MATH 334 - DIFFERENTIAL EQUATIONS Method of undetermined coefficients Question Bank Set 1

MATH 334 - DIFFERENTIAL

EQUATIONS - Method of

undetermined coefficients

Question Bank - Set 1

Liberty University

Question 1

Question

Solve the following non-homogeneous second-order differential equation using

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x−3x

Solution

To solve the given non-homogeneous second-order differential equation using the

method of undetermined coefficients, we first find the complementary solution

to the associated homogeneous equation. Then, we guess a particular solution

that accounts for the form of the non-homogeneous terms and solve for the

undetermined coefficients.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous equation y′′ −3y′+ 2y= 0 is r2−3r+ 2 = 0, which

factors to (r−1)(r−2) = 0. Therefore, the roots are r1= 1 and r2= 2. The

complementary solution is given by yc=c1er1x+c2er2x, where c1and c2are

arbitrary constants.

Thus, the complementary solution is yc=c1ex+c2e2x.

Step 2: Find the particular solution We guess the particular solution

has the form yp=Ae2x+Bx +C, where A,B, and Care undetermined

coefficients to be determined.

Substitute ypinto the original differential equation:

y′′ −3y′+ 2y= 6e2x−3x

(Ae2x)′′ −3(Ae2x)′+ 2(Ae2x) + B−3B+ 2(Ae2x) + C= 6e2x−3x

4Ae2x−6Ae2x+ 2Ae2x+B−3B+ 2Ae2x+C= 6e2x−3x

Matching coefficients of like terms, we get: A=−1, B = 0, C = 0

Therefore, the particular solution is yp=−e2x.

Step 3: Combine solutions The general solution to the non-homogeneous

differential equation is the sum of the complementary and particular solutions:

y=yc+yp=c1ex+c2e2x−e2x.

Question 2

Question

Find the particular solution to the differential equation y′′ −4y′+ 4y= 10e2x

using the method of undetermined coefficients.

Solution

Step 1: First, find the complementary solution to the homogeneous equation

y′′ −4y′+ 4y= 0. The characteristic equation is r2−4r+ 4 = 0, which factors

to (r−2)2= 0. So, the complementary solution is yc=c1e2x+c2xe2x.

Step 2: Assume the particular solution to the non-homogeneous equation

has the form yp=Ae2x, where Ais a constant to be determined.

Step 3: Calculate the first and second derivatives of yp:

y′

p= 2Ae2x, y′′

p= 4Ae2x

Step 4: Substitute ypand its derivatives into the differential equation to get:

4Ae2x−4(2Ae2x) + 4(Ae2x) = 10e2x

Step 5: Simplify the equation:

4Ae2x−8Ae2x+ 4Ae2x= 10e2x

0 = 10e2x

Since 0 = 10e2x, our assumption for ypwas incorrect.

Step 6: Assume a new form for the particular solution: yp=Ax2e2x. Cal-

culate the first and second derivatives of this form.

Step 7: Substitute ypand its derivatives into the differential equation and

solve for A.

Step 8: Once you have found the value of A, the particular solution is

yp=Ax2e2x.

Therefore, the particular solution to the given differential equation is yp=

Ax2e2x.

Question 3

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −2y′+y=ex+ sin(x).

Solution

To solve the given nonhomogeneous differential equation using the method of

undetermined coefficients, we will first find the complementary function yc(x)

and then determine the particular solutions for exand sin(x).

Step 1: Find the complementary function We first find the comple-

mentary function by solving the corresponding homogeneous equation:

y′′ −2y′+y= 0.

The characteristic equation is r2−2r+ 1 = 0, which factors to (r−1)2= 0.

Thus, we have a repeated root r= 1, and the complementary function is:

yc(x) = c1ex+c2xex.

Step 2: Find the particular solution for exFor y(1)

p=Aex, substitute

into the original differential equation:

(Aex)′′ −2(Aex)′+Aex=ex.

Differentiate and substitute back:

Aex−2Aex+Aex=ex⇒ −Aex=ex.

This implies A=−1, so a particular solution for exis y(1)

p=−ex.

Step 3: Find the particular solution for sin(x) For y(2)

p=Bsin(x) +

Ccos(x), substitute into the original differential equation:

(Bsin(x) + Ccos(x))′′ −2(Bsin(x) + Ccos(x))′+Bsin(x) + Ccos(x) = sin(x)

⇒(B−2C) sin(x) + (2B+C) cos(x)−2(Bcos(x)−Csin(x)) + Bsin(x) + Ccos(x) = sin(x).

Solving the system of equations, we get B= 0 and C=−1

2, so a particular

solution for sin(x) is y(2)

p=−1

2cos(x).

Step 4: Form the general solution The general solution to the nonho-

mogeneous differential equation is the sum of the complementary function and

the particular solutions:

y(x) = yc(x) + y(1)

p+y(2)

p=c1ex+c2xex−ex−1

2cos(x).

Therefore, the solution to the differential equation y′′ −2y′+y=ex+ sin(x) is

y(x) = c1ex+c2xex−ex−1

2cos(x).

Question 4

Question

Solve the following non-homogeneous differential equation using the method of

undetermined coefficients:

y′′ + 4y′+ 4y= 2e−2x+x2

Solution

Step 1: First, solve the homogeneous equation y′′ + 4y′+ 4y= 0 to find the

complementary function yc(x). The characteristic equation is r2+ 4r+ 4 = 0.

This factors to (r+ 2)2= 0, so we have a repeated root r=−2. Therefore, the

complementary function is yc(x)=(c1+c2x)e−2x.

Step 2: Now, for the particular solution yp(x), we guess a form for yp(x)

based on the terms on the right side of the non-homogeneous equation. Since

2e−2xcorresponds to Ce−2xwhere Cis a constant, and x2corresponds to

Ax2+Bx +Cwhere A,B, and Care constants, we make the guess:

yp(x) = Ae−2x+Bx2+Cx +D

Step 3: Find the first and second derivatives of yp(x):

y′

p(x) = −2Ae−2x+ 2Bx +C

y′′

p(x)=4Ae−2x+ 2B

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the non-homogeneous equa-

tion and simplify:

4Ae−2x+2B+ 4(−2Ae−2x+2Bx +C) +4(Ae−2x+Bx2+Cx+D)=2e−2x+x2

Step 5: Equate the coefficients of like terms on both sides of the equation:

yp(x) = −2Bx + (2B+C)x+ (4A+ 4C+D)e−2x= 2e−2x+x2

Step 6: From the previous step, equate coefficients to determine the values

of A,B,C, and D:

4A+ 4C+D= 2

2B+C= 0 =⇒C=−2B

−2B= 1 =⇒B=−1

4A−8B+D= 0 =⇒D= 4A−8(−1

Step 7: Solve the system of equations to find the values of A,B,C, and D.











4A+ 4C+D= 2

C=−2B

4A−8B+D= 0

Solving this system will give the values of A,B,C, and D.

Question 5

Question

Find the particular solution to the differential equation y′′ −3y′+ 2y= 4ex+

5 sin(x) using the method of undetermined coefficients.

Solution

To find the particular solution using the method of undetermined coefficients,

we assume that the particular solution has the form:

yp=Aex+Bsin(x) + Cx

where A,B, and Care constants to be determined.

Step 1: Find the derivatives of yp

y′

p=Aex+Bcos(x) + C

y′′

p=Aex−Bsin(x)

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation

y′′

p−3y′

p+ 2yp= (Aex−Bsin(x)) −3(Aex+Bcos(x) + C) + 2(Aex+Bsin(x) + Cx)

= (A−3A+ 2A)ex+ (−B+ 3B+ 2Bsin(x)) + (2C)

= 0ex+ 2Bsin(x)+2C

Step 3: Equate coefficients of the terms on both sides of the equa-

tion (2B= 4

2C= 5

Step 4: Solve for Band C

B= 2, C =5

Step 5: Write the particular solution

yp=Aex+ 2 sin(x) + 5

Therefore, the particular solution to the given differential equation is:

yp=Aex+ 2 sin(x) + 5

Question 6

Question

Find the particular solution to the non-homogeneous differential equation y′′ −

4y= 6 sin(2x) using the method of undetermined coefficients.

Solution

To find the particular solution using the method of undetermined coefficients,

we first need to find the complementary solution.

Step 1: Find the complementary solution The complementary solution

is the solution to the homogeneous equation y′′ −4y= 0. The characteristic

equation is r2−4 = 0, which gives us r=±2. Therefore, the complementary

solution is yc=c1e2x+c2e−2x.

Step 2: Guess the form of the particular solution Since the right-hand

side of the non-homogeneous equation is 6 sin(2x), we guess that the particular

solution has the form yp=asin(2x) + bcos(2x).

Step 3: Find the derivatives of yp

yp=asin(2x) + bcos(2x)

y′

p= 2acos(2x)−2bsin(2x)

y′′

p=−4asin(2x)−4bcos(2x)

Step 4: Substitute ypand its derivatives into the differential equa-

tion Substitute yp,y′

p, and y′′

pinto the differential equation y′′ −4y= 6 sin(2x):

−4asin(2x)−4bcos(2x)−4(asin(2x) + bcos(2x)) = 6 sin(2x)

Step 5: Solve for aand bBy comparing coefficients, we get:

−4a−4b= 0

−4b= 6

Solving the system of equations, we find a=−3

2and b=−3

Step 6: Form the particular solution Therefore, the particular solution

is yp=−3

2sin(2x)−3

2cos(2x).

Step 7: Write the general solution The general solution is the sum of

the complementary and particular solutions:

y=yc+yp=c1e2x+c2e−2x−3

2sin(2x)−3

2cos(2x)

Question 7

Question

Solve the following second-order non-homogeneous differential equation using

the method of undetermined coefficients:

y′′ −2y′+y=ex

Solution

1. First, find the complementary solution for the associated homogeneous

equation y′′ −2y′+y= 0.



The characteristic equation is r2−2r+ 1 = 0.



Factoring gives us (r−1)2= 0, which has a repeated root r= 1.



Therefore, the complementary solution is yc=c1ex+c2xex.

2. Next, find a particular solution ypthat satisfies the non-homogeneous

equation y′′ −2y′+y=ex.



Assume yp=Aexwhere Ais a constant to be determined.



Find y′

pand y′′

y′

p=Aexand y′′

p=Aex



Substitute these into the differential equation:

Aex−2Aex+Aex=ex

−Aex=ex



We obtain A=−1.



Therefore, the particular solution is yp=−ex.

3. The general solution to the non-homogeneous equation is the sum of the

complementary and particular solutions:

y=yc+yp=c1ex+c2xex−ex

Question 8

Question

Solve the following differential equation using the method of undetermined co-

efficients:

y′′ −4y′+ 4y= 2e2x.

Solution

Step 1: First, find the complementary solution to the homogeneous equation

y′′ −4y′+ 4y= 0. The characteristic equation is r2−4r+ 4 = 0, which

simplifies to (r−2)2= 0. Thus, the solution to the homogeneous equation is

yc(x)=(c1+c2x)e2x.

Step 2: Now, let’s determine the particular solution using the method of

undetermined coefficients. Since the right-hand side of the differential equation

is 2e2x, we try a particular solution of the form yp(x) = Ae2x. Substitute yp(x)

and its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 2e2x.

4Ae2x−8Ae2x+ 4Ae2x= 2e2x.

Step 3: Solve for Aby equating the coefficients of e2xon both sides of the

equation:

4A−8A+ 4A= 2.

0=2.

Since this equation has no solution, we need to modify our guess for the

particular solution. We should include a factor of xin our guess:

yp(x) = Axe2x.

Step 4: Substitute yp(x) and its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 2e2x.

4Axe2x+ 4Ae2x−8Axe2x+ 8Ae2x+ 4Axe2x= 2e2x.

Step 5: Solve for Aby equating the coefficients of e2xon both sides of the

equation:

4A−8A+ 4A= 2.

0=2.

Since this equation also has no solution, we need to modify our guess again.

We should include a factor of x2in our guess:

yp(x) = Bx2e2x.

Step 6: Substitute yp(x) and its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 2e2x.

4Bx2e2x+ 8Bxe2x+ 2Be2x−4Bx2e2x−8Bxe2x+ 4Bx2e2x= 2e2x.

Step 7: Solve for Bby equating the coefficients of e2xon both sides of the

equation:

2B= 2.

B= 1.

Therefore, the particular solution is yp(x) = x2e2x.

Step 8: The general solution to the differential equation is the sum of the

complementary and particular solutions:

y(x) = yc(x) + yp(x)=(c1+c2x)e2x+x2e2x.

Question 9

Question

Solve the following non-homogeneous linear differential equation using the method

of undetermined coefficients:

y′′ −y′−6y= 3e2x

Solution

Step 1: Find the complementary solution by solving the associated homogeneous

equation:

y′′ −y′−6y= 0

The characteristic equation is r2−r−6 = 0, which factors to (r−3)(r+2) = 0.

So the complementary solution is:

yc=c1e3x+c2e−2x

Step 2: Assume the particular solution has the form:

yp=Ae2x

Step 3: Calculate the derivatives of yp:

y′

p= 2Ae2x

y′′

p= 4Ae2x

Step 4: Substitute yp,y′

p, and y′′

pinto the original differential equation:

4Ae2x−2Ae2x−6Ae2x= 3e2x

Step 5: Simplify the equation:

2Ae2x= 3e2x

A=3

Step 6: Therefore, the particular solution is:

yp=3

2e2x

Step 7: The general solution is the sum of the complementary and particular

solutions:

y=yc+yp=c1e3x+c2e−2x+3

2e2x

So, the solution to the differential equation is:

y=c1e3x+c2e−2x+3

2e2x

Question 10

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −4y′+ 4y= 3e2x−5x2

with the initial conditions y(0) = 1 and y′(0) = 2.

Solution

To solve the nonhomogeneous differential equation using the method of unde-

termined coefficients, we first find the complementary function yc(x) by solving

the associated homogeneous equation:

y′′ −4y′+ 4y= 0

Characteristics equation: r2−4r+ 4 = 0

This factors as: (r−2)2= 0

The repeated root is r= 2, so the complementary function is:

yc(x)=(C1+C2x)e2x

Next, we find the particular solutions for the nonhomogeneous terms 3e2x

and −5x2:



For 3e2x, we try the particular solution y(1)

p(x) = Ae2x.

Since 2 is a repeated root of the characteristic equation, we multiply y(1)

p(x)

by xto make it linearly independent of yc(x):

y(1)

p(x) = Ax2e2x

Taking derivatives and substituting into the differential equation, we find

A=3



For −5x2, we try the particular solution y(2)

p(x) = Bx2.

Taking derivatives and substituting into the differential equation, we find

B=−5

Therefore, the particular solution yp(x) is:

yp(x) = 3

2x2e2x−5

4x2

The general solution is the sum of the complementary function and the

particular solution:

y(x) = yc(x) + yp(x)=(C1+C2x)e2x+3

2x2e2x−5

4x2

Applying the initial conditions y(0) = 1 and y′(0) = 2:

y(0) = C1= 1

y′(0) = 2C1+ 2C2= 2

Solving these equations gives C1= 1 and C2= 0.

Therefore, the solution to the initial value problem is:

y(x) = (1)e2x+3

2x2e2x−5

4x2

Question 11

Question

Find the particular solution to the differential equation y′′ −4y= 2e2xusing

the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary solution to the homogeneous equation

y′′ −4y= 0. The characteristic equation is r2−4 = 0, which gives us r=±2.

Therefore, the complementary solution is yc=c1e2x+c2e−2x.

Step 2: Next, we guess the form of the particular solution to the given

non-homogeneous equation. Since the right-hand side is 2e2x, a good guess is

yp=Ae2x.

Step 3: Plug ypinto the differential equation to find A. We have:

y′

p= 2Ae2x,

y′′

p= 4Ae2x.

Substitute these into the differential equation:

4Ae2x−4(Ae2x)=2e2x.

Step 4: Simplify and solve for A:

e2x(4A−4A)=2e2x,

0=2e2x.

This equation has no solution for A, so we modify our guess for the particular

solution. Since the complementary solution contains e2x, the correct form for

ypis yp=Axe2x.

Step 5: Plug ypinto the differential equation and solve for A:

y′

p=Ae2x+ 2Axe2x,

y′′

p= 2Ae2x+ 4Axe2x.

Substitute these into the differential equation:

2Ae2x+ 4Axe2x−4(Axe2x)=2e2x,

2Ae2x= 2e2x.

Step 6: Solve for A:

2A= 2,

A= 1.

Therefore, the particular solution is yp=xe2x. The general solution to the

differential equation is y=yc+yp=c1e2x+c2e−2x+xe2x.

Question 12

Question

Consider the following differential equation:

y′′ −3y′+ 2y= 4ex+ 3xsin(x)

Using the method of undetermined coefficients, find the particular solution to

this differential equation.

Solution

Step 1: Find the complementary solution ycof the homogeneous part of the

differential equation. The characteristic equation is r2−3r+ 2 = 0, which has

roots r1= 1 and r2= 2. Therefore, the complementary solution is:

yc=c1ex+c2e2x

Step 2: Find the form of the particular solution ypbased on the forcing term.

Since the forcing term contains terms of the form exand xsin(x), we assume

the particular solution has the form:

yp=Aex+Bx sin(x) + Cx cos(x)

Step 3: Calculate the derivatives of ypneeded to substitute back into the

differential equation.

y′

p=Aex+ (Bsin(x) + Bx cos(x)) + C(cos(x)−xsin(x))

y′′

p=Aex+ 2Bcos(x)+2C(−sin(x)−xcos(x))

Step 4: Substitute yp,y′

p, and y′′

pback into the differential equation.

(A−3A+2A)ex+(2B−3(A+B)+2C(B−2C)) cos(x)+(2C−3(B+C)−2(B+2Cx)) sin(x)=4ex+3xsin(x)

Step 5: Equate the coefficients of ex, cos(x), and sin(x) on both sides of the

equation to solve for A,B, and C.

A−3A+ 2A= 4 =⇒0 = 4 =⇒infeasible

2B−3(A+B)+2C(B−2C) = 0

2C−3(B+C)−2(B+ 2Cx)=3x

Step 6: Solve the system of equations to find Band C. After solving, we

find B=−3

2and C=−3

10 .

Therefore, the particular solution to the differential equation is:

yp=−3

2xsin(x)−3

10xcos(x)

Step 7: Finally, the general solution to the differential equation is the sum

of the complementary and particular solutions.

y=yc+yp=c1ex+c2e2x−3

2xsin(x)−3

10xcos(x)

Question 13

Question

Use the method of undetermined coefficients to solve the following nonhomoge-

neous differential equation:

y′′ −3y′+ 2y= 4ex

Solution

Let’s first find the general solution to the corresponding homogeneous equation:

y′′ −3y′+ 2y= 0

Step 1: Find the characteristic roots The characteristic equation is

r2−3r+ 2 = 0. Solving this equation gives us r1= 1 and r2= 2.

Step 2: Write down the complementary function The complementary

function is given by:

yc(x) = c1ex+c2e2x

Step 3: Guess the form of the particular solution Since the nonhomo-

geneous term is 4ex, we guess the particular solution has the form yp(x) = Aex

where Ais a constant to be determined.

Step 4: Find the derivatives of the particular solution

yp(x) = Aex

y′

p(x) = Aex

y′′

p(x) = Aex

Step 5: Substitute the particular solution into the differential

equation Substituting yp,y′

p, and y′′

pinto the differential equation gives:

Aex−3Aex+ 2Aex= 4ex

Step 6: Solve for the constant ASolving the equation Aex−3Aex+

2Aex= 4exgives A= 4.

Step 7: Write down the particular solution The particular solution is:

yp(x)=4ex

Step 8: Write down the general solution The general solution is the

sum of the complementary function and the particular solution:

y(x) = yc(x) + yp(x) = c1ex+c2e2x+ 4ex

Question 14

Question

Solve the following non-homogeneous second-order differential equation using

the method of undetermined coefficients:

y′′ −3y′+ 2y=ex+xsin x.

Solution

To solve the non-homogeneous differential equation using the method of unde-

termined coefficients, we will first solve the associated homogeneous equation

and then find particular solutions for each term on the right-hand side of the

equation.

Step 1: Find the solutions to the homogeneous equation

The associated homogeneous equation is:

y′′ −3y′+ 2y= 0.

The characteristic equation is r2−3r+2 = 0, which factors to (r−1)(r−2) = 0.

So, the roots are r1= 1 and r2= 2.

Therefore, the general solution to the homogeneous equation is:

yh=c1ex+c2e2x.

Step 2: Find a particular solution for ex

We guess a particular solution of the form yp=Aex, where Ais a constant

to be determined.

Substitute ypinto the original differential equation:

(Aex)′′ −3(Aex)′+ 2(Aex) = ex.

This simplifies to (A−3A+ 2A)ex=ex, giving A= 1.

Therefore, a particular solution for exis yp1=ex.

Step 3: Find a particular solution for xsin x

We guess a particular solution of the form yq= (Cx +D) sin x+ (Ex +

F) cos x, where C,D,E, and Fare constants to be determined.

Substitute yqinto the original differential equation and equate coefficients

of like terms to find C,D,E, and F.

After solving for the constants, we find that C=−1

2,D=1

4,E=1

2, and

F= 0.

Therefore, a particular solution for xsin xis yp2= (−1

2x+1

4) sin x+(1

2x) cos x.

Step 4: Assemble the general solution

The general solution to the non-homogeneous differential equation is the

sum of the general solution to the homogeneous equation and the particular

solutions:

y=yh+yp=c1ex+c2e2x+ex+ (−1

2x+1

4) sin x+ (1

2x) cos x.

Question 15

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −y= 2ex+ 3 cos x

Solution

We begin by finding the complementary solution to the homogeneous differential

equation y′′ −y= 0. The characteristic equation is r2−1 = 0, which has roots

r=±1. Thus, the complementary solution is given by yc(x) = c1ex+c2e−x,

where c1and c2are arbitrary constants.

Now we find the particular solution to the non-homogeneous differential

equation y′′ −y= 2ex+ 3 cos x. Since the right hand side contains terms with

exand cos x, we seek a particular solution of the form yp(x) = Aex+Bcos x+

Csin x, where A,B, and Care constants to be determined.

Step 1: Find yp(x) Substitute yp(x) = Aex+Bcos x+Csin xinto the

differential equation to find the values of A,B, and C.

y′′

p−yp= 2ex+ 3 cos x

(Aex+Bcos x+Csin x)′′ −(Aex+Bcos x+Csin x)=2ex+ 3 cos x

Aex−Bex−Csin x= 2ex+ 3 cos x

Equating coefficients of like terms gives A= 2, B=−3, and C= 0.

Therefore, the particular solution is yp(x)=2ex−3 cos x.

Step 2: Find the general solution The general solution to the non-

homogeneous differential equation is the sum of the complementary and partic-

ular solutions:

y(x) = yc(x) + yp(x) = c1ex+c2e−x+ 2ex−3 cos x

Thus, the general solution to the given differential equation is y(x) = c1ex+

c2e−x+ 2ex−3 cos x.

Question 16

Question

Solve the following second-order non-homogeneous differential equation using

the method of undetermined coefficients: y′′ −3y′+ 2y=e2x+x2.

Solution

To solve the non-homogeneous differential equation using the method of un-

determined coefficients, we first find the complementary function by solving

the corresponding homogeneous equation y′′ −3y′+ 2y= 0, then we guess a

particular solution for the non-homogeneous equation.

Step 1: Find the complementary function The characteristic equation

for the homogeneous equation y′′ −3y′+ 2y= 0 is r2−3r+ 2 = 0, which factors

as (r−1)(r−2) = 0. Therefore, the complementary function is yc=c1ex+c2e2x.

Step 2: Guess the particular solution For the non-homogeneous part

e2x, we guess a particular solution of the form yp=Ae2x. For the non-

homogeneous term x2, we guess a particular solution of the form yp=Bx2+

Cx +D.

Step 3: Evaluate derivatives and substitute into the differential

equation The derivatives of ypare: - y′

p= 2Ae2x-y′′

p= 4Ae2xSubstitute

these derivatives into the differential equation y′′ −3y′+ 2y=e2x+x2.

4Ae2x−3(2Ae2x)+2Ae2x=e2x. Simplifying, we get 5Ae2x=e2x, which

gives A=1

For the x2term, substitute yp=Bx2+Cx +Dinto the differential equation

and simplify to get 2B−3(2B) + 2(Bx2+Cx +D) = x2.

This gives 2x2−6x2+ 2Bx2+ 2Cx −6Bx + 2D=x2. Equating coefficients,

we get 2B−6 = 1, so B=7

Step 4: Write the general solution The general solution to the non-

homogeneous differential equation is the sum of the complementary function

and the particular solution: y=yc+yp.

Therefore, the general solution is y=c1ex+c2e2x+1

5e2x+7

2x2+Cx +D,

where c1,c2,C, and Dare constants.

Question 17

Question

Solve the following second order non-homogeneous differential equation using

the method of undetermined coefficients:

y′′ −4y′+ 4y= 5e2x.

Solution

Step 1: Find the complementary function (CF) by solving the corresponding

homogeneous equation: y′′ −4y′+ 4y= 0. The characteristic equation is r2−

4r+4 = 0, which factors into (r−2)2= 0. Thus, the CF is yCF =c1e2x+c2xe2x.

Step 2: Determine the form of the particular solution (PS) based on the

form of the non-homogeneous term. Since the non-homogeneous term is 5e2x,

and 2xalready appears in the CF, we try Ae2xas the form of the PS.

Step 3: Substitute the form of the PS into the original differential equation to

find the values of the undetermined coefficients. Differentiate yP S and substitute

into the differential equation:

y′′

P S −4y′

P S + 4yP S = 5e2x.

This leads to 4Ae2x−8Ae2x+ 4Ae2x= 5e2x. Solving, we get A=5

Therefore, the particular solution is yP S =5

4e2x.

Step 4: The general solution is the sum of the complementary function and

the particular solution:

y=yCF +yP S =c1e2x+c2xe2x+5

4e2x.

Question 18

Question

Consider the second-order differential equation y′′ −4y′+ 4y= 8e2x. Solve the

differential equation using the method of undetermined coefficients.

Solution

1. First, find the complementary solution to the homogeneous equation y′′ −

4y′+ 4y= 0.

The characteristic equation is r2−4r+4 = 0, which simplifies to (r−2)2=

0. Thus the complementary solution is of the form yc=c1e2x+c2xe2x.

2. Now consider the particular solution yp=Ae2xwhere Ais a constant to

be determined.

3. Calculate the first and second derivatives of yp:

y′

p= 2Ae2xand y′′

p= 4Ae2x

Substitute yp,y′

p, and y′′

pback into the original differential equation:

4Ae2x−4(2Ae2x) + 4(Ae2x)=8e2x

Simplify to find the value of A:

4Ae2x−8Ae2x+ 4Ae2x= 8e2x=⇒A= 2

4. Therefore, the particular solution is yp= 2e2x.

5. The general solution to the original differential equation is given by the

sum of the complementary and particular solutions:

y=yc+yp=c1e2x+c2xe2x+ 2e2x

Question 19

Question

Solve the following second-order homogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y=e2x+x.

Solution

To solve this differential equation using the method of undetermined coefficients,

we first need to find the homogeneous solution to the given differential equation

by solving the characteristic equation.

Step 1: Find the homogeneous solution The characteristic equation is

r2−4r+ 4 = 0. Factoring this quadratic equation, we get (r−2)2= 0, so

we have a repeated root r= 2. Thus, the homogeneous solution is of the form

yh= (c1+c2x)e2x.

Step 2: Find the particular solution Now, we assume a particular so-

lution of the form yp=Ae2x+Bx +Cfor the nonhomogeneous part of the

differential equation.

Step 3: Calculate the derivatives Find y′

pand y′′

y′

p= 2Ae2x+B

y′′

p= 4Ae2x

Step 4: Substitute into the differential equation and solve for the

coefficients Substitute yp, y′

p, y′′

pinto the differential equation:

4Ae2x−4(2Ae2x+B) + 4(Ae2x+Bx +C) = e2x+x

Simplify the equation:

(4A−8A+ 4A)e2x−4B+ 4B+ 4C=e2x+x

0+4C=x

Matching coefficients, we get:

4C=x=⇒C=1

So, the particular solution is yp=Ae2x+Bx +1

Step 5: Combine the homogeneous and particular solutions The

general solution is the sum of the homogeneous solution and the particular

solution:

y=yh+yp= (c1+c2x)e2x+Ae2x+Bx +1

Therefore, the solution to the differential equation is:

y= (c1+A)e2x+ (c2+B)x+1

Question 20

Question

Find the particular solution to the nonhomogeneous differential equation y′′ −

4y′+ 5y= 18e2xsin(3x) using the method of undetermined coefficients.

Solution

1. First step is to find the complementary solution by solving the homoge-

neous equation y′′ −4y′+5y= 0. Let’s assume the complementary solution

to be yc=erx.

2. The characteristic equation is r2−4r+ 5 = 0. Solving this quadratic

equation gives r= 2 ±i. Therefore, the complementary solution is yc=

e2x(Acos(3x) + Bsin(3x)), where Aand Bare constants.

3. Next, we need to find the particular solution ypof the nonhomogeneous

equation.



For e2xsin(3x), we assume the particular solution takes the form

yp=e2x(Ccos(3x) + Dsin(3x)).



Calculate the first and second derivatives of ypto substitute back

into the differential equation.



Substitute ypand its derivatives back into the differential equation

y′′ −4y′+ 5y= 18e2xsin(3x) and solve for Cand D.

4. After finding the particular solution yp, the general solution is given by

y=yc+yp.

Question 21

Question

Solve the following differential equation using the method of undetermined co-

efficients:

y′′ −3y′+ 2y=ex

Solution

Step 1: First, find the complementary solution by solving the homogeneous

equation

y′′

h−3y′

h+ 2yh= 0

The characteristic equation is r2−3r+2 = 0, which factors to (r−1)(r−2) = 0.

So, the complementary solution is

yh(x) = c1ex+c2e2x

Step 2: Assume the particular solution has the form

yp(x) = Aex

where Ais a constant to be determined.

Step 3: Calculate the first and second derivatives of yp(x):

y′

p(x) = Aex

y′′

p(x) = Aex

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential equa-

tion:

Aex−3Aex+ 2Aex=ex

Step 5: Simplify the equation and solve for A:

−Aex=ex

A=−1

Step 6: The particular solution is now found to be:

yp(x) = −ex

Step 7: The general solution is the sum of the complementary and particular

solutions:

y(x) = yh(x) + yp(x)

y(x) = c1ex+c2e2x−ex

So, the solution to the differential equation is

y(x) = c1ex+c2e2x−ex

Question 22

Question

Find the particular solution to the following differential equation using the

method of undetermined coefficients: y′′ +y= sin(3x).

Solution

To find the particular solution to the given differential equation, we will first

find the complementary solution and then use the method of undetermined

coefficients to find the particular solution.

Step 1: Find the Complementary Solution The complementary so-

lution is the solution to the homogeneous version of the differential equation,

which is y′′ +y= 0. Solving this, we get the characteristic equation r2+ 1 = 0,

which gives us r=±i.

Therefore, the complementary solution is yc(x) = c1cos(x) + c2sin(x).

Step 2: Assume a Form for the Particular Solution The particular

solution to the non-homogeneous differential equation y′′ +y= sin(3x) has the

form yp(x) = Asin(3x) + Bcos(3x). Here, Aand Bare constants we need to

determine.

Step 3: Find yp(x)and its derivatives



yp(x) = Asin(3x) + Bcos(3x)



y′

p(x) = 3Acos(3x)−3Bsin(3x)



y′′

p(x) = −9Asin(3x)−9Bcos(3x)

Step 4: Substitute into the Original Differential Equation Substitute

yp(x), y′

p(x), and y′′

p(x) into the original differential equation:

−9Asin(3x)−9Bcos(3x) + Asin(3x) + Bcos(3x) = sin(3x)

Step 5: Solve for the Coefficients Equating coefficients of sin(3x) and

cos(3x) on both sides, we get:

−9A+A= 0 =⇒A= 0

−9B+B= 1 =⇒B=−1

Step 6: Form the Particular Solution Therefore, the particular solution

is yp(x) = −1

8cos(3x).

Step 7: Write the General Solution The general solution is given by

combining the complementary and particular solutions:

y(x) = yc(x) + yp(x) = c1cos(x) + c2sin(x)−1

8cos(3x)

So, the particular solution to the differential equation y′′ +y= sin(3x) is

y(x) = c1cos(x) + c2sin(x)−1

8cos(3x).

Question 23

Question

Find a particular solution to the differential equation y′′ −3y′+ 2y= 2 sin(x) +

3x2using the method of undetermined coefficients.

Solution

To find a particular solution to the given differential equation, we will first find

the complementary solution and then guess a particular form for the particular

solution.

Step 1: Find the complementary solution The characteristic equation

of the differential equation y′′ −3y′+ 2y= 0 is r2−3r+ 2 = 0. Factoring, we

get (r−1)(r−2) = 0, so the complementary solution is yc=c1ex+c2e2xwhere

c1and c2are constants.

Step 2: Guess a form for the particular solution For a particular

solution to the given differential equation, we guess the form yp=Asin(x) +

Bx2+Cx +D, where A,B,C, and Dare constants to be determined.

Step 3: Find the derivatives of ypy′

p=Acos(x)+2Bx +C

y′′

p=−Asin(x)+2B

Step 4: Substitute ypand its derivatives into the differential equa-

tion Substitute yp,y′

p, and y′′

pinto the original differential equation:

(−Asin(x) + 2B)−3(Acos(x) + 2Bx +C) + 2(Asin(x) + Bx2+Cx +D) =

2 sin(x)+3x2

Grouping like terms, we get:

(2B−3C+2D)+((2B−3A+2C)x)+(−A−3A)sin(x)+(B)x2= 2 sin(x)+3x2

Step 5: Match coefficients Since the terms should match, we have the

following system of equations:

2B−3C+ 2D= 0

2B−3A+ 2C= 0

−A−3A= 2

B= 3

Solving these equations, we find A=−2

4=−1

2,B= 3, C=9

4, and D=−3.

Step 6: Write down the particular solution Therefore, the particular

solution is yp=−1

2sin(x)+3x2+9

4x−3.

Step 7: Write down the general solution The general solution of the

differential equation is y=yc+yp=c1ex+c2e2x−1

2sin(x)+3x2+9

4x−3.

Question 24

Question

Find the particular solution to the non-homogeneous differential equation y′′ −

3y′+ 2y= 2e3xusing the method of undetermined coefficients.

Solution

Step 1: First, find the complementary solution by solving the associated

homogeneous equation: y′′ −3y′+ 2y= 0. The characteristic equation is

r2−3r+ 2 = 0, which factors as (r−1)(r−2) = 0. So the complementary

solution is yc=c1ex+c2e2x, where c1and c2are constants.

Step 2: Assume the particular solution has the form yp=Ae3x, where Ais a

constant to be determined. Substitute ypand its derivatives into the differential

equation: (9A−9A+ 2A)e3x= 2e3x. This simplifies to 2Ae3x= 2e3x.

Step 3: Solve for Aby equating coefficients: 2A= 2 =⇒A= 1. Therefore,

the particular solution is yp=e3x.

Step 4: The general solution to the non-homogeneous differential equation

is the sum of the complementary and particular solutions: y=yc+yp=

c1ex+c2e2x+e3x, where c1and c2are arbitrary constants.

Thus, the particular solution to the given non-homogeneous differential equa-

tion is y=c1ex+c2e2x+e3x.

Question 25

Question

Consider the following second-order homogeneous differential equation:

y′′ + 4y= sec(2x).

Determine the particular solution using the method of undetermined coeffi-

cients.

Solution

To solve the given differential equation using the method of undetermined co-

efficients, we must first find the homogeneous solution.

Step 1: Find the homogeneous solution The homogeneous part of the

differential equation is given by y′′ + 4y= 0. The characteristic equation is

r2+ 4 = 0, which has roots r=±2i. Therefore, the homogeneous solution is

yh(x) = c1cos(2x) + c2sin(2x).

Step 2: Find the particular solution To find the particular solution, we

assume the form yp(x) = Asec(2x), where Ais a constant to be determined.

Now, we need to find the first and second derivatives of yp(x):

y′

p(x)=2Asec(2x) tan(2x) and y′′

p(x)=2Asec(2x) tan(2x)·sec(2x)+2Asec2(2x).

Step 3: Substitute into the differential equation Substitute yp(x),

y′

p(x), and y′′

p(x) into the original differential equation:

2Asec(2x) tan(2x)·sec(2x)+2Asec2(2x) + 4(Asec(2x)) = sec(2x).

Step 4: Solve for the constant After simplifying the equation above, we

find that A=1

2. So the particular solution is yp(x) = 1

2sec(2x).

Step 5: Write the general solution The general solution to the differ-

ential equation is the sum of the homogeneous and particular solutions:

y(x) = yh(x) + yp(x) = c1cos(2x) + c2sin(2x) + 1

2sec(2x).

Question 26

Question

Solve the following second-order linear homogeneous differential equation using

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6ex

Solution

To solve this differential equation using the method of undetermined coefficients,

we first find the general solution to the associated homogeneous equation:

y′′ −3y′+ 2y= 0

The characteristic equation is r2−3r+2 = 0, which factors as (r−1)(r−2) =

0. So, the roots are r1= 1 and r2= 2.

Therefore, the general solution to the homogeneous equation is:

yh=c1er1x+c2er2x=c1ex+c2e2x

To find the particular solution to the given non-homogeneous equation, we

make an educated guess for the form of yp(x) based on the form of the forcing

term 6ex. Since the forcing term is in the form of Ceax, we try the particular

solution in the form of yp=Aex, where Ais a constant to be determined.

Substitute yp=Aexinto the original differential equation:

y′′ −3y′+ 2y= 6ex

(Aex)′′ −3(Aex)′+ 2(Aex)=6ex

A(ex)′′ −3A(ex)′+ 2Aex= 6ex

Aex−3Aex+ 2Aex= 6ex

0=6ex

This is a contradiction. Therefore, we need to modify our guess for the particular

solution by including a factor of xin the exponential term.

Let’s try our new guess yp=Ax2exand substitute it into the differential

equation.

Step 1: Find the first and second derivatives of yp:

yp=Ax2ex

y′

p= (2Ax +Ax2)ex

y′′

p= (2A+ 4Ax +Ax2)ex

Step 2: Substitute yp=Ax2ex,y′

p= (2Ax+Ax2)ex, and y′′

p= (2A+4Ax+

Ax2)exinto the differential equation:

(2A+ 4Ax +Ax2)ex−3(2Ax +Ax2)ex+ 2(Ax2ex)=6ex

Step 3: Simplify the equation:

(2A+ 4Ax +Ax2−6Ax −3Ax2+ 2Ax2)ex= 6ex

(2A−6Ax +Ax2)ex= 6ex

Step 4: Equate coefficients to find A:

2A−6Ax +Ax2= 6

Ax2−6Ax + 2A= 6

Compare coefficients:

A= 6

So, the particular solution is yp= 6x2ex.

Step 5: The general solution to the original differential equation is the sum

of the homogeneous and particular solutions:

y=yh+yp=c1ex+c2e2x+ 6x2ex

Therefore, the solution to the differential equation y′′ −3y′+ 2y= 6exis

y=c1ex+c2e2x+ 6x2ex.

Question 27

Question

Solve the following non-homogeneous differential equation using the method of

undetermined coefficients: y′′ −3y′+ 2y= 6e2x+ 5x2+ 4x+ 1.

Solution

Step 1: First, we solve the corresponding homogeneous equation y′′ −3y′+2y=

0. The characteristic equation is r2−3r+2 = 0, which factors to (r−1)(r−2) =

0. So, the two roots are r1= 1 and r2= 2. Thus, the complementary solution

is yc=c1ex+c2e2x, where c1and c2are constants.

Step 2: Next, we find the particular solution for the non-homogeneous equa-

tion. Since the right side of the equation consists of a combination of terms,

we will take the form of the particular solution to be the sum of the particular

solutions for each term individually:

yp=Ay1+Bx2+Cx +D·e2x

Step 3: We calculate the derivatives of ypneeded to substitute into the

differential equation:

y′

p= 2Bx +C+ 2D·e2x

y′′

p= 2B+ 4D·e2x

Step 4: Substitute yp,y′

p, and y′′

pback into the differential equation and

simplify:

2B+4D·e2x−3(2Bx+C+2D·e2x)+2(Ay1+Bx2+Cx+D·e2x)=6e2x+5x2+4x+1

Step 5: Equate coefficients of like terms to determine the values of A, B, C,

and D. After solving for these constants, substitute them back into the partic-

ular solution.

Step 6: The general solution is the sum of the complementary and particular

solutions:

y=yc+yp=c1ex+c2e2x+Ax2+Bx +C+De2x

This is the complete solution to the non-homogeneous differential equation

y′′ −3y′+ 2y= 6e2x+ 5x2+ 4x+ 1.

Question 28

Question

Find the particular solution to the differential equation y′′ −2y′−3y= 2ex+

5 sin(2x) using the method of undetermined coefficients.

Solution

Step 1: First, find the complementary solution by solving the characteristic

equation m2−2m−3 = 0. The characteristic equation factors as (m−3)(m+1) =

0, so the roots are m= 3 and m=−1. Therefore, the complementary solution

is yc=c1e3x+c2e−x.

Step 2: To find the particular solution, we look for a solution of the form

yp=Aex+Bsin(2x)+Ccos(2x). Differentiating twice: y′

p=Aex+2Bcos(2x)−

2Csin(2x), y′′

p=Aex−4Bsin(2x)−4Ccos(2x).

Step 3: Substitute ypand its derivatives into the differential equation and

simplify. y′′ −2y′−3y= (Aex−4Bsin(2x)−4Ccos(2x))−2(Aex+2Bcos(2x)−

2Csin(2x)) −3(Aex+Bsin(2x) + Ccos(2x)).

This simplifies to: −2A+ 6B+ 3C)ex+ (−4A−10B) sin(2x)+(−4A−

10C) cos(2x)=2ex+ 5 sin(2x).

Step 4: Equate the coefficients of like terms on both sides of the equation.

This leads to the following system of equations: −2A+ 6B+ 3C= 2, −4A−

10B= 0, −4A−10C= 5.

Solving this system, we find A=−5

8, B =−1

4, C =−1

Step 5: The particular solution is then yp=−5

8ex−1

4sin(2x)−1

8cos(2x).

Step 6: The general solution to the differential equation is the sum of the

complementary and particular solutions: y=yc+yp=c1e3x+c2e−x−5

8ex−

4sin(2x)−1

8cos(2x).

Question 29

Question

Find the particular solution of the non-homogeneous linear differential equation:

y′′ −3y′+ 2y= 4ex+ 3 cos(x)

using the method of undetermined coefficients.

Solution

To find the particular solution of the differential equation using the method of

undetermined coefficients, we first find the complementary solution by solving

the associated homogeneous equation:

y′′ −3y′+ 2y= 0

The characteristic equation is given by r2−3r+ 2 = 0, which factors as (r−

1)(r−2) = 0. So, the complementary solution is given by:

yc(x) = c1ex+c2e2x

Now, to find the particular solution yp(x), we look at the form of the forc-

ing term 4ex+ 3 cos(x). Since these terms will not match any term in the

complementary solution, we assume the particular solution has the form:

yp(x) = Aex+Bcos(x) + Csin(x)

where A,B, and Care constants to be determined.

We then find the derivatives of yp(x):

y′

p(x) = Aex−Bsin(x) + Ccos(x)

y′′

p(x) = Aex−Bcos(x)−Csin(x)

Substitute yp(x) and its derivatives back into the original differential equa-

tion and match coefficients of the terms with same functions:

(A−3A)ex+ (−B−3B) sin(x)+(C−3C) cos(x)=4ex+ 3 cos(x)

Solving for A,B, and C:

−2Aex−4Bsin(x)−2Ccos(x)=4ex+ 3 cos(x)

Equating coefficients:











−2A= 4

−4B= 0

−2C= 3

⇒









A=−2

B= 0

C=−3

Therefore, the particular solution is:

yp(x) = −2ex−3

2sin(x)

The general solution of the differential equation is then:

y(x) = yc(x) + yp(x) = c1ex+c2e2x−2ex−3

2sin(x)

Question 30

Question

Find the particular solution to the differential equation y′′ −3y′+2y=e2xcos x

using the method of undetermined coefficients.

Solution

To solve this differential equation, we will first find the complementary solution

by solving the characteristic equation. After that, we will use the method of

undetermined coefficients to find the particular solution.

Step 1: Find the complementary solution The characteristic equation

is r2−3r+ 2 = 0. Solving this equation, we get r= 1 and r= 2. Therefore,

the complementary solution is given by yc=c1ex+c2e2x.

Step 2: Find the particular solution We assume the particular solution

has the form yp=Ae2xcos x+Be2xsin x.

Step 3: Calculate the derivatives

y′

p= (2Ae2xcos x−Ae2xsin x)−(2Be2xsin x+Be2xcos x)

=e2x(2Acos x−2Bsin x)

y′′

p=e2x(2Acos x−2Bsin x)−e2x(2Asin x+ 2Bcos x)

= 2e2x(Acos x−Bsin x)

Step 4: Substitute into the differential equation Substitute yp,y′

and y′′

pinto the differential equation and equating coefficients:

−2Acos x−2Bsin x−3(2Acos x−2Bsin x) + 2(Acos x−Bsin x) = e2xcos x

Step 5: Solve for coefficients Comparing coefficients on both sides of the

equation: - Simplify the left side and set it equal to the right side. - Solve for

Aand B.

Step 6: Write the general solution The general solution to the differ-

ential equation is the sum of the complementary and particular solutions:

y=yc+yp=c1ex+c2e2x+Ae2xcos x+Be2xsin x

Question 31

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −5y′+ 6y= 3e2x.

Solution

Step 1: First, find the complementary solution yc(x) by solving the correspond-

ing homogeneous equation:

y′′ −5y′+ 6y= 0.

The characteristic equation is r2−5r+6 = 0, which factors to (r−2)(r−3) = 0.

So the complementary solution is

yc(x) = c1e2x+c2e3x,

where c1and c2are arbitrary constants.

Step 2: To find the particular solution, we assume that yp(x) has the form

yp(x) = Ax2e2x.

Now, calculate the first and second derivatives of yp(x):

First derivative:

y′

p(x) = (2Ax +Ax2)e2x=e2x(Ax2+ 2Ax),

Second derivative:

y′′

p(x) = e2x(2A+ 4Ax + 2A) = e2x(2A+ 4Ax + 2A).

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential

equation and simplify:

e2x(2A+ 4Ax + 2A)−5e2x(Ax2+ 2Ax)+6Ax2e2x= 3e2x,

2A+ 4Ax + 2A−5Ax2−10Ax + 6Ax2= 3,

(2A+ 2A) + (4A−10A)x+ (2 −6)x2= 3,

4A−6x2= 3.

Step 4: Equate coefficients to find the value of A:

−6A= 3,

A=−1

Step 5: Now, the particular solution is

yp(x) = −1

2x2e2x.

Step 6: Therefore, the general solution to the non-homogeneous differential

equation is

y(x) = yc(x) + yp(x) = c1e2x+c2e3x−1

2x2e2x,

where c1and c2are arbitrary constants.

Question 32

Question

Find the particular solution of the differential equation y′′ −4y′+ 4y= 16e2x+

20xe2xusing the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary solution of the homogeneous part

of the differential equation: y′′ −4y′+ 4y= 0. The characteristic equation is

r2−4r+ 4 = 0, which factors to (r−2)2= 0. So, the complementary solution

is yc=c1e2x+c2xe2x.

Step 2: Next, we assume the particular solution has the form yp=Ae2x+

Bxe2x+Cx. We substitute this into the differential equation.

Step 3: Calculate the first and second derivatives of yp:y′

p= 2Ae2x+

Be2x+ 2Bxe2x+C y′′

p= 4Ae2x+ 2Be2x+ 4Bxe2x

Step 4: Plug yp,y′

p, and y′′

pinto the differential equation: (4A−8B)e2x=

16e2x(4A−8B+ 2B)e2x+C= 20xe2x

Step 5: Solve the system of equations: 4A−8B= 16 =⇒A= 4

4A−6B= 20 =⇒B=−2C= 0

Step 6: Therefore, the particular solution is yp= 4e2x−2xe2x, and the

general solution is y=yc+yp=c1e2x+c2xe2x+ 4e2x−2xe2x.

Question 33

Question

Find the general solution to the differential equation y′′ −4y′+ 4y= 2e2xusing

the method of undetermined coefficients.

Solution

Step 1: Find the complementary solution by solving the characteristic equation

r2−4r+ 4 = 0. The characteristic equation has a repeated root r= 2, so the

complementary solution is of the form yc= (C1+C2x)e2x.

Step 2: Assume the particular solution yphas the form Axe2x, where Ais a

constant.

Step 3: Compute the first and second derivatives of yp.yp=Axe2xy′

(2A+Ax)e2xy′′

p= (4A+ 4Ae2x+Ax)e2x

Step 4: Substitute yp,y′

p, and y′′

pinto the differential equation. (4A+4Ae2x+

Ax)e2x−4(2A+Ax)e2x+ 4Axe2x= 2e2x

Step 5: Simplify and solve for A. 4A+ 4Ae2x+Ax −8A−4Ax + 4Axe2x= 2

(−4A−4Ax)e2x= 2 A=−1

Therefore, the particular solution is yp=−1

2xe2x.

Step 6: Write the general solution as the sum of the complementary and

particular solutions. y= (C1+C2x)e2x−1

2xe2xy= (C1−1

2x)e2x+C2xe2x

So, the general solution to the differential equation is y= (C1−1

2x)e2x+

C2xe2x.

Question 34

Question

Find the particular solution to the following differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 2e2x.

Solution

To find the particular solution, we first need to find the complementary solution

to the homogeneous equation y′′ −4y′+ 4y= 0. The characteristic equation

is r2−4r+ 4 = 0, which simplifies to (r−2)2= 0. Thus, the complementary

solution is yc=c1e2x+c2xe2x.

Next, we guess a particular solution of the form yp=ae2x, where ais a

constant to be determined.

Step 1: Calculate the derivatives of yp.

y′

p= 2ae2x, y′′

p= 4ae2x.

Step 2: Substitute ypand its derivatives into the original differential equa-

tion.

(4a−8a+ 4ae2x)−4(2ae2x)+4ae2x= 2e2x.

This simplifies to −4ae2x= 2e2x.

Step 3: Equate the coefficients of e2xon both sides.

−4a= 2 ⇒a=−1

Step 4: Form the particular solution. Since a=−1

2, the particular solution

is yp=−1

2e2x.

Step 5: Form the general solution. The general solution is the sum of the

complementary and particular solutions:

y=yc+yp=c1e2x+c2xe2x−1

2e2x.

Therefore, the particular solution to the differential equation is y=c1e2x+

c2xe2x−1

2e2x.

Question 35

Question

Determine the particular solution for the following non-homogeneous differential

equation:

y′′ −4y′+ 4y= 3e2x

Solution

To find the particular solution for the given non-homogeneous differential equa-

tion using the method of undetermined coefficients, we assume the particular

solution has the form:

yp=Ae2x

where Ais a constant to be determined.

(Ae2x)′′ −3(Ae2x)′+ 2(Ae2x) + B−3B+ 2(Ae2x) + C= 6e2x−3x

4Ae2x−6Ae2x+ 2Ae2x+B−3B+ 2Ae2x+C= 6e2x−3x

Matching coefficients of like terms, we get: A=−1, B = 0, C = 0

Therefore, the particular solution is yp=−e2x.

Step 3: Combine solutions The general solution to the non-homogeneous

differential equation is the sum of the complementary and particular solutions:

y=yc+yp=c1ex+c2e2x−e2x.

Question 2

Question

Find the particular solution to the differential equation y′′ −4y′+ 4y= 10e2x

using the method of undetermined coefficients.

Solution

Step 1: First, find the complementary solution to the homogeneous equation

y′′ −4y′+ 4y= 0. The characteristic equation is r2−4r+ 4 = 0, which factors

to (r−2)2= 0. So, the complementary solution is yc=c1e2x+c2xe2x.

Step 2: Assume the particular solution to the non-homogeneous equation

has the form yp=Ae2x, where Ais a constant to be determined.

Step 3: Calculate the first and second derivatives of yp:

y′

p= 2Ae2x, y′′

p= 4Ae2x

Step 4: Substitute ypand its derivatives into the differential equation to get:

4Ae2x−4(2Ae2x) + 4(Ae2x) = 10e2x

Step 5: Simplify the equation:

4Ae2x−8Ae2x+ 4Ae2x= 10e2x

0 = 10e2x

Since 0 = 10e2x, our assumption for ypwas incorrect.

Step 6: Assume a new form for the particular solution: yp=Ax2e2x. Cal-

culate the first and second derivatives of this form.

Step 7: Substitute ypand its derivatives into the differential equation and

solve for A.

Step 8: Once you have found the value of A, the particular solution is

yp=Ax2e2x.

Therefore, the particular solution to the given differential equation is yp=

Ax2e2x.

Question 3

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −2y′+y=ex+ sin(x).

Solution

To solve the given nonhomogeneous differential equation using the method of

undetermined coefficients, we will first find the complementary function yc(x)

and then determine the particular solutions for exand sin(x).

Step 1: Find the complementary function We first find the comple-

mentary function by solving the corresponding homogeneous equation:

y′′ −2y′+y= 0.

The characteristic equation is r2−2r+ 1 = 0, which factors to (r−1)2= 0.

Thus, we have a repeated root r= 1, and the complementary function is:

yc(x) = c1ex+c2xex.

Step 2: Find the particular solution for exFor y(1)

p=Aex, substitute

into the original differential equation:

(Aex)′′ −2(Aex)′+Aex=ex.

Differentiate and substitute back:

Aex−2Aex+Aex=ex⇒ −Aex=ex.

This implies A=−1, so a particular solution for exis y(1)

p=−ex.

Step 3: Find the particular solution for sin(x) For y(2)

p=Bsin(x) +

Ccos(x), substitute into the original differential equation:

(Bsin(x) + Ccos(x))′′ −2(Bsin(x) + Ccos(x))′+Bsin(x) + Ccos(x) = sin(x)

⇒(B−2C) sin(x) + (2B+C) cos(x)−2(Bcos(x)−Csin(x)) + Bsin(x) + Ccos(x) = sin(x).

Solving the system of equations, we get B= 0 and C=−1

2, so a particular

solution for sin(x) is y(2)

p=−1

2cos(x).

Step 4: Form the general solution The general solution to the nonho-

mogeneous differential equation is the sum of the complementary function and

the particular solutions:

y(x) = yc(x) + y(1)

p+y(2)

p=c1ex+c2xex−ex−1

2cos(x).

Therefore, the solution to the differential equation y′′ −2y′+y=ex+ sin(x) is

y(x) = c1ex+c2xex−ex−1

2cos(x).

Question 4

Question

Solve the following non-homogeneous differential equation using the method of

undetermined coefficients:

y′′ + 4y′+ 4y= 2e−2x+x2

Solution

Step 1: First, solve the homogeneous equation y′′ + 4y′+ 4y= 0 to find the

complementary function yc(x). The characteristic equation is r2+ 4r+ 4 = 0.

This factors to (r+ 2)2= 0, so we have a repeated root r=−2. Therefore, the

complementary function is yc(x)=(c1+c2x)e−2x.

Step 2: Now, for the particular solution yp(x), we guess a form for yp(x)

based on the terms on the right side of the non-homogeneous equation. Since

2e−2xcorresponds to Ce−2xwhere Cis a constant, and x2corresponds to

Ax2+Bx +Cwhere A,B, and Care constants, we make the guess:

yp(x) = Ae−2x+Bx2+Cx +D

Step 3: Find the first and second derivatives of yp(x):

y′

p(x) = −2Ae−2x+ 2Bx +C

y′′

p(x)=4Ae−2x+ 2B

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the non-homogeneous equa-

tion and simplify:

4Ae−2x+2B+ 4(−2Ae−2x+2Bx +C) +4(Ae−2x+Bx2+Cx+D)=2e−2x+x2

Step 5: Equate the coefficients of like terms on both sides of the equation:

yp(x) = −2Bx + (2B+C)x+ (4A+ 4C+D)e−2x= 2e−2x+x2

Step 6: From the previous step, equate coefficients to determine the values

of A,B,C, and D:

4A+ 4C+D= 2

2B+C= 0 =⇒C=−2B

−2B= 1 =⇒B=−1

4A−8B+D= 0 =⇒D= 4A−8(−1

Step 7: Solve the system of equations to find the values of A,B,C, and D.











4A+ 4C+D= 2

C=−2B

4A−8B+D= 0

Solving this system will give the values of A,B,C, and D.

Question 5

Question

Find the particular solution to the differential equation y′′ −3y′+ 2y= 4ex+

5 sin(x) using the method of undetermined coefficients.

Solution

To find the particular solution using the method of undetermined coefficients,

we assume that the particular solution has the form:

yp=Aex+Bsin(x) + Cx

where A,B, and Care constants to be determined.

Step 1: Find the derivatives of yp

y′

p=Aex+Bcos(x) + C

y′′

p=Aex−Bsin(x)

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation

y′′

p−3y′

p+ 2yp= (Aex−Bsin(x)) −3(Aex+Bcos(x) + C) + 2(Aex+Bsin(x) + Cx)

= (A−3A+ 2A)ex+ (−B+ 3B+ 2Bsin(x)) + (2C)

= 0ex+ 2Bsin(x)+2C

Step 3: Equate coefficients of the terms on both sides of the equa-

tion (2B= 4

2C= 5

Step 4: Solve for Band C

B= 2, C =5

Step 5: Write the particular solution

yp=Aex+ 2 sin(x) + 5

Therefore, the particular solution to the given differential equation is:

yp=Aex+ 2 sin(x) + 5

Question 6

Question

Find the particular solution to the non-homogeneous differential equation y′′ −

4y= 6 sin(2x) using the method of undetermined coefficients.

Solution

To find the particular solution using the method of undetermined coefficients,

we first need to find the complementary solution.

Step 1: Find the complementary solution The complementary solution

is the solution to the homogeneous equation y′′ −4y= 0. The characteristic

equation is r2−4 = 0, which gives us r=±2. Therefore, the complementary

solution is yc=c1e2x+c2e−2x.

Step 2: Guess the form of the particular solution Since the right-hand

side of the non-homogeneous equation is 6 sin(2x), we guess that the particular

solution has the form yp=asin(2x) + bcos(2x).

Step 3: Find the derivatives of yp

yp=asin(2x) + bcos(2x)

y′

p= 2acos(2x)−2bsin(2x)

y′′

p=−4asin(2x)−4bcos(2x)

Step 4: Substitute ypand its derivatives into the differential equa-

tion Substitute yp,y′

p, and y′′

pinto the differential equation y′′ −4y= 6 sin(2x):

−4asin(2x)−4bcos(2x)−4(asin(2x) + bcos(2x)) = 6 sin(2x)

Step 5: Solve for aand bBy comparing coefficients, we get:

−4a−4b= 0

−4b= 6

Solving the system of equations, we find a=−3

2and b=−3

Step 6: Form the particular solution Therefore, the particular solution

is yp=−3

2sin(2x)−3

2cos(2x).

Step 7: Write the general solution The general solution is the sum of

the complementary and particular solutions:

y=yc+yp=c1e2x+c2e−2x−3

2sin(2x)−3

2cos(2x)

Question 7

Question

Solve the following second-order non-homogeneous differential equation using

the method of undetermined coefficients:

y′′ −2y′+y=ex

Solution

1. First, find the complementary solution for the associated homogeneous

equation y′′ −2y′+y= 0.



The characteristic equation is r2−2r+ 1 = 0.



Factoring gives us (r−1)2= 0, which has a repeated root r= 1.



Therefore, the complementary solution is yc=c1ex+c2xex.

2. Next, find a particular solution ypthat satisfies the non-homogeneous

equation y′′ −2y′+y=ex.



Assume yp=Aexwhere Ais a constant to be determined.



Find y′

pand y′′

y′

p=Aexand y′′

p=Aex



Substitute these into the differential equation:

Aex−2Aex+Aex=ex

−Aex=ex



We obtain A=−1.



Therefore, the particular solution is yp=−ex.

3. The general solution to the non-homogeneous equation is the sum of the

complementary and particular solutions:

y=yc+yp=c1ex+c2xex−ex

Question 8

Question

Solve the following differential equation using the method of undetermined co-

efficients:

y′′ −4y′+ 4y= 2e2x.

Solution

Step 1: First, find the complementary solution to the homogeneous equation

y′′ −4y′+ 4y= 0. The characteristic equation is r2−4r+ 4 = 0, which

simplifies to (r−2)2= 0. Thus, the solution to the homogeneous equation is

yc(x)=(c1+c2x)e2x.

Step 2: Now, let’s determine the particular solution using the method of

undetermined coefficients. Since the right-hand side of the differential equation

is 2e2x, we try a particular solution of the form yp(x) = Ae2x. Substitute yp(x)

and its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 2e2x.

4Ae2x−8Ae2x+ 4Ae2x= 2e2x.

Step 3: Solve for Aby equating the coefficients of e2xon both sides of the

equation:

4A−8A+ 4A= 2.

0=2.

Since this equation has no solution, we need to modify our guess for the

particular solution. We should include a factor of xin our guess:

yp(x) = Axe2x.

Step 4: Substitute yp(x) and its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 2e2x.

4Axe2x+ 4Ae2x−8Axe2x+ 8Ae2x+ 4Axe2x= 2e2x.

Step 5: Solve for Aby equating the coefficients of e2xon both sides of the

equation:

4A−8A+ 4A= 2.

0=2.

Since this equation also has no solution, we need to modify our guess again.

We should include a factor of x2in our guess:

yp(x) = Bx2e2x.

Step 6: Substitute yp(x) and its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 2e2x.

4Bx2e2x+ 8Bxe2x+ 2Be2x−4Bx2e2x−8Bxe2x+ 4Bx2e2x= 2e2x.

Step 7: Solve for Bby equating the coefficients of e2xon both sides of the

equation:

2B= 2.

B= 1.

Therefore, the particular solution is yp(x) = x2e2x.

Step 8: The general solution to the differential equation is the sum of the

complementary and particular solutions:

y(x) = yc(x) + yp(x)=(c1+c2x)e2x+x2e2x.

Question 9

Question

Solve the following non-homogeneous linear differential equation using the method

of undetermined coefficients:

y′′ −y′−6y= 3e2x

Solution

Step 1: Find the complementary solution by solving the associated homogeneous

equation:

y′′ −y′−6y= 0

The characteristic equation is r2−r−6 = 0, which factors to (r−3)(r+2) = 0.

So the complementary solution is:

yc=c1e3x+c2e−2x

Step 2: Assume the particular solution has the form:

yp=Ae2x

Step 3: Calculate the derivatives of yp:

y′

p= 2Ae2x

y′′

p= 4Ae2x

Step 4: Substitute yp,y′

p, and y′′

pinto the original differential equation:

4Ae2x−2Ae2x−6Ae2x= 3e2x

Step 5: Simplify the equation:

2Ae2x= 3e2x

A=3

Step 6: Therefore, the particular solution is:

yp=3

2e2x

Step 7: The general solution is the sum of the complementary and particular

solutions:

y=yc+yp=c1e3x+c2e−2x+3

2e2x

So, the solution to the differential equation is:

y=c1e3x+c2e−2x+3

2e2x

Question 10

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −4y′+ 4y= 3e2x−5x2

with the initial conditions y(0) = 1 and y′(0) = 2.

Solution

To solve the nonhomogeneous differential equation using the method of unde-

termined coefficients, we first find the complementary function yc(x) by solving

the associated homogeneous equation:

y′′ −4y′+ 4y= 0

Characteristics equation: r2−4r+ 4 = 0

This factors as: (r−2)2= 0

The repeated root is r= 2, so the complementary function is:

yc(x)=(C1+C2x)e2x

Next, we find the particular solutions for the nonhomogeneous terms 3e2x

and −5x2:



For 3e2x, we try the particular solution y(1)

p(x) = Ae2x.

Since 2 is a repeated root of the characteristic equation, we multiply y(1)

p(x)

by xto make it linearly independent of yc(x):

y(1)

p(x) = Ax2e2x

Taking derivatives and substituting into the differential equation, we find

A=3



For −5x2, we try the particular solution y(2)

p(x) = Bx2.

Taking derivatives and substituting into the differential equation, we find

B=−5

Therefore, the particular solution yp(x) is:

yp(x) = 3

2x2e2x−5

4x2

The general solution is the sum of the complementary function and the

particular solution:

y(x) = yc(x) + yp(x)=(C1+C2x)e2x+3

2x2e2x−5

4x2

Applying the initial conditions y(0) = 1 and y′(0) = 2:

y(0) = C1= 1

y′(0) = 2C1+ 2C2= 2

Solving these equations gives C1= 1 and C2= 0.

Therefore, the solution to the initial value problem is:

y(x) = (1)e2x+3

2x2e2x−5

4x2

Question 11

Question

Find the particular solution to the differential equation y′′ −4y= 2e2xusing

the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary solution to the homogeneous equation

y′′ −4y= 0. The characteristic equation is r2−4 = 0, which gives us r=±2.

Therefore, the complementary solution is yc=c1e2x+c2e−2x.

Step 2: Next, we guess the form of the particular solution to the given

non-homogeneous equation. Since the right-hand side is 2e2x, a good guess is

yp=Ae2x.

Step 3: Plug ypinto the differential equation to find A. We have:

y′

p= 2Ae2x,

y′′

p= 4Ae2x.

Substitute these into the differential equation:

4Ae2x−4(Ae2x)=2e2x.

Step 4: Simplify and solve for A:

e2x(4A−4A)=2e2x,

0=2e2x.

This equation has no solution for A, so we modify our guess for the particular

solution. Since the complementary solution contains e2x, the correct form for

ypis yp=Axe2x.

Step 5: Plug ypinto the differential equation and solve for A:

y′

p=Ae2x+ 2Axe2x,

y′′

p= 2Ae2x+ 4Axe2x.

Substitute these into the differential equation:

2Ae2x+ 4Axe2x−4(Axe2x)=2e2x,

2Ae2x= 2e2x.

Step 6: Solve for A:

2A= 2,

A= 1.

Therefore, the particular solution is yp=xe2x. The general solution to the

differential equation is y=yc+yp=c1e2x+c2e−2x+xe2x.

Question 12

Question

Consider the following differential equation:

y′′ −3y′+ 2y= 4ex+ 3xsin(x)

Using the method of undetermined coefficients, find the particular solution to

this differential equation.

Solution

Step 1: Find the complementary solution ycof the homogeneous part of the

differential equation. The characteristic equation is r2−3r+ 2 = 0, which has

roots r1= 1 and r2= 2. Therefore, the complementary solution is:

yc=c1ex+c2e2x

Step 2: Find the form of the particular solution ypbased on the forcing term.

Since the forcing term contains terms of the form exand xsin(x), we assume

the particular solution has the form:

yp=Aex+Bx sin(x) + Cx cos(x)

Step 3: Calculate the derivatives of ypneeded to substitute back into the

differential equation.

y′

p=Aex+ (Bsin(x) + Bx cos(x)) + C(cos(x)−xsin(x))

y′′

p=Aex+ 2Bcos(x)+2C(−sin(x)−xcos(x))

Step 4: Substitute yp,y′

p, and y′′

pback into the differential equation.

(A−3A+2A)ex+(2B−3(A+B)+2C(B−2C)) cos(x)+(2C−3(B+C)−2(B+2Cx)) sin(x)=4ex+3xsin(x)

Step 5: Equate the coefficients of ex, cos(x), and sin(x) on both sides of the

equation to solve for A,B, and C.

A−3A+ 2A= 4 =⇒0 = 4 =⇒infeasible

2B−3(A+B)+2C(B−2C) = 0

2C−3(B+C)−2(B+ 2Cx)=3x

Step 6: Solve the system of equations to find Band C. After solving, we

find B=−3

2and C=−3

10 .

Therefore, the particular solution to the differential equation is:

yp=−3

2xsin(x)−3

10xcos(x)

Step 7: Finally, the general solution to the differential equation is the sum

of the complementary and particular solutions.

y=yc+yp=c1ex+c2e2x−3

2xsin(x)−3

10xcos(x)

Question 13

Question

Use the method of undetermined coefficients to solve the following nonhomoge-

neous differential equation:

y′′ −3y′+ 2y= 4ex

Solution

Let’s first find the general solution to the corresponding homogeneous equation:

y′′ −3y′+ 2y= 0

Step 1: Find the characteristic roots The characteristic equation is

r2−3r+ 2 = 0. Solving this equation gives us r1= 1 and r2= 2.

Step 2: Write down the complementary function The complementary

function is given by:

yc(x) = c1ex+c2e2x

Step 3: Guess the form of the particular solution Since the nonhomo-

geneous term is 4ex, we guess the particular solution has the form yp(x) = Aex

where Ais a constant to be determined.

Step 4: Find the derivatives of the particular solution

yp(x) = Aex

y′

p(x) = Aex

y′′

p(x) = Aex

Step 5: Substitute the particular solution into the differential

equation Substituting yp,y′

p, and y′′

pinto the differential equation gives:

Aex−3Aex+ 2Aex= 4ex

Step 6: Solve for the constant ASolving the equation Aex−3Aex+

2Aex= 4exgives A= 4.

Step 7: Write down the particular solution The particular solution is:

yp(x)=4ex

Step 8: Write down the general solution The general solution is the

sum of the complementary function and the particular solution:

y(x) = yc(x) + yp(x) = c1ex+c2e2x+ 4ex

Question 14

Question

Solve the following non-homogeneous second-order differential equation using

the method of undetermined coefficients:

y′′ −3y′+ 2y=ex+xsin x.

Solution

To solve the non-homogeneous differential equation using the method of unde-

termined coefficients, we will first solve the associated homogeneous equation

and then find particular solutions for each term on the right-hand side of the

equation.

Step 1: Find the solutions to the homogeneous equation

The associated homogeneous equation is:

y′′ −3y′+ 2y= 0.

The characteristic equation is r2−3r+2 = 0, which factors to (r−1)(r−2) = 0.

So, the roots are r1= 1 and r2= 2.

Therefore, the general solution to the homogeneous equation is:

yh=c1ex+c2e2x.

Step 2: Find a particular solution for ex

We guess a particular solution of the form yp=Aex, where Ais a constant

to be determined.

Substitute ypinto the original differential equation:

(Aex)′′ −3(Aex)′+ 2(Aex) = ex.

This simplifies to (A−3A+ 2A)ex=ex, giving A= 1.

Therefore, a particular solution for exis yp1=ex.

Step 3: Find a particular solution for xsin x

We guess a particular solution of the form yq= (Cx +D) sin x+ (Ex +

F) cos x, where C,D,E, and Fare constants to be determined.

Substitute yqinto the original differential equation and equate coefficients

of like terms to find C,D,E, and F.

After solving for the constants, we find that C=−1

2,D=1

4,E=1

2, and

F= 0.

Therefore, a particular solution for xsin xis yp2= (−1

2x+1

4) sin x+(1

2x) cos x.

Step 4: Assemble the general solution

The general solution to the non-homogeneous differential equation is the

sum of the general solution to the homogeneous equation and the particular

solutions:

y=yh+yp=c1ex+c2e2x+ex+ (−1

2x+1

4) sin x+ (1

2x) cos x.

Question 15

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −y= 2ex+ 3 cos x

Solution

We begin by finding the complementary solution to the homogeneous differential

equation y′′ −y= 0. The characteristic equation is r2−1 = 0, which has roots

r=±1. Thus, the complementary solution is given by yc(x) = c1ex+c2e−x,

where c1and c2are arbitrary constants.

Now we find the particular solution to the non-homogeneous differential

equation y′′ −y= 2ex+ 3 cos x. Since the right hand side contains terms with

exand cos x, we seek a particular solution of the form yp(x) = Aex+Bcos x+

Csin x, where A,B, and Care constants to be determined.

Step 1: Find yp(x) Substitute yp(x) = Aex+Bcos x+Csin xinto the

differential equation to find the values of A,B, and C.

y′′

p−yp= 2ex+ 3 cos x

(Aex+Bcos x+Csin x)′′ −(Aex+Bcos x+Csin x)=2ex+ 3 cos x

Aex−Bex−Csin x= 2ex+ 3 cos x

Equating coefficients of like terms gives A= 2, B=−3, and C= 0.

Therefore, the particular solution is yp(x)=2ex−3 cos x.

Step 2: Find the general solution The general solution to the non-

homogeneous differential equation is the sum of the complementary and partic-

ular solutions:

y(x) = yc(x) + yp(x) = c1ex+c2e−x+ 2ex−3 cos x

Thus, the general solution to the given differential equation is y(x) = c1ex+

c2e−x+ 2ex−3 cos x.

Question 16

Question

Solve the following second-order non-homogeneous differential equation using

the method of undetermined coefficients: y′′ −3y′+ 2y=e2x+x2.

Solution

To solve the non-homogeneous differential equation using the method of un-

determined coefficients, we first find the complementary function by solving

the corresponding homogeneous equation y′′ −3y′+ 2y= 0, then we guess a

particular solution for the non-homogeneous equation.

Step 1: Find the complementary function The characteristic equation

for the homogeneous equation y′′ −3y′+ 2y= 0 is r2−3r+ 2 = 0, which factors

as (r−1)(r−2) = 0. Therefore, the complementary function is yc=c1ex+c2e2x.

Step 2: Guess the particular solution For the non-homogeneous part

e2x, we guess a particular solution of the form yp=Ae2x. For the non-

homogeneous term x2, we guess a particular solution of the form yp=Bx2+

Cx +D.

Step 3: Evaluate derivatives and substitute into the differential

equation The derivatives of ypare: - y′

p= 2Ae2x-y′′

p= 4Ae2xSubstitute

these derivatives into the differential equation y′′ −3y′+ 2y=e2x+x2.

4Ae2x−3(2Ae2x)+2Ae2x=e2x. Simplifying, we get 5Ae2x=e2x, which

gives A=1

For the x2term, substitute yp=Bx2+Cx +Dinto the differential equation

and simplify to get 2B−3(2B) + 2(Bx2+Cx +D) = x2.

This gives 2x2−6x2+ 2Bx2+ 2Cx −6Bx + 2D=x2. Equating coefficients,

we get 2B−6 = 1, so B=7

Step 4: Write the general solution The general solution to the non-

homogeneous differential equation is the sum of the complementary function

and the particular solution: y=yc+yp.

Therefore, the general solution is y=c1ex+c2e2x+1

5e2x+7

2x2+Cx +D,

where c1,c2,C, and Dare constants.