the Cauchy-Euler equation. The resulting equation is a
Cauchy-Euler equation with the substitution x=et. Let z=y(x) = y(et),
then we can rewrite the equation as:
e2tz′′ −2etz′+ 3z= 0
This is a second order linear differential equation with constant coefficients,
which can be solved using standard methods.
Question 9
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the method of
undetermined coefficients.
Solution
1. Let’s first find the characteristic equation of the Cauchy-Euler equation
x2y′′ −5xy′+ 6y= 0 by assuming y=xm.
2. Step 1: Substitute y=xminto the equation to get m(m−1)xm−5mxm+
6xm= 0.
3. Step 2: Simplify the equation to get m2−6m+ 6 = 0.
4. Step 3: Solve the quadratic equation m2−6m+ 6 = 0 to find the roots
m1= 3 and m2= 2.
5. Therefore, the solution to the homogeneous Cauchy-Euler equation is
yh(x) = c1x3+c2x2.
8
6. Step 4: Assume a particular solution of the form yp=Ax2+Bx and
substitute it back into the original equation.
7. Step 5: Calculate y′
pand y′′
pand substitute into the differential equation
to get 2A−5(Ax +B) + 6(Ax2+Bx) = 0.
8. Step 6: Simplify the equation and solve for A and B to find A= 1 and
B= 2.
9. So, the particular solution is yp(x) = x2+ 2x.
10. Step 7: The general solution to the Cauchy-Euler equation is y(x) =
yh(x) + yp(x) = c1x3+c2x2+x2+ 2x.
Question 10
Question
Solve the Cauchy-Euler differential equation:
x2y′′ −5xy′+ 8y= 0
Solution
Step 1: Let’s assume a solution of the form y=xrand find the derivatives.
y′=rxr−1and 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)+8xr= 0
Step 3: Simplify the equation.
r(r−1)xr−5rxr+ 8xr= 0
Step 4: Factor out xr.
xr(r2−6r+ 8) = 0
Step 5: Set r2−6r+ 8 = 0 to find the characteristic equation.
r2−6r+ 8 = (r−2)(r−4) = 0
r1= 2 and r2= 4
Step 6: Form the general solution.
y(x) = c1xr1+c2xr2=c1x2+c2x4
where c1and c2are arbitrary constants.
9
Question 11
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.
Step 3: Substitute y,y′, and y′′ into the differential equation.
Step 4: Simplify the equation using the assumption that y=xr.
Step 5: Solve the resulting equation for rto obtain the characteristic equa-
tion.
Step 6: Solve the characteristic equation to find the values of r.
Step 7: Form the general solution using the roots of the characteristic equa-
tion.
Step 8: If roots are complex, use Euler’s formula to rewrite the general
solution in terms of sine and cosine functions. Otherwise, proceed to the next
step.
Step 9: Substitute the values of rinto the general solution to obtain the
final solution.
Step 10: Simplify the final solution if necessary.
Therefore, the solution to the Cauchy-Euler equation x2y′′ + 3xy′−4y= 0
is y(x) = c1x+c2x−4.
Question 12
Question
Solve the Cauchy-Euler equation: x2y′′ −5xy′+ 6y=x3
Solution
Step 1: We first assume the solution has the form y=xrfor some constant r.
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′′ back into the Cauchy-Euler equation:
x2(r(r−1)xr−2)−5x(rxr−1)+6xr=x3
r(r−1)xr−5rxr+ 6xr=x3
Step 4: Combine like terms and simplify:
r2xr−rxr−5rxr+ 6xr=x3
10
(r2−6r+ 6)xr=x3
Step 5: Equate the coefficients of like powers of xon both sides:
r2−6r+ 6 = 0
Step 6: Solve the quadratic equation r2−6r+ 6 = 0 to find the roots r1and
r2using the quadratic formula:
r=6±√62−4∗1∗6
2
r=6±√12
2
r= 3 ±√3
Step 7: Write down the general solution to the Cauchy-Euler equation:
y=c1x3+√3+c2x3−√3
where c1and c2are arbitrary constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −5xy′+
6y=x3is y=c1x3+√3+c2x3−√3.
Question 13
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 3y=x2ln x.
Solution
Step 1: Assume a solution of the form y=xr. Then, find the first and sec-
ond derivatives of y. Step 2: Substitute y,y′, and y′′ into the Cauchy-Euler
equation. Step 3: Simplify the equation by collecting like terms and setting the
non-homogeneous term equal to zero. Step 4: Solve the resulting homogeneous
equation x2y′′−3xy′+3y= 0 by substituting y=xrand finding the characteris-
tic equation. Step 5: Find the roots of the characteristic equation and determine
the general solution of the homogeneous equation. Step 6: Use the method of
undetermined coefficients to find a particular solution to the non-homogeneous
equation. Step 7: Combine the general solution of the homogeneous equation
with the particular solution to obtain the complete solution.
11
Question 14
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
Solution
To solve the Cauchy-Euler equation, we will assume a solution of the form
y=xr. We will then substitute this into the given differential equation and
find the values of rthat satisfy it.
Step 1: Substitute y=xrinto the differential equation. We have:
x2y′′ −3xy′+ 4y= 0
Substitute y=xr:
x2r(r−1)xr−2−3xrxr−1+ 4xr= 0
Step 2: Simplify the equation. Simplify the equation by combining like
terms:
r(r−1)xr−3rxr+ 4xr= 0
r2xr−rxr−3rxr+ 4xr= 0
r2−4r+ 1 = 0
Step 3: Find the roots of the characteristic equation. Solve the
characteristic equation r2−4r+ 1 = 0 to find the roots:
r=4±p42−4(1)(1)
2(1)
r=4±√16 −4
2
r=4±√12
2
r=4±2√3
2
r= 2 ±√3
Step 4: Write the general solution. Since the roots are distinct, the
general solution is:
y(x) = c1x2+√3+c2x2−√3
where c1and c2are arbitrary constants.
12
Question 15
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y=x3.
Solution
Step 1: Assume a solution of the form y=xm.
Step 2: Find the first and second derivatives of y.
y′=mxm−1and y′′ =m(m−1)xm−2.
Step 3: Substitute yand its derivatives back into the Cauchy-Euler equation.
x2(m(m−1)xm−2)−3x(mxm−1)+4xm=x3.
Step 4: Simplify the equation.
m(m−1)xm−3mxm+ 4xm=x3.
Step 5: Combine like terms.
m2−4m+ 4 = 1.
Step 6: Solve the quadratic equation m2−4m+ 4 = 1.
m2−4m+ 3 = 0.
(m−1)(m−3) = 0.
m= 1 or m= 3.
Step 7: Form the general solution.
y=c1x+c2x3,
where c1and c2are arbitrary constants.
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y=x3is y=c1x+c2x3.
Question 16
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y=x4.
13
Solution
Step 1: Let’s first find the characteristic equation by substituting y=xminto
the differential equation. Then, we have
y′=mxm−1and y′′ =m(m−1)xm−2.
Substitute these expressions into the Cauchy-Euler equation to get
x2(m(m−1)xm−2)−5x(mxm−1)+6xm=x4.
Step 2: Simplify the equation to get
m(m−1)xm−5mxm+ 6xm=x4.
Step 3: Combine the terms to obtain
m2−m−5m+ 6 = 1.
Step 4: Simplify the equation to get
m2−6m+ 6 = 0.
Step 5: Solve the quadratic equation to find the roots:
m1= 3 + √3 and m2= 3 −√3.
Step 6: The general solution to the Cauchy-Euler equation is given by
y(x) = c1xm1+c2xm2,
where c1and c2are constants.
Step 7: Therefore, the solution to the Cauchy-Euler equation x2y′′ −5xy′+
6y=x4is
y(x) = c1x3+√3+c2x3−√3.
Question 17
Question
Find the general solution of the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
To solve the Cauchy-Euler equation, we make the substitution y=xmand find
the roots of the characteristic equation.
Step 1: Find the characteristic equation The characteristic equation
corresponding to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 is given by:
m(m−1) −3m+ 4 = 0
14
Simplifying, we get:
m2−4m+ 4 = 0
(m−2)2= 0
The characteristic equation has a repeated root m= 2.
Step 2: Find the solution Since m= 2 is a repeated root, the general
solution of the Cauchy-Euler equation is given by:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Therefore, the general solution of the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is:
y(x) = c1x2+c2x2ln(x)
Question 18
Question
Solve the Cauchy-Euler equation: (x2−3x)y′′ + (4x−6)y′−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′′ into the Cauchy-Euler equation:
(x2−3x)(r(r−1)xr−2) + (4x−6)(rxr−1)−4xr= 0
.
Step 4: Simplify the equation:
r(r−1)xr−3r(r−1)xr+ 4rxr−6rxr−4xr= 0
.
Step 5: Combine like terms:
r2xr−3r2xr+ 4rxr−6rxr−4xr= 0
.
Step 6: Factor out xr:
(r2−3r2+ 4r−6r−4)xr= 0
15
.
Step 7: Simplify the expression:
(r2−3r+ 4r−6r−4)xr= 0
.
Step 8: Combine like terms:
(r2+r−4)xr= 0
.
Step 9: Solve the characteristic equation r2+r−4 = 0 to find the roots r1
and r2.
Step 10: The roots of the characteristic equation are r1= 1 and r2=−4.
Step 11: The general solution is given by:
y=c1x1+c2x−4
, where c1and c2are constants.
Question 19
Question
Find the general solution to the Cauchy-Euler equation: x2y′′ −5xy′+ 6y= 0.
Solution
Step 1: Assume the solution has the form y=xm. Step 2: Take derivatives
of y. Step 3: Substitute derivatives back into the differential equation. Step
4: Solve the resulting characteristic equation. Step 5: Determine the general
solution.
Step 1: Assume the solution has the form y=xm.
Let y=xm.
Step 2: Take derivatives of y.
Then, y′=mxm−1and y′′ =m(m−1)xm−2.
Step 3: Substitute derivatives back into the differential equation.
The Cauchy-Euler equation becomes: x2(m(m−1)xm−2)−5x(mxm−1) +
6xm= 0
This simplifies to: m(m−1)xm−5mxm+ 6xm= 0
Step 4: Solve the resulting characteristic equation.
Dividing by xm, we get the characteristic equation: m(m−1) −5m+ 6 = 0
Solving this equation, we find m= 2 and m= 3.
Step 5: Determine the general solution.
Thus, the general solution to the Cauchy-Euler equation x2y′′−5xy′+6y= 0
is: y=c1x2+c2x3, where c1and c2are arbitrary constants.
16
Question 20
Question
Solve the Cauchy-Euler equation x2y′′−3xy′+7y= 0 by making the substitution
y=xm.
Solution
Step 1: Substitute y=xminto the Cauchy-Euler equation x2y′′−3xy′+7y= 0.
x2(m(m−1)xm−2)−3x(mxm−1)+7xm= 0
Step 2: Simplify the equation by grouping the terms with x.
m(m−1)xm−3mxm+ 7xm= 0
m(m−1) −3m+ 7 = 0
Step 3: Solve the equation m(m−1) −3m+ 7 = 0 for m.
m2−m−3m+ 7 = 0
m2−4m+ 7 = 0
Step 4: Use the quadratic formula to solve for m.
m=4±p(−4)2−4(1)(7)
2(1)
m=4±√16 −28
2
m=4±√−12
2
m=4±2i√3
2
m= 2 ±i√3
Step 5: Write down the general solution using the roots m= 2 + i√3 and
m= 2 −i√3.
y(x) = c1x2+i√3+c2x2−i√3
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
7y= 0 is y(x) = c1x2+i√3+c2x2−i√3.
Question 21
Question
Solve the Cauchy-Euler equation x2y′′ −4xy′+ 6y= 0 using the substitution
y=xm.
17
Solution
Step 1: Compute the first and second derivatives of y.
y=xm⇒y′=mxm−1and y′′ =m(m−1)xm−2
Step 2: Substitute y,y′, and y′′ back into the Cauchy-Euler equation.
x2y′′ −4xy′+ 6y= 0 ⇒x2(m(m−1)xm−2)−4x(mxm−1)+6xm= 0
Step 3: Simplify the equation.
m(m−1)xm−4mxm+ 6xm= 0
Step 4: Factor out xmand solve for m.
m(m−1) −4m+ 6 = 0 ⇒m2−m−4m+ 6 = 0
Step 5: Simplify and solve for m.
m2−5m+ 6 = 0 ⇒(m−2)(m−3) = 0 ⇒m= 2 or m= 3
Step 6: Write down the general form of the solution using both values of m.
y=c1x2+c2x3where c1and c2are arbitrary constants.
Therefore, the general solution of the Cauchy-Euler equation x2y′′ −4xy′+
6y= 0 is y=c1x2+c2x3.
Question 22
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0 using the method of
undetermined coefficients.
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:
x2y′′ −3xy′+ 4y=x2[r(r−1)xr−2]−3x[rxr−1]+4xr= 0.
Step 4: Simplify the equation to obtain:
r(r−1)xr−3rxr+ 4xr=xr[r(r−1) −3r+ 4] = 0.
18
Step 5: Solve the characteristic equation r2−4r+ 4 = (r−2)2= 0 to find
the repeated root r1=r2= 2.
Step 6: The general solution to the Cauchy-Euler equation is given by y(x) =
c1xr1+c2xr2=c1x2+c2x2= (c1+c2)x2.
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is y(x) = c1x2+c2x2= (c1+c2)x2, where c1and c2are arbitrary constants.
Question 23
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y(x) = xrand differentiate to find y′and
y′′.
Let y(x) = xr
y′(x) = rxr−1
y′′(x) = r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the differential equation.
x2y′′ −3xy′+ 4y= 0
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Factor out xrand simplify the equation.
xr(r(r−1) −3r+ 4) = 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 4: Find the roots of the characteristic equation.
r= 2
Step 5: Form the general solution using the roots found.
y(x) = c1x2+c2x2ln(x)
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y(x) = c1x2+c2x2ln(x).
19
Question 24
Question
Solve the Cauchy-Euler equation:
x2y′′ −xy′+y= 0
Solution
Step 1: Assume a solution of the form y=xmand find the derivatives. Substi-
tute into the differential equation to get an auxiliary equation.
Step 2: Differentiate y=xmtwice to find y′and y′′:
y′=mxm−1and y′′ =m(m−1)xm−2
Step 3: Substitute y,y′, and y′′ into the Cauchy-Euler equation to get:
x2(m(m−1)xm−2)−x(mxm−1) + xm= 0
Step 4: Simplify the equation to get the auxiliary equation:
m(m−1)xm−mxm+xm= 0
Step 5: Factor out xmfrom the equation:
xm(m(m−1) −m+ 1) = 0
Step 6: Solve the auxiliary equation m(m−1) −m+ 1 = 0 to find the roots.
Step 7: The roots of the auxiliary equation are m1= 1, m2= 1, which are
equal and real.
Step 8: Write down the general solution for equal and real roots:
y(x) = c1x1ln(x) + c2x1
where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler equation is y(x) = c1xln(x) +
c2x.
Question 25
Question
Solve the Cauchy-Euler equation x2y′′ −5xy′+ 6y= 0 using the substitution
y=xm.
20
Solution
Step 1: Find the derivatives of y=xm.
Let y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute the derivatives into the Cauchy-Euler equation.
x2y′′ −5xy′+ 6y= 0
x2(m(m−1)xm−2)−5x(mxm−1)+6xm= 0
m(m−1)xm−5mxm+ 6xm= 0
m(m−1) −5m+ 6 = 0
Step 3: Solve for m.
m2−m−5m+ 6 = 0
m2−6m+ 6 = 0
(m−3)(m−2) = 0
This gives two possible solutions: m= 3 or m= 2.
Step 4: Form the general solution using the roots m1= 3 and m2= 2.
For m=3:y1=x3
For m=2:y2=x2
Therefore, the general solution is
y=c1x3+c2x2
where c1and c2are arbitrary constants.
Question 26
Question
Solve the Cauchy-Euler equation x2y′′−3xy′+3y= 0 by making the substitution
y(x) = xr.
Solution
Step 1: Write the Cauchy-Euler equation in terms of the new variable r. Let
y(x) = xr. Then, we can find y′and y′′ as follows:
y′=rxr−1and y′′ =r(r−1)xr−2
21
Substitute y(x) = xr,y′, and y′′ into the Cauchy-Euler equation to get:
x2(r(r−1)xr−2)−3x(rxr−1)+3xr= 0
Step 2: Simplify the equation. Simplify the equation to get: r(r−1)xr−
3rxr+ 3xr= 0
Step 3: Factor out xr. Factor out xrfrom the equation to get: xr(r(r−
1) −3r+ 3) = 0
Step 4: Solve for r. Set the expression inside the parentheses equal to zero:
r(r−1) −3r+ 3 = 0
Solve this quadratic equation to find the roots r1and r2.
Step 5: Write the general solution. The general solution to the Cauchy-
Euler equation is:
y(x) = c1xr1+c2xr2
where c1and c2are arbitrary constants.
Question 27
Question
Solve the Cauchy-Euler equation: x2y′′ −3xy′+ 4y= 0.
Solution
To solve this Cauchy-Euler equation, we will assume y=xmand find the values
of mthat satisfy the equation.
Step 1: Find the derivatives of yLet’s find the first and second deriva-
tives of y=xm:
y′=mxm−1and y′′ =m(m−1)xm−2
Step 2: Substitute yand its derivatives into the equation Substitute
y, y′,and y′′ into the Cauchy-Euler equation:
x2(m(m−1)xm−2)−3x(mxm−1)+4xm= 0
Simplify the equation:
m(m−1)xm−3mxm+ 4xm= 0
Step 3: Solve for mCombining like terms, we get:
m2−m−3m+ 4 = 0
Simplify further:
m2−4m+ 4 = 0
22
(m−2)2= 0
Step 4: Find the solutions for mThe solution to the quadratic equation
is:
m= 2
Step 5: Form the general solution Since m= 2, we have one solution
y1=x2.
Step 6: Find the second linearly independent solution To find the
second linearly independent solution, we use the method of reduction of order.
We can assume y2=vx2, where vis a function of x.
Differentiating y2:
y′
2=v′x2+ 2vx
y′′
2=v′′x2+ 4v′x+ 2v
Substitute y2, y′
2, and y′′
2back into the Cauchy-Euler equation:
x2(v′′x2+ 4v′x+ 2v)−3x(v′x2+ 2vx)+4vx2= 0
Simplify the equation and solve for v(x).
Step 7: Write the general solution The general solution to the Cauchy-
Euler equation x2y′′ −3xy′+ 4y= 0 is:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Question 28
Question
Solve the Cauchy-Euler equation:
x2y′′ −9xy′+ 25y= 0.
Solution
Step 1: Write the characteristic equation associated with the Cauchy-Euler
equation by assuming y=xr. Step 2: Find the roots of the characteristic
equation. Step 3: Form the general solution using the roots found in step 2.
Step 1: Write the characteristic equation by assuming y=xr:
r(r−1)xr−9rxr+ 25xr= 0
Step 2: Find the roots of the characteristic equation: Simplify the equation
to get:
r2−10r+ 25 = 0
23
Solve this quadratic equation to find the roots, r1and r2:
r1=r2= 5
Step 3: Form the general solution: Since we have a repeated root (r1=
r2= 5), the general solution is:
y(x) = c1x5+c2x5ln(x)
where c1and c2are arbitrary constants.
Question 29
Question
Solve the Cauchy-Euler differential equation (x2−3x+ 2)y′′ −2xy′+ 2y= 0
using the method of undetermined coefficients.
Solution
To solve the given Cauchy-Euler equation (x2−3x+ 2)y′′ −2xy′+ 2y= 0, we
first assume a solution of the form y=xr.
Step 1: Find the derivatives Calculate the first and second derivatives
of y:
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute into the differential equation Substitute y,y′, and
y′′ into the differential equation:
(x2−3x+ 2)(r(r−1)xr−2)−2x(rxr−1)+2xr= 0
Step 3: Simplify the equation Simplify the equation by combining like
terms:
r(r−1)xr−3(rxr)+2xr−2rxr+ 2xr= 0
r(r−1)xr−3rxr+ 2xr−2rxr+ 2xr= 0
r(r−1)xr−3rxr+ 2xr−2rxr+ 2xr= 0
r2xr−5rxr+ 2xr= 0
Step 4: Solve for rSet the expression equal to zero and solve for r:
r2−5r+ 2 = 0
Using the quadratic formula, we find the roots to be r1= 2 and r2=1
2.
Step 5: General solution The general solution will be in the form of
y=c1xr1+c2xr2, where c1and c2are arbitrary constants.
Therefore, the solution to the Cauchy-Euler differential equation (x2−3x+
2)y′′ −2xy′+ 2y= 0 is y=c1x2+c2x1/2.
24
Question 30
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Assume a solution of the form y=xr, where ris a constant to be
determined. Then we have:
y′=rxr−1and y′′ =r(r−1)xr−2
Step 2: Substitute y,y′, and y′′ into the original differential equation:
x2(r(r−1)xr−2)−3x(rxr−1)+4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
Step 3: Simplify the equation:
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 4: Since we have a repeated root r= 2, we have the solution:
y=c1x2+c2x2ln(x)
where c1and c2are constants.
Question 31
Question
Solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0.
Solution
Step 1: Write the characteristic equation by assuming y=xr.
Characteristic equation: x2r(r−1) −3xr + 4 = 0
r(r−1) −3r+ 4 = 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
r= 2 (double root)
25
Step 2: Write the general solution using the double root. Since r= 2 is a
double root, the general solution is given by:
y(x) = c1x2+c2x2ln(x)
where c1and c2are arbitrary constants.
Step 3: Find the specific solution by substituting the general solution back
into the differential equation.
x2y′′ −3xy′+ 4y=x2(2c2x+c2)−3x(2c1x+ 2c2) + 4(c1x2+c2x2ln(x))
= 2c2x3+c2x2−6c1x2−6c2x+ 4c1x2+ 4c2x2ln(x)
= (2c2−6c1+ 4c1)x2+ (c2−6c2)x+ 4c2x2ln(x)
Step 4: Equate the coefficients to zero. From the last step, we have:
(2c2−6c1+ 4c1= 0
c2−6c2= 0
Solving these equations gives c1=−1
2and c2= 0.
Therefore, the solution to the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0
is:
y(x) = −1
2x2
Question 32
Question
Solve the Cauchy-Euler equation x2y′′ + 3xy′+y= 0 using the substitution
y=xr.
Solution
Step 1: Compute the first and second derivatives of y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the Cauchy-Euler equation.
x2(r(r−1)xr−2)+3x(rxr−1) + xr= 0
Step 3: Simplify the equation by combining like terms.
r(r−1)xr+ 3rxr+xr= 0
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Step 4: Factor out xrfrom the equation.
xr(r2−r+ 3r+ 1) = 0
Step 5: Simplify the equation further.
xr(r2+ 2r+ 1) = 0
Step 6: Find the roots of the characteristic equation r2+ 2r+ 1 = 0.
r2+ 2r+ 1 = (r+ 1)2= 0
r=−1
Step 7: Write down the general solution using the roots r=−1.
y(x) = c1x−1+c2x−1ln(x)
Question 33
Question
Solve the Cauchy-Euler equation (x−1)y′′−2xy′+2y= 0 using the substitution
y=xm.
Solution
Step 1: We start by substituting y=xminto the differential equation.
(x−1)[m(m−1)xm−2]−2x[mxm−1]+2xm= 0
Step 2: Simplify the expression by expanding and collecting like terms.
m(m−1)xm−m(2x)xm−1+ 2xm= 0
Step 3: Divide through by xmto simplify the equation.
m(m−1) −2mx + 2 = 0
Step 4: Rewrite the equation in a standard form by rearranging the terms.
m2−m−2mx + 2 = 0
Step 5: This is a quadratic equation in m, so we can solve for mby using
the quadratic formula.
m=1±√1+8x
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Step 6: There are two roots, m1and m2. Substituting these back into the
equation y=xm, we get the following two solution functions.
y1=x(1+√1+8x)/2and y2=x(1−√1+8x)/2
Step 7: Therefore, the general solution to the Cauchy-Euler equation (x−
1)y′′ −2xy′+ 2y= 0 is given by
y(x) = c1x(1+√1+8x)/2+c2x(1−√1+8x)/2
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Question 34
Question
Solve the Cauchy-Euler equation: x2y′′ −xy′+ 6y= 0.
Solution
Step 1: Write the Cauchy-Euler equation in standard form. Assume y=xr.
x2y′′ −xy′+ 6y= 0
x2(r)(r−1)xr−2−x(r)xr−1+ 6xr= 0
r(r−1)xr−rxr+ 6xr= 0
r(r−1) −r+ 6 = 0
r2−2r+ 6 = 0
(r−3)(r+ 2) = 0
r= 3,−2
Step 2: Write the general solution using r1= 3 and r2=−2.
y=c1x3+c2x−2
Step 3: Check linear independence by verifying that the Wronskian is
nonzero.
W=
x3x−2
3x2−2x−3
=x3(−2x−3)−x−2(3x2)
=−2−3x2= 0
Since W= 0, the functions are linearly independent.
Step 4: The general solution to the Cauchy-Euler equation is:
y=c1x3+c2x−2
Question 35
Question
Solve the Cauchy-Euler equation:
x2y′′ −3xy′+ 4y= 0
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Solution
To solve the Cauchy-Euler equation x2y′′ −3xy′+ 4y= 0, we make the substi-
tution y=xmfor some real number m.
Step 1: Find the first and second derivatives of y=xm.
y=xm
y′=mxm−1
y′′ =m(m−1)xm−2
Step 2: Substitute y,y′, and y′′ back 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
m(m−1) −3m+ 4 = 0 (divide by xm)
Step 3: Solve the characteristic equation m(m−1) −3m+ 4 = 0.
m2−m−3m+ 4 = 0
m2−4m+ 4 = 0
(m−2)2= 0
m= 2
Step 4: Form the general solution with m= 2.
y=c1x2+c2x2ln x
Therefore, the general solution to the Cauchy-Euler equation x2y′′ −3xy′+
4y= 0 is y=c1x2+c2x2ln x.
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