MATH 334 - DIFFERENTIAL EQUATIONS Method of undetermined coefficients Question Bank Set 2

MATH 334 - DIFFERENTIAL

EQUATIONS - Method of

undetermined coefficients

Question Bank - Set 2

Liberty University

Question 1

Question

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

using the method of undetermined coefficients.

Solution

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

to the homogeneous equation y′′

h−2y′

h−3yh= 0.

Step 1: Find the complementary solution The characteristic equation

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

solution is yh=c1e3x+c2e−x.

Step 2: Assume the form of the particular solution Since the right-

hand side of the differential equation is 3e−x, we assume the particular solution

has the form yp=Ae−x.

Step 3: Find the derivatives We have y′

p=−Ae−xand y′′

p=Ae−x.

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

and y′′

pinto the differential equation y′′ −2y′−3y= 3e−x:

Ae−x−2(−Ae−x)−3(Ae−x) = 3e−x

Step 5: Solve for A Simplify the equation: Ae−x+2Ae−x−3Ae−x= 3e−x.

This simplifies to 0 = 3e−x, which is not true for all x.

Step 6: Modify the assumed form Since e−xis also a solution to the

homogeneous equation, we need to modify our assumed form. We multiply by

xto get yp=Axe−x.

Step 7: Find the derivatives Now we have y′

p= (Ax −A)e−xand y′′

(A−2Ax +A)e−x.

Step 8: Substitute into the differential equation Substitute yp,y′

and y′′

pinto the differential equation:

(A−2Ax +A)e−x−2(Ax −A)e−x−3Axe−x= 3e−x

Step 9: Solve for A Simplify the equation and solve for A. After simpli-

fying, you should find A=−1.

Step 10: Write the particular solution The particular solution is yp=

−xe−x.

Step 11: Write the general solution The general solution is the sum of

the complementary and particular solutions:

y=yh+yp=c1e3x+c2e−x−xe−x

Question 2

Question

Find the particular solution to the differential equation y′′ + 2y′−3y= 6x2+ 4

using the method of undetermined coefficients.

Solution

To find the particular solution to the given differential equation, we need to

follow these steps:

Step 1: Find the complementary solution by solving the associated homoge-

neous equation y′′ + 2y′−3y= 0. The characteristic equation is r2+ 2r−3 = 0,

which factors as (r+ 3)(r−1) = 0. So the complementary solution is yc=

c1e−3x+c2ex.

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

side of the given differential equation is a quadratic polynomial, we choose the

particular solution to be of the form yp=Ax2+Bx +C.

Step 3: Substitute ypinto the differential equation and solve for the coeffi-

cients A,B, and C.

y′′

p+ 2y′

p−3yp= 6x2+ 4

y′

p= 2Ax +B,

y′′

p= 2A.

Substitute yp,y′

p, and y′′

pinto the differential equation:

2A+ 2(2Ax +B)−3(Ax2+Bx +C)=6x2+ 4

−3Ax2+ 5Ax −3Bx + 2A+ 2B−3C= 6x2+ 4

Equating coefficients on both sides:

−3A= 6,

5A−3B= 0,

−3B+ 2 = 4.

Solving these equations, we find A=−2, B=−10, and C=−12.

Step 4: Combine the complementary solution and the particular solution

to get the general solution:

y=yc+yp=c1e−3x+c2ex−2x2−10x−12.

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

6x2+ 4 is y=c1e−3x+c2ex−2x2−10x−12.

Question 3

Question

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

undetermined coefficients:

y′′ + 4y= sin(2x)

Solution

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

termined coefficients, we need to find a particular solution and combine it with

the complementary solution of the associated homogeneous equation.

Step 1: Find the complementary solution

The associated homogeneous equation is y′′ + 4y= 0. The characteristic

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

complementary solution is of the form yc(x) = c1cos(2x) + c2sin(2x).

Step 2: Find the particular solution

Since the right-hand side of the non-homogeneous equation is sin(2x), we

assume a particular solution of the form yp(x) = Asin(2x) + Bcos(2x).

Step 3: Find the derivatives

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

y′

p(x)=2Acos(2x)−2Bsin(2x)

y′′

p(x) = −4Asin(2x)−4Bcos(2x)

Step 4: Substitute into the differential equation

Substitute yp(x), y′

p(x), and y′′

p(x) into the non-homogeneous differential

equation:

−4Asin(2x)−4Bcos(2x) + 4(Asin(2x) + Bcos(2x)) = sin(2x)

Step 5: Solve for coefficients

Equating coefficients of like terms gives the following system of equations:

−4A+ 4B= 0

−4B= 1

Solving this system of equations gives A=−1/4 and B=−1/4.

Step 6: Write the general solution

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

y(x) = yc(x) + yp(x) = c1cos(2x) + c2sin(2x)−1

4sin(2x)−1

4cos(2x)

Question 4

Question

Find the particular solution of the nonhomogeneous differential equation:

y′′ −5y′+ 6y= 2e3x

Solution

To find the particular solution of the nonhomogeneous differential equation, we

will use the method of undetermined coefficients.

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

homogeneous equation:

y′′ −5y′+ 6y= 0

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

0. Therefore, the complementary solution is:

yc(x) = c1e2x+c2e3x

Step 2: Next, guess the form of the particular solution. Since the nonho-

mogeneous term is 2e3x, a good guess is:

yp(x) = Ae3x

where Ais a constant to be determined.

Step 3: Substitute yp(x) and its derivatives into the original differential

equation to find A:

yp(x) = Ae3x

y′

p(x)=3Ae3x

y′′

p(x)=9Ae3x

Substitute these into the differential equation:

9Ae3x−5(3Ae3x) + 6(Ae3x)=2e3x

Simplify:

(9A−15A+ 6A)e3x= 2e3x

Solve for A:

0A= 2

This implies that there is no solution for A, so we need to adjust our guess.

Step 4: Adjust the guess for yp(x) to account for the overlap with the

complementary solution:

yp(x) = Ax2e3x

where Ais a new constant to be determined.

Step 5: Substitute yp(x) and its derivatives into the original differential

equation:

yp(x) = Ax2e3x

y′

p(x) = (2Ax + 3Ax2)e3x

y′′

p(x) = (6Ax + 6Ax + 9Ax2)e3x

Substitute these into the differential equation:

(6Ax + 6Ax + 9Ax2)e3x−5(2Ax + 3Ax2)e3x+ 6(Ax2e3x)=2e3x

Simplify:

(18Ax2)e3x= 2e3x

Compare coefficients:

18A= 2

A=1

Step 6: Form the particular solution using the determined value of A:

yp(x) = 1

9x2e3x

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

the particular solution:

y(x) = yc(x) + yp(x) = c1e2x+c2e3x+1

9x2e3x

Therefore, the particular solution of the differential equation is y(x) =

c1e2x+c2e3x+1

9x2e3x.

Question 5

Question

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

the method of undetermined coefficients:

y′′ −y= 2ex+x2.

Solution

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

termined coefficients, we first need to find the general solution to the comple-

mentary equation.

Step 1: Find the Complementary Solution The complementary equa-

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

r= 1 and r=−1. Therefore, the complementary solution is yc=c1ex+c2e−x.

Step 2: Find the Particular Solution Now, let’s find the particular

solution for the non-homogeneous part of the equation. We try a particular

solution of the form:

yp=Aex+Bx2+Cx +D,

where A,B,C, and Dare constants to be determined.

Substitute ypand its derivatives into the differential equation:

y′′

p−yp= 2ex+x2.

Step 3: Substitute ypinto the Differential Equation Substitute yp

and its derivatives into the differential equation. Then, solve for the unknown

coefficients A,B,C, and D.

yp=Aex+Bx2+Cx +D,

y′

p=Aex+ 2Bx +C,

y′′

p=Aex+ 2B.

Substitute yp,y′

p, and y′′

pback into the differential equation:

(Aex+ 2B)−(Aex+Bx2+Cx +D)=2ex+x2.

Step 4: Solve for the Constants Comparing coefficients, we get the

following equations:

A−A= 2,

2B−B= 0,

−C= 0,

−D= 2.

Solving these equations gives A= 2, B= 0, C= 0, and D=−2. Therefore,

the particular solution is yp= 2ex−2.

Step 5: Write the General Solution The general solution is the sum of

the complementary solution and the particular solution:

y=yc+yp=c1ex+c2e−x+ 2ex−2.

Thus, the solution to the differential equation is y=c1ex+c2e−x+ 2ex−2.

Question 6

Question

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

by the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x.

Solution

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

efficients, we first find the complementary solution by solving the associated

homogeneous equation y′′ −3y′+ 2y= 0.

Step 1: Find the complementary solution. The characteristic equation

is r2−3r+ 2 = 0, which factors as (r−1)(r−2) = 0. Thus, we have r1= 1

and r2= 2.

Therefore, the complementary solution is:

yc(x) = c1ex+c2e2x,

where c1and c2are arbitrary constants.

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

has the form yp(x) = Ae2x, since 6e2xis of the same form.

Taking derivatives, we have:

y′

p(x) = 2Ae2xand y′′

p(x)=4Ae2x.

Substitute these derivatives back into the differential equation, we get:

4Ae2x−3(2Ae2x) + 2(Ae2x)=6e2x.

Simplifying, we get A= 2.

Therefore, the particular solution is yp(x) = 2e2x.

Step 3: Find the general solution. The general solution to the non-

homogeneous differential equation is the sum of the complementary solution

and the particular solution:

y(x) = yc(x) + yp(x) = c1ex+c2e2x+ 2e2x.

Thus, the general solution to the given differential equation is:

y(x) = c1ex+ (c2+ 2)e2x,

where c1and c2are arbitrary constants.

Question 7

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 2e2x

Solution

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

we first find the complementary function of the corresponding homogeneous

equation y′′ −4y′+ 4y= 0:

Step 1: Find the complementary function

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

Thus, we have a repeated root r= 2.

Therefore, the complementary function is given by:

yc(x) = c1e2x+c2xe2x

Step 2: Find the particular solution

Since the right-hand side of the given equation is 2e2x, we try the particular

solution in the form:

yp(x) = Ae2x

Now, we differentiate yp(x) to find y′

pand y′′

y′

p= 2Ae2xand y′′

p= 4Ae2x

Substitute yp,y′

p, and y′′

pback into the original equation:

4Ae2x−4(2Ae2x) + 4(Ae2x)=2e2x

Simplifying, we get: 4Ae2x= 2e2x

4A= 2

A=1

So, the particular solution is:

yp(x) = 1

2e2x

Step 3: Write the general solution

Now, the general solution is the sum of the complementary function and par-

ticular solution:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x+1

2e2x

Therefore, the solution to the differential equation is:

y(x) = c1e2x+c2xe2x+1

2e2x

Question 8

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −5y′+ 6y= 20e2x

Solution

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

termined coefficients, we first need to find the complementary function (the

solution to the associated homogeneous equation) and the particular solution.

Step 1: Find the complementary function

The associated homogeneous equation is:

y′′ −5y′+ 6y= 0

The characteristic equation is:

r2−5r+ 6 = 0

Factoring the characteristic equation gives:

(r−2)(r−3) = 0

So, the complementary function is:

yc(x) = c1e2x+c2e3x

Step 2: Find the particular solution

We assume the particular solution has the form:

yp(x) = Ae2x

where Ais a constant.

Substitute yp(x) and its derivatives into the original equation:

y′′

p−5y′

p+ 6yp= 20e2x

(4A−10A+ 6A)e2x= 20e2x

Simplifying gives:

A=−5

So, the particular solution is:

yp(x) = −5e2x

Step 3: Find the general solution

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

particular solution:

y(x) = yc(x) + yp(x)

y(x) = c1e2x+c2e3x−5e2x

y(x)=(c1−5)e2x+c2e3x

Therefore, the general solution to the differential equation is:

y(x)=(c1−5)e2x+c2e3x

Question 9

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 2e2x+ 3x

Solution

We first find the complementary function by solving the associated homogeneous

equation:

y′′

h−4y′

h+ 4yh= 0

The characteristic equation is r2−4r+ 4 = 0, which gives a repeated root of

r= 2. Therefore, the complementary function is given by:

yc= (c1+c2x)e2x

Next, we find the particular solution through the method of undetermined

coefficients as follows:

yp=A1e2x+A2x+A3

where A1e2xaccounts for the term 2e2xand A2x+A3accounts for the term 3x.

Step 1: Calculate the first and second derivatives of yp.

y′

p= 2A1e2x+A2, y′′

p= 4A1e2x

Step 2: Substitute yp,y′

p, and y′′

pback into the original differential equation

and simplify.

4A1e2x−4(2A1e2x+A2) + 4(A1e2x+A2x+A3)=2e2x+ 3x

Step 3: Equate coefficients of e2x,x, and the constant term on both sides

of the equation to find A1,A2, and A3.

A1=1

4, A2=−1

8, A3=3

Therefore, the particular solution is:

yp=1

4e2x−1

8x+3

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

the particular solution.

y=yc+yp= (c1+c2x)e2x+1

4e2x−1

8x+3

Question 10

Question

Solve the differential equation y′′ −4y=e2x+ 3x2using the method of unde-

termined coefficients.

Solution

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

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

(r−2)(r+ 2) = 0. So the solutions are yc=c1e2x+c2e−2x.

Step 2: Assume the particular solution has the form yp=Ae2x+Bx2+

Cx +D. Then, calculate y′

pand y′′

p:y′

p= 2Ae2x+ 2Bx +C y′′

p= 4Ae2x+ 2B

Step 3: Substitute yp,y′

p, and y′′

pinto the original differential equation and

simplify: (4Ae2x+ 2B)−4(Ae2x+Bx2+Cx +D) = e2x+ 3x2

Step 4: Equate coefficients to solve for A,B,C, and D: For e2xterms:

4A−4A= 1 =⇒0 = 1 (Not possible, so adjust yp) For x2terms: 0 −4B=

3 =⇒B=−3

4For xterms: 0 −4C= 0 =⇒C= 0 For constant terms:

0−4D= 0 =⇒D= 0

Step 5: Adjust ypusing the new values of B,C, and D:yp=Ae2x−3

4x2

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

solutions: y=yc+yp=c1e2x+c2e−2x+Ae2x−3

4x2

Question 11

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 12e2x.

Solution

The characteristic equation is r2−4r+ 4 = 0, which has a repeated root of

r= 2. Therefore, the complementary function is yc(x) = c1e2x+c2xe2x.

To find the particular solution, we assume a particular solution of the form

yp(x) = Ae2x. Since 2 is a repeated root of the characteristic equation, we

multiply by xto avoid overlap. Thus, we assume yp(x) = Ax2e2x.

Step 1: Calculate the derivatives of yp(x).

y′

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

y′′

p(x) = (2A+ 4Ax + 2Ax2)e2x

Step 2: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation.

(2A+ 4Ax + 2Ax2)e2x−4(2Ax +Ax2)e2x+ 4Ax2e2x= 12e2x

Step 3: Simplify the equation and solve for A.

2A+ 4Ax + 2Ax2−8Ax −4Ax2+ 4Ax2= 12

2A= 12

A= 6

Step 4: The particular solution is yp(x)=6x2e2x.

Step 5: The general solution is the sum of the complementary function and

the particular solution.

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x+ 6x2e2x

Question 12

Question

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

undetermined coefficients:

y′′ + 4y′+ 4y= 2e−2x+xcos x

Solution

We first find the complementary solution yc(x) by solving the homogeneous part

of the differential equation, which is y′′ + 4y′+ 4y= 0.

Step 1: Find the characteristic equation The characteristic equation

is given by r2+ 4r+ 4 = 0, which simplifies to (r+ 2)2= 0. Thus, we have a

repeated root r=−2.

Step 2: Write down the complementary solution Since we have a

repeated root, the complementary solution will be in the form:

yc(x) = c1e−2x+c2xe−2x

Now, we apply the method of undetermined coefficients to find the particular

solution yp(x).

Step 3: Assume a form for the particular solution We assume the

particular solution has the form:

yp(x) = Ae−2x+Bx cos x+Cx sin x

Step 4: Find the derivatives of yp(x)

y′

p(x) = −2Ae−2x+Bcos x−Bx sin x+Csin x+Cx cos x

y′′

p(x) = 4Ae−2x+ 4Bcos x−2Bsin x−2Csin x+ 2Cx cos x

Step 5: Substitute yp(x)and its derivatives into the differential

equation

4Ae−2x+4Bcos x−2Bsin x−2Csin x+2Cx cos x+4(−2Ae−2x+Bcos x−Bx sin x+Csin x+Cx cos x)+4(Ae−2x+Bx cos x+Cx sin x)=2e−2x+xcos x

Step 6: Solve for unknown coefficients Matching coefficients:

4Ae−2x−4Ae−2x+ 4Ae−2x= 2e−2x=⇒A=1

4Bcos x−2Bsin x+ 4Bcos x=xcos x=⇒B=1

−2Csin x+ 2Cx cos x+ 4Csin x= 0 =⇒C= 0

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

yp(x) = 1

4e−2x+1

4xcos x

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

sum of the complementary and particular solutions:

y(x) = yc(x) + yp(x) = c1e−2x+c2xe−2x+1

4e−2x+1

4xcos x

Question 13

Question

Find the particular solution to the differential equation y′′ −6y′+ 9y= 6x2e3x

using the method of undetermined coefficients.

Solution

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

we first find the complementary solution by solving the associated homogeneous

equation y′′ −6y′+ 9y= 0.

Step 1: Find the complementary solution The characteristic equation

is r2−6r+ 9 = 0, which factors as (r−3)2= 0. So, the complementary solution

is of the form yc=c1e3x+c2xe3x, where c1and c2are constants.

Step 2: Find the particular solution Assume the particular solution

has the form yp=Ax2e3x. Then calculate its first and second derivatives:

y′

p= (2Ax+3Ax2)e3xy′′

p= (2A+6Ax+6Ax+9Ax2)e3x= (9Ax2+12Ax+2A)e3x

Substitute ypand its derivatives into the differential equation: (9Ax2+

12Ax + 2A)e3x−6(2Ax + 3Ax2)e3x+ 9(Ax2)e3x= 6x2e3x

Equating coefficients of like terms gives: (9A−6·2A+ 9A)x2e3x+ (12A−

18A)x+ 2A= 6x2Simplify this equation to solve for A.

Step 3: Solve for ASolving the equation obtained in Step 2, we get:

12A−18A= 0 =⇒ −6A= 0 =⇒A= 0

Step 4: Write down the particular solution Since A= 0, the particular

solution is yp= 0.

Step 5: Write down the general solution The general solution to the

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

y=yc+yp=c1e3x+c2xe3x

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

c1e3x+c2xe3x.

Question 14

Question

Find the particular solution to the following differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 5e2x

Solution

To find the particular solution to the differential equation, we will assume the

particular solution has the form yp=Ae2x, where Ais a constant to be deter-

mined.

Step 1: Find y′

pand y′′

y′

p= 2Ae2x

y′′

p= 4Ae2x

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation

y′′ −4y′+ 4y= 5e2x

4Ae2x−4(2Ae2x) + 4(Ae2x)=5e2x

4Ae2x−8Ae2x+ 4Ae2x= 5e2x

0=5e2x

Step 3: Determine the value of AThe equation 0 = 5e2xis clearly not

true for all values of x, which means our assumption was incorrect. To find the

correct particular solution, we need to include a term of the form Bxe2xwhere

Bis a constant to be determined.

Step 4: Modify our assumption for ypWe will now assume the partic-

ular solution has the form yp=Ae2x+Bxe2x, where Aand Bare constants to

be determined.

Step 5: Find y′

pand y′′

y′

p= 2Ae2x+ 2Be2x+Bxe2x

y′′

p= 4Ae2x+ 4Be2x+ 4Be2x+ 2Be2x

Step 6: Substitute yp,y′

p, and y′′

pinto the differential equation

y′′ −4y′+ 4y= 5e2x

(4Ae2x+ 4Be2x+ 4Be2x+ 2Be2x)−4(2Ae2x+ 2Be2x+Bxe2x) + 4(Ae2x+Bxe2x)=5e2x

Step 7: Solve for Aand BCompare coefficients of e2xand xe2xto

determine the values of Aand B.

Question 15

Question

Find the particular solution to the differential equation: y′′ −4y′+ 4y= 3e2x+

5xe2x, where y(0) = 1 and y′(0) = 2.

Solution

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

method of undetermined coefficients.

Step 1: Guess the form of the particular solution based on the form of the

non-homogeneous terms. We will assume the particular solution has the form:

yp=Ae2x+Bxe2x

where Aand Bare undetermined coefficients to be found.

Step 2: Calculate the first and second derivatives of yp.

y′

p= 2Ae2x+ (B+ 2Bx)e2x

y′′

p= 4Ae2x+ 4Be2x+ 2Be2x+ 2Bxe2x

Step 3: Substitute yp,y′

p, and y′′

pinto the differential equation y′′ −4y′+4y=

3e2x+ 5xe2x.

4Ae2x+ 4Be2x+ 2Be2x+ 2Bxe2x−4(2Ae2x+ 2Be2x+B(1 + 2x)e2x) + 4(Ae2x+Bxe2x)

= 3e2x+ 5xe2x

Step 4: Simplify and equate coefficients.

(2B−8B+ 4B)e2x+ (2B−8B)xe2x= 3e2x+ 5xe2x

Equate coefficients of e2xand xe2xseparately:

−2B= 3 for e2x

−6B= 5 for xe2x

Solving the above two equations gives B=−3

Step 5: Substitute Bback into the equation −2B= 3 to find A.

−2(−3

2)=3

3=3

This is true for all A, so Ais arbitrary.

Step 6: Write the particular solution using the values of Aand B.

yp=Ae2x−3

2xe2x

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

where ycis the complementary function and ypis the particular solution.

Question 16

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 4e2x+ 5x2.

Solution

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

determined coefficients, we need to find the particular solution yp(x) to the

equation.

Step 1: Find the complementary function yc(x) by solving the correspond-

ing homogeneous equation:

y′′ −3y′+ 2y= 0.

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

Therefore, the complementary function is

yc(x) = C1ex+C2e2x,

where C1and C2are constants.

Step 2: Find the particular solution yp(x) by assuming the form of the

solution based on the non-homogeneous terms:

yp(x) = Ae2x+Bx2+Cx +D.

Here, A,B,C, and Dare coefficients to be determined.

Step 3: Substitute the particular solution yp(x) back into the original dif-

ferential equation and solve for the coefficients:

y′′ −3y′+ 2y= 4e2x+ 5x2,

⇒y′′

p−3y′

p+ 2yp= 4e2x+ 5x2,

⇒4A−3(2A)+2Ae2x+ 2B−3(2B+C) + 2(Bx2+Cx +D)=4e2x+ 5x2.

Step 4: Equate like terms and solve the system of equations to find the

values of A,B,C, and D:

2A−6A= 4 ⇒ −4A= 4 ⇒A=−1,

2B−6B= 5 ⇒ −4B= 5 ⇒B=−5

−3(2B+C)=0 ⇒ −3−5

2+C= 0 ⇒C=−5

2D−3C= 0 ⇒2D−3−5

2= 0 ⇒D=15

Step 5: The particular solution is then

yp(x) = −e2x−5

4x2−5

2x+15

Step 6: The general solution to the non-homogeneous differential equation

y(x) = yc(x) + yp(x) = C1ex+C2e2x−e2x−5

4x2−5

2x+15

Question 17

Question

Solve the following nonhomogeneous linear differential equation using the method

of undetermined coefficients:

y′′ −2y′+y= 3ex+ 2x2.

Solution

Step 1: Find the complementary solution ycby solving the associated homoge-

neous equation y′′ −2y′+y= 0. The characteristic equation is r2−2r+ 1 = 0,

which factors as (r−1)2= 0. So, we have a repeated root r= 1. Hence, the

complementary solution is yc=c1ex+c2xex.

Step 2: Find the particular solution ypby guessing a form based on the

nonhomogeneous terms. Since the right-hand side contains terms of the form ex

and x2, we can guess the particular solution of the form yp=Aex+Bx2+Cx+D.

Step 3: Substitute ypinto the differential equation and solve for the coeffi-

cients A,B,C, and D.

yp=Aex+Bx2+Cx +D

y′

p=Aex+ 2Bx +C

y′′

p=Aex+ 2B

Substitute these into the differential equation:

(Aex+ 2B)−2(Aex+ 2Bx +C)+(Aex+Bx2+Cx +D)=3ex+ 2x2.

Step 4: Collect like terms and solve for A,B,C, and D. Matching coeffi-

cients of ex,x2, and constant terms on both sides, we get:

A−2A+A= 3 ⇒A= 3

−2B−2B= 2 ⇒B=−1

−2C= 0 ⇒C= 0

2D= 0 ⇒D= 0

Thus, the particular solution is yp= 3ex−1

2x2.

Step 5: The general solution is the sum of the complementary and particular

solutions. Therefore, the general solution to the given differential equation is:

y=yc+yp=c1ex+c2xex+ 3ex−1

2x2.

Question 18

Question

Find the general solution to the following non-homogeneous differential equa-

tion:

y′′ −4y′+ 4y= 2e2x+ 3x

where y(0) = 1 and y′(0) = 0.

Solution

Step 1: First, we find the general solution to the corresponding homogeneous

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

gives a repeated root r= 2. So the homogeneous solution is yh= (c1+c2x)e2x.

Step 2: Next, we find a particular solution to the non-homogeneous equation

using the method of undetermined coefficients. Assume a particular solution of

the form yp=Ae2x+Bx +C. Substitute this into the differential equation and

solve for A,B, and C.

Step 3: Differentiate yptwice to find y′

pand y′′

y′

p= 2Ae2x+Band y′′

p= 4Ae2x.

Step 4: Substitute yp,y′

p, and y′′

pback into the non-homogeneous equation.

(4A−4B+ 4(Ae2x+Bx +C)) −4(2Ae2x+B) + 4(Ae2x+Bx +C) = 2e2x+ 3x.

Step 5: Simplify the equation and solve for A,B, and Cto get the particular

solution.

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

solutions.

y=yh+yp

Question 19

Question

Consider the differential equation y′′ −4y′+ 4y= 2e2x. Solve the differential

equation using the method of undetermined coefficients.

Solution

Step 1: 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. This gives us a repeated root r= 2, so the complementary

solution is yc=C1e2x+C2xe2x.

Step 2: Assume the particular solution has the form yp=Ae2x.

Step 3: Calculate the derivatives of ypand substitute into the differential

equation to find the values of A.

Substitute ypand its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 2e2x

4Ae2x−8Ae2x+ 4Ae2x= 2e2x

−4Ae2x= 2e2x

Step 4: Solve for A.

A=−1

Step 5: The particular solution is yp=−1

2e2x.

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

solutions.

y=yc+yp=C1e2x+C2xe2x−1

2e2x

Therefore, the solution to the differential equation y′′ −4y′+4y= 2e2xusing

the method of undetermined coefficients is y=C1e2x+C2xe2x−1

2e2x.

Question 20

Question

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

using the method of undetermined coefficients.

Solution

Step 1: First, we find the characteristic equation of the homogeneous equation.

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

So, the homogeneous solution is yh(x)=(c1+c2x)ex.

Step 2: Now, we guess a particular solution of the form yp(x) = Aexcos x+

Bexsin xfor the non-homogeneous equation.

Step 3: Find y′

p(x) and y′′

p(x):

y′

p(x) = Aexcos x−Aexsin x+Bexsin x+Bexcos x

y′′

p(x) = Aexcos x−2Aexsin x−Aexsin x+Aexcos x

+Bexsin x+ 2Bexcos x−Bexcos x+Bexsin x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation

and simplify:

−2Aexsin x+ 2Bexcos x=excos x

By comparing coefficients, we find A=−1

2and B= 0.

Step 5: The particular solution is yp(x) = −1

2excos x.

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

solutions:

y(x)=(c1+c2x)ex−1

2excos x

Therefore, the particular solution to the given differential equation is y(x) =

(c1+c2x)ex−1

2excos x.

Question 21

Question

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

the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary function by solving the homogeneous

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

factors as (r−2)2= 0, giving a repeated root r= 2. Hence, the complementary

function is of the form yc=c1e2x+c2xe2x.

Step 2: Now, we guess the form of the particular solution for the nonho-

mogeneous equation y′′ −4y′+ 4y=e2x. Since the right side of the equation is

e2x, we try a particular solution of the form yp=Ae2x, where Ais a constant

to be determined.

Step 3: Compute the derivatives of yp:

y′

p= 2Ae2x, y′′

p= 4Ae2x

Step 4: Substitute ypand its derivatives back into the differential equation:

4Ae2x−4(2Ae2x) + 4(Ae2x) = e2x

Step 5: Simplify the equation:

4Ae2x−8Ae2x+ 4Ae2x=e2x⇒0 = e2x

Step 6: Since 0 =e2x, there is no solution for A. To resolve this, we

multiply the guess by x:yp=Axe2x.

Step 7: Compute the derivatives of the new guess:

y′

p= (Ax +A)e2x, y′′

p= (Ax + 2A)e2x

Step 8: Substitute the new guess and its derivatives back into the differential

equation:

(Ax + 2A)e2x−4(Ax +A)e2x+ 4Axe2x=e2x

Step 9: Simplify the equation:

Ae2x=e2x

Step 10: Comparing coefficients, we find that A= 1. Therefore, the par-

ticular solution is yp=xe2x.

Step 11: The general solution is the sum of the complementary function

and the particular solution:

y=yc+yp=c1e2x+c2xe2x+xe2x

Question 22

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −3y′+ 2y= 8e4x.

Solution

To solve the given nonhomogeneous differential equation, we will first find

the complementary function, then determine the particular solution using the

method of undetermined coefficients, and finally combine both solutions.

Step 1: Find the complementary function The complementary func-

tion yc(x) satisfies the associated homogeneous differential equation, y′′ −3y′+

2y= 0. We assume the solution has the form yc(x) = emx and substitute into

the homogeneous equation:

Plugging yc(x) = emx into y′′ −3y′+ 2y= 0 gives:

(m2−3m+ 2)emx = 0.

The characteristic equation is m2−3m+2 = 0, which factors to (m−1)(m−

2) = 0. Thus, m1= 1 and m2= 2.

Therefore, the complementary function is:

yc(x) = C1ex+C2e2x,

where C1and C2are constants.

Step 2: Find the particular solution We seek a particular solution in the

form yp(x) = Ae4x, since 8e4xmatches the form of the forcing term. Substitute

yp(x) = Ae4xinto the differential equation y′′ −3y′+ 2y= 8e4x:

Differentiating yp(x) twice and substituting back into the differential equa-

tion gives:

(16A−12A+ 2A)e4x= 8e4x.

Solving 6A= 8 ⇒A=4

Therefore, the particular solution is:

yp(x) = 4

3e4x.

Step 3: Combine solutions The general solution is the sum of the com-

plementary function and the particular solution:

y(x) = yc(x) + yp(x) = C1ex+C2e2x+4

3e4x,

where C1and C2are constants that can be determined from initial conditions.

Question 23

Question

Find the particular solution to the differential equation y′′ −4y′+ 4y=xe2x

using the method of undetermined coefficients.

Solution

To find the particular solution to the differential equation y′′ −4y′+ 4y=xe2x,

we will first find the complementary solution and then the particular solution.

Step 1: Find the complementary solution The characteristic equation

for the given homogeneous differential equation y′′ −4y′+4y= 0 is r2−4r+4 = 0.

This equation has a repeated root of r= 2.

Thus, the complementary solution is yc(x)=(c1+c2x)e2x, where c1and c2

are arbitrary constants.

Step 2: Find the particular solution Assume the particular solution

has the form yp(x) = Ax2e2x. We then find the first and second derivatives:

y′

p(x) = (2Ax + 2A)e2xand y′′

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

Substitute these into the differential equation to find A: (2A+ 4A)x+

4A)e2x−4((2Ax+2A)e2x)+4(Ax2e2x) = xe2x. This simplifies to 4Ae2x=xe2x,

so solving for Agives A=1

Therefore, the particular solution is yp(x) = 1

4x2e2x.

Step 3: Find the general solution The general solution to the differential

equation is the sum of the complementary solution and the particular solution:

y(x) = yc(x) + yp(x) = (c1+c2x)e2x+1

4x2e2x, where c1and c2are constants

to be determined by initial conditions, if provided.

Question 24

Question

Find a particular solution to the differential equation y′′ −4y′+ 4y= 5e2x−

4x2+ 3x−2 using the method of undetermined coefficients.

Solution

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

assume the form of the particular solution based on the terms on the right-hand

side of the equation.

Step 1: Find the particular solution for ypto the homogeneous equation

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

repeated root r= 2. Therefore, the solution to yhis yh=C1e2x+C2xe2x.

Step 2: Evaluate the terms on the right-hand side: 5e2x,−4x2, 3x, and −2.

We assume the form of ypto be:

yp=Ae2x+Bx2+Cx +D

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

solve for the coefficients A,B,C, and D.

y′′

p−4y′

p+ 4yp= 5e2x−4x2+ 3x−2

Step 4: Calculate the derivatives:

y′

p= 2Ae2x+ 2Bx +C

y′′

p= 4Ae2x+ 2B

Step 5: Substitute yp,y′

p, and y′′

pinto the differential equation and simplify.

(4Ae2x+ 2B)−4(2Ae2x+ 2Bx +C) + 4(Ae2x+Bx2+Cx +D)

= 5e2x−4x2+ 3x−2

Step 6: Match coefficients of like terms and solve the resulting system of

equations to find A,B,C, and D.

Step 7: Substitute the values of A,B,C, and Dback into ypto obtain the

particular solution.

Therefore, the particular solution to the given differential equation is

yp=5

2e2x−4x2+ 3x−9

Question 25

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x+ 5x2.

Solution

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

efficients, we first find the complementary function by solving the associated

homogeneous equation:

y′′ −3y′+ 2y= 0.

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

0, giving us roots r1= 1 and r2= 2. Therefore, the complementary function is

yc(x) = c1ex+c2e2x,

where c1and c2are arbitrary constants.

Next, we find the particular solution for the nonhomogeneous equation. We

guess the form of the particular solution to be:

yp(x) = Ae2x+Bx2+Cx +D.

We then find the derivatives needed to substitute into the original differential

equation:

y′

p(x)=2Ae2x+ 2Bx +C,

y′′

p(x)=4Ae2x+ 2B.

Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential equation:

(4Ae2x+ 2B)−3(2Ae2x+ 2Bx +C) + 2(Ae2x+Bx2+Cx +D)=6e2x+ 5x2.

This simplifies to the following equation:

2Ae2x+Bx2+ (2A−2B+C)x+ (4A−3C+ 2D)=6e2x+ 5x2.

Matching coefficients, we get the system of equations:

2A= 6,

B= 5,

2A−2B+C= 0,

4A−3C+ 2D= 0.

Solving this system of equations gives A= 3, B= 5, C= 7, and D= 9.

Therefore, the particular solution is:

yp(x)=3e2x+ 5x2+ 7x+ 9.

Thus, the general solution to the nonhomogeneous differential equation is:

y(x) = yc(x) + yp(x) = c1ex+c2e2x+ 3e2x+ 5x2+ 7x+ 9.

Question 26

Question

Find the particular solution of the nonhomogeneous differential equation: y′′ −

y= sin(x) + ex, given that the complementary solution is yc=c1cos(x) +

c2sin(x).

Solution

To find the particular solution of the nonhomogeneous differential equation using

the method of undetermined coefficients, we assume the particular solution has

the form yp=Asin(x) + Bex.

Step 1: Calculate the first and second derivatives:

yp=Asin(x) + Bex

y′

p=Acos(x) + Bex

y′′

p=−Asin(x) + Bex

Step 2: Substitute into the differential equation:

−Asin(x) + Bex−(Asin(x) + Bex) = sin(x) + ex

Step 3: Equate coefficients and solve for A and B:

−2A= sin(x) (coefficient of sin(x) on both sides)

2B=ex(coefficient of exon both sides)

⇒A=−1

2sin(x), B =1

2ex

Step 4: Write the particular solution:

yp=−1

2sin(x) + 1

2ex

Step 5: Write the general solution:

y=yc+yp=c1cos(x) + c2sin(x)−1

2sin(x) + 1

2ex

Therefore, the particular solution of the nonhomogeneous differential equa-

tion is y=c1cos(x) + c2sin(x)−1

2sin(x) + 1

2ex.

Question 27

Question

Use the method of undetermined coefficients to find a particular solution to the

nonhomogeneous differential equation:

y′′ −4y′+ 4y= 12e2x.

Solution

To find a particular solution to the nonhomogeneous differential equation, we

assume a particular solution of the form yp=Ae2x, where Ais a constant to

be determined.

Step 1: Find ypand its derivatives

yp=Ae2x, y′

p= 2Ae2x, y′′

p= 4Ae2x

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

tion Substitute ypand its derivatives into the differential equation:

y′′ −4y′+ 4y= 12e2x

4Ae2x−4(2Ae2x) + 4(Ae2x) = 12e2x

Step 3: Simplify the equation and solve for A

4Ae2x−8Ae2x+ 4Ae2x= 12e2x

0 = 12e2x

Since there is no solution to this equation, we must try a different form for

the particular solution.

Step 4: Try a different form for ypSince our assumed form Ae2xis a

solution to the homogeneous equation, we need to multiply by xto get linearly

independent solutions. Let’s try yp=Axe2x.

Step 5: Find ypand its derivatives

yp=Axe2x, y′

p= (2Axe2x+Ae2x), y′′

p= (4Axe2x+ 4Ae2x+ 2Ae2x)

Step 6: Substitute ypand its derivatives into the differential equa-

tion Substitute ypand its derivatives into the differential equation:

(4Axe2x+ 4Ae2x+ 2Ae2x)−4(2Axe2x+Ae2x)+4Axe2x= 12e2x

Step 7: Simplify the equation and solve for A

4Axe2x+ 4Ae2x+ 2Ae2x−8Axe2x−4Ae2x+ 4Axe2x= 12e2x

2Ae2x= 12e2x

A= 6

Therefore, a particular solution to the differential equation is yp= 6xe2x.

Question 28

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 3e2x+ 2x2

Solution

Step 1: First, we find the general solution to the corresponding homogeneous

equation:

y′′ −4y′+ 4y= 0

The characteristic equation is r2−4r+ 4 = 0, which factors as (r−2)2= 0.

Thus, we have a repeated root r= 2, and the solution to the homogeneous

equation is of the form:

yh(x) = c1e2x+c2xe2x

Step 2: Now, we find a particular solution yp(x) to the non-homogeneous

equation. For the term 3e2x, we try a particular solution of the form Ae2x, and

for the term 2x2, we try a particular solution of the form Bx2+Cx +D.

Step 3: Substitute yp(x) = Ae2x+Bx2+Cx +Dback into the original

differential equation:

(Ae2x)′′ −4(Ae2x)′+ 4(Ae2x)+2x2= 3e2x+ 2x2

Step 4: Compute the derivatives and plug them back into the equation:

4Ae2x−4Ae2x+ 4Ae2x+ 2x2= 3e2x+ 2x2

Step 5: Equate like terms:

2x2= 2x2=⇒there are no restrictions on the values of A, B, C, and D

Step 6: Therefore, a particular solution is yp(x) = Ae2x+Bx2+Cx +D.

Combining the particular and homogeneous solutions, the general solution to

the non-homogeneous equation is:

y(x) = yh(x) + yp(x) = c1e2x+c2xe2x+Ae2x+Bx2+Cx +D

Question 29

Question

Consider the differential equation y′′ −5y′+ 6y=e2x. Find the particular

solution using the method of undetermined coefficients.

Solution

To find the particular solution to the given differential equation using the

method of undetermined coefficients, we first need to find the complementary

solution by solving the associated homogeneous equation y′′ −5y′+ 6y= 0.

Step 1: Find the complementary solution. The characteristic equation

is r2−5r+ 6 = (r−2)(r−3) = 0. Thus, the complementary solution is given

by yc=c1e2x+c2e3x.

Step 2: Assume the form of the particular solution. Since the right-

hand side is e2x, and 2 is the root of the characteristic equation with multiplicity

1, we assume the particular solution has the form yp=Ae2x.

Step 3: Find the derivatives of yp.We have yp=Ae2x,y′

p= 2Ae2x,

and y′′

p= 4Ae2x.

Step 4: Substitute back into the differential equation. Substitute

yp,y′

p, and y′′

pback into the differential equation y′′ −5y′+ 6y=e2x:

4Ae2x−5(2Ae2x) + 6(Ae2x) = e2x

Step 5: Solve for the unknown coefficient. Solving the equation above,

we get:

A(4 −10 + 6)e2x=e2x=⇒A=1

Therefore, the particular solution is yp=1

2e2x.

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=c1e2x+c2e3x+1

2e2x

where c1and c2are arbitrary constants.

Question 30

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 6xe2x

Solution

Step 1: First, find the general solution to the associated homogeneous equation:

y′′

h−4y′

h+ 4yh= 0

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

Thus, the characteristic root is r= 2 with multiplicity 2. Therefore, the general

solution to the associated homogeneous equation is:

yh(x) = c1e2x+c2xe2x

Step 2: Next, find a particular solution for the non-homogeneous equation

by assuming:

yp(x)=(ax +b)e2x

Then find y′

pand y′′

y′

p=ae2x+ 2axe2x+be2x

y′′

p= 2ae2x+ 2ae2x+ 4axe2x+ 2be2x

Substitute yp,y′

p, and y′′

pinto the original differential equation:

(2ae2x+2ae2x+4axe2x+2be2x)−4(ae2x+2axe2x+be2x)+4(ax+b)e2x= 6xe2x

This simplifies to:

2ae2x+ 4axe2x+ 2be2x= 6xe2x

Step 3: Equate coefficients to solve for aand b:

2a= 0

4a= 6

2b= 0

Solving these equations gives a= 3 and b= 0.

Step 4: Find the particular solution:

yp(x) = (3x)e2x

Step 5: The general solution to the non-homogeneous equation is the sum of

the general solution to the associated homogeneous equation and the particular

solution:

y(x) = yh(x) + yp(x) = c1e2x+c2xe2x+ 3xe2x

So, the solution to the differential equation y′′ −4y′+ 4y= 6xe2xis y(x) =

c1e2x+c2xe2x+ 3xe2x.

Question 31

Question

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

undetermined coefficients:

y′′ −6y′+ 9y= 2e3x.

Solution

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

termined coefficients, we will first find the general solution to the corresponding

homogeneous equation, then determine the particular solution.

Step 1: Find the general solution to the homogeneous equation

The characteristic equation for the homogeneous equation y′′ −6y′+ 9y= 0 is

r2−6r+ 9 = 0, which factors to (r−3)2= 0. Thus, the characteristic roots

are r= 3 (a repeated root).

Therefore, the general solution to the homogeneous equation is

yh(x) = (c1+c2x)e3x,

where c1and c2are arbitrary constants.

Step 2: Guess the particular solution Since the non-homogeneous term

is of the form aebx, we guess the particular solution to be of the form yp(x) =

Ae3x, where Ais a coefficient to be determined.

Step 3: Find the derivatives of the particular solution We have

yp(x) = Ae3x, so the first and second derivatives are

y′

p(x) = 3Ae3xand y′′

p(x)=9Ae3x.

Step 4: Substitute the particular solution into the differential

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

9Ae3x−6(3Ae3x) + 9(Ae3x)=2e3x.

Step 5: Solve for the unknown coefficient Solving the equation 9Ae3x−

18Ae3x+ 9Ae3x= 2e3xgives A=1

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

is yp(x) = 1

9e3x.

Step 7: Write the general solution The general solution to the non-

homogeneous differential equation is

y(x) = yh(x) + yp(x)=(c1+c2x)e3x+1

9e3x,

where c1and c2are arbitrary constants.

Question 32

Question

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

undetermined coefficients:

y′′ −4y′+ 4y= 6e2x.

Solution

1. First, we solve the homogeneous equation y′′ −4y′+ 4y= 0 to find the

complementary function. The characteristic equation is r2−4r+ 4 = 0,

which factors as (r−2)2= 0. So, the complementary function is yc=

c1e2x+c2xe2x.

2. Next, we need to find a particular solution to the non-homogeneous equa-

tion. We propose a particular solution of the form yp=Ae2x, where Ais

a constant to be determined.

3. Substitute ypand its derivatives back into the differential equation:

y′

p= 2Ae2x,

y′′

p= 4Ae2x.

Substituting into the original equation gives:

4Ae2x−4(2Ae2x) + 4(Ae2x)=6e2x.

4. Simplifying the equation gives 4Ae2x−8Ae2x+4Ae2x= 6e2x. This reduces

to 0 = 6e2x. This equation has no solution, so we need to adjust our

particular solution.

5. Since e2xis already present in the homogeneous solution, we try a new

particular solution of the form yp=Ax2e2x, where Ais a new constant

to be determined.

6. Calculate the first and second derivatives of ypand substitute them back

into the original differential equation.

7. Solving for the value of Aby substituting ypinto the original differential

equation gives A= 3/2. Therefore, the particular solution is yp=3

2x2e2x.

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

sum of the complementary function and the particular solution, i.e.,

y=yc+yp=c1e2x+c2xe2x+3

2x2e2x.

Question 33

Question

Find the particular solution to the differential equation y′′ −4y′+ 4y= 2xe2x

using the method of undetermined coefficients.

Solution

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

we will assume that the particular solution has the form:

yp(x) = (Ax2+Bx +C)e2x

Step 1: Find the derivatives

y′

p(x) = (2Ax +B)e2x+ (Ax2+Bx +C)2e2x

y′′

p(x)=2Ae2x+ 2Axe2x+ 2Be2x+ 2Ax2e2x+ 2Be2x+ 2Ce2x

= (2A+ 2B)e2x+ (4A+ 2B)xe2x+ 2Cx2e2x

Step 2: Substitute derivatives into the differential equation Substi-

tute yp(x), y′

p(x), and y′′

p(x) into the differential equation:

y′′

p−4y′

p+4yp= (2A+2B)e2x+(4A+2B)xe2x+2Cx2e2x−4((2A+2B)e2x+(4A+2B)xe2x+2Cx2e2x)+4((Ax2+Bx+C)e2x)

Step 3: Solve for coefficients Match the terms on both sides of the

equation and solve for the coefficients A,B, and C. From the constant terms,

we get C=1

From the coefficient of e2x, we get 2A+ 2B−4(2A+ 2B) + 4A= 0 =⇒

−6A−6B= 0 =⇒A=−B.

From the coefficient of xe2x, we get 4A+ 2B−4(4A+ 2B) + 4B= 2 =⇒

−14A−6B= 2 =⇒A=−1

14 .

Therefore, A=−1

14 ,B=1

14 , and C=1

Step 4: Write the particular solution The particular solution is:

yp(x) = −1

14x2+1

14x+1

4e2x

Question 34

Question

Solve the following second order linear nonhomogeneous differential equation

using the method of undetermined coefficients:

y′′ −3y′+ 2y= 3ex+ 2x.

Solution

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

ficients, we will first find the general solution to the corresponding homogeneous

equation, and then find a particular solution to the nonhomogeneous equation.

Step 1: Find the general solution to the homogeneous equation

The homogeneous equation corresponding to the given differential equation

is:

y′′ −3y′+ 2y= 0.

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

0. Thus, the roots are r= 1 and r= 2.

Therefore, the general solution to the homogeneous equation is:

yh(x) = c1ex+c2e2x,

where c1and c2are arbitrary constants.

Step 2: Find a particular solution to the nonhomogeneous equation

For the nonhomogeneous part 3ex+2x, we make an educated guess based on

the form of this term. We hypothesize that a particular solution has the form:

yp(x) = Aex+Bx +C.

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

y′

p(x) = Aex+B,

y′′

p(x) = Aex.

Substitute yp(x), y′

p(x), and y′′

p(x) back into the differential equation:

−Aex+ 3(Aex+B) + 2(Aex+Bx +C)=3ex+ 2x.

By comparing coefficients, we can find A= 1, B=−2, and C= 0.

Therefore, a particular solution to the nonhomogeneous equation is:

yp(x) = ex−2x.

Step 3: Find the general solution

The general solution to the nonhomogeneous equation is the sum of the

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

nonhomogeneous equation:

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

y(x) = c1ex+c2e2x+ex−2x,

where c1and c2are arbitrary constants.

Question 35

Question

Find the particular solution of the nonhomogeneous differential equation

y′′ −3y′+ 2y= 6xex.

Solution

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

we assume the particular solution has the form yp=Ax2ex+Bxex.

yp=Ax2ex+Bxex,

y′

p= (2Ax +Aex+B)ex,

y′′

p= (2A+ 2Aex+ 3Bex).

(1)

Step 7: Find the derivatives Now we have y′

p= (Ax −A)e−xand y′′

(A−2Ax +A)e−x.

Step 8: Substitute into the differential equation Substitute yp,y′

and y′′

pinto the differential equation:

(A−2Ax +A)e−x−2(Ax −A)e−x−3Axe−x= 3e−x

Step 9: Solve for A Simplify the equation and solve for A. After simpli-

fying, you should find A=−1.

Step 10: Write the particular solution The particular solution is yp=

−xe−x.

Step 11: Write the general solution The general solution is the sum of

the complementary and particular solutions:

y=yh+yp=c1e3x+c2e−x−xe−x

Question 2

Question

Find the particular solution to the differential equation y′′ + 2y′−3y= 6x2+ 4

using the method of undetermined coefficients.

Solution

To find the particular solution to the given differential equation, we need to

follow these steps:

Step 1: Find the complementary solution by solving the associated homoge-

neous equation y′′ + 2y′−3y= 0. The characteristic equation is r2+ 2r−3 = 0,

which factors as (r+ 3)(r−1) = 0. So the complementary solution is yc=

c1e−3x+c2ex.

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

side of the given differential equation is a quadratic polynomial, we choose the

particular solution to be of the form yp=Ax2+Bx +C.

Step 3: Substitute ypinto the differential equation and solve for the coeffi-

cients A,B, and C.

y′′

p+ 2y′

p−3yp= 6x2+ 4

y′

p= 2Ax +B,

y′′

p= 2A.

Substitute yp,y′

p, and y′′

pinto the differential equation:

2A+ 2(2Ax +B)−3(Ax2+Bx +C)=6x2+ 4

−3Ax2+ 5Ax −3Bx + 2A+ 2B−3C= 6x2+ 4

Equating coefficients on both sides:

−3A= 6,

5A−3B= 0,

−3B+ 2 = 4.

Solving these equations, we find A=−2, B=−10, and C=−12.

Step 4: Combine the complementary solution and the particular solution

to get the general solution:

y=yc+yp=c1e−3x+c2ex−2x2−10x−12.

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

6x2+ 4 is y=c1e−3x+c2ex−2x2−10x−12.

Question 3

Question

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

undetermined coefficients:

y′′ + 4y= sin(2x)

Solution

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

termined coefficients, we need to find a particular solution and combine it with

the complementary solution of the associated homogeneous equation.

Step 1: Find the complementary solution

The associated homogeneous equation is y′′ + 4y= 0. The characteristic

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

complementary solution is of the form yc(x) = c1cos(2x) + c2sin(2x).

Step 2: Find the particular solution

Since the right-hand side of the non-homogeneous equation is sin(2x), we

assume a particular solution of the form yp(x) = Asin(2x) + Bcos(2x).

Step 3: Find the derivatives

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

y′

p(x)=2Acos(2x)−2Bsin(2x)

y′′

p(x) = −4Asin(2x)−4Bcos(2x)

Step 4: Substitute into the differential equation

Substitute yp(x), y′

p(x), and y′′

p(x) into the non-homogeneous differential

equation:

−4Asin(2x)−4Bcos(2x) + 4(Asin(2x) + Bcos(2x)) = sin(2x)

Step 5: Solve for coefficients

Equating coefficients of like terms gives the following system of equations:

−4A+ 4B= 0

−4B= 1

Solving this system of equations gives A=−1/4 and B=−1/4.

Step 6: Write the general solution

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

y(x) = yc(x) + yp(x) = c1cos(2x) + c2sin(2x)−1

4sin(2x)−1

4cos(2x)

Question 4

Question

Find the particular solution of the nonhomogeneous differential equation:

y′′ −5y′+ 6y= 2e3x

Solution

To find the particular solution of the nonhomogeneous differential equation, we

will use the method of undetermined coefficients.

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

homogeneous equation:

y′′ −5y′+ 6y= 0

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

0. Therefore, the complementary solution is:

yc(x) = c1e2x+c2e3x

Step 2: Next, guess the form of the particular solution. Since the nonho-

mogeneous term is 2e3x, a good guess is:

yp(x) = Ae3x

where Ais a constant to be determined.

Step 3: Substitute yp(x) and its derivatives into the original differential

equation to find A:

yp(x) = Ae3x

y′

p(x)=3Ae3x

y′′

p(x)=9Ae3x

Substitute these into the differential equation:

9Ae3x−5(3Ae3x) + 6(Ae3x)=2e3x

Simplify:

(9A−15A+ 6A)e3x= 2e3x

Solve for A:

0A= 2

This implies that there is no solution for A, so we need to adjust our guess.

Step 4: Adjust the guess for yp(x) to account for the overlap with the

complementary solution:

yp(x) = Ax2e3x

where Ais a new constant to be determined.

Step 5: Substitute yp(x) and its derivatives into the original differential

equation:

yp(x) = Ax2e3x

y′

p(x) = (2Ax + 3Ax2)e3x

y′′

p(x) = (6Ax + 6Ax + 9Ax2)e3x

Substitute these into the differential equation:

(6Ax + 6Ax + 9Ax2)e3x−5(2Ax + 3Ax2)e3x+ 6(Ax2e3x)=2e3x

Simplify:

(18Ax2)e3x= 2e3x

Compare coefficients:

18A= 2

A=1

Step 6: Form the particular solution using the determined value of A:

yp(x) = 1

9x2e3x

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

the particular solution:

y(x) = yc(x) + yp(x) = c1e2x+c2e3x+1

9x2e3x

Therefore, the particular solution of the differential equation is y(x) =

c1e2x+c2e3x+1

9x2e3x.

Question 5

Question

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

the method of undetermined coefficients:

y′′ −y= 2ex+x2.

Solution

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

termined coefficients, we first need to find the general solution to the comple-

mentary equation.

Step 1: Find the Complementary Solution The complementary equa-

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

r= 1 and r=−1. Therefore, the complementary solution is yc=c1ex+c2e−x.

Step 2: Find the Particular Solution Now, let’s find the particular

solution for the non-homogeneous part of the equation. We try a particular

solution of the form:

yp=Aex+Bx2+Cx +D,

where A,B,C, and Dare constants to be determined.

Substitute ypand its derivatives into the differential equation:

y′′

p−yp= 2ex+x2.

Step 3: Substitute ypinto the Differential Equation Substitute yp

and its derivatives into the differential equation. Then, solve for the unknown

coefficients A,B,C, and D.

yp=Aex+Bx2+Cx +D,

y′

p=Aex+ 2Bx +C,

y′′

p=Aex+ 2B.

Substitute yp,y′

p, and y′′

pback into the differential equation:

(Aex+ 2B)−(Aex+Bx2+Cx +D)=2ex+x2.

Step 4: Solve for the Constants Comparing coefficients, we get the

following equations:

A−A= 2,

2B−B= 0,

−C= 0,

−D= 2.

Solving these equations gives A= 2, B= 0, C= 0, and D=−2. Therefore,

the particular solution is yp= 2ex−2.

Step 5: Write the General Solution The general solution is the sum of

the complementary solution and the particular solution:

y=yc+yp=c1ex+c2e−x+ 2ex−2.

Thus, the solution to the differential equation is y=c1ex+c2e−x+ 2ex−2.

Question 6

Question

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

by the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x.

Solution

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

efficients, we first find the complementary solution by solving the associated

homogeneous equation y′′ −3y′+ 2y= 0.

Step 1: Find the complementary solution. The characteristic equation

is r2−3r+ 2 = 0, which factors as (r−1)(r−2) = 0. Thus, we have r1= 1

and r2= 2.

Therefore, the complementary solution is:

yc(x) = c1ex+c2e2x,

where c1and c2are arbitrary constants.

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

has the form yp(x) = Ae2x, since 6e2xis of the same form.

Taking derivatives, we have:

y′

p(x) = 2Ae2xand y′′

p(x)=4Ae2x.

Substitute these derivatives back into the differential equation, we get:

4Ae2x−3(2Ae2x) + 2(Ae2x)=6e2x.

Simplifying, we get A= 2.

Therefore, the particular solution is yp(x) = 2e2x.

Step 3: Find the general solution. The general solution to the non-

homogeneous differential equation is the sum of the complementary solution

and the particular solution:

y(x) = yc(x) + yp(x) = c1ex+c2e2x+ 2e2x.

Thus, the general solution to the given differential equation is:

y(x) = c1ex+ (c2+ 2)e2x,

where c1and c2are arbitrary constants.

Question 7

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 2e2x

Solution

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

we first find the complementary function of the corresponding homogeneous

equation y′′ −4y′+ 4y= 0:

Step 1: Find the complementary function

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

Thus, we have a repeated root r= 2.

Therefore, the complementary function is given by:

yc(x) = c1e2x+c2xe2x

Step 2: Find the particular solution

Since the right-hand side of the given equation is 2e2x, we try the particular

solution in the form:

yp(x) = Ae2x

Now, we differentiate yp(x) to find y′

pand y′′

y′

p= 2Ae2xand y′′

p= 4Ae2x

Substitute yp,y′

p, and y′′

pback into the original equation:

4Ae2x−4(2Ae2x) + 4(Ae2x)=2e2x

Simplifying, we get: 4Ae2x= 2e2x

4A= 2

A=1

So, the particular solution is:

yp(x) = 1

2e2x

Step 3: Write the general solution

Now, the general solution is the sum of the complementary function and par-

ticular solution:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x+1

2e2x

Therefore, the solution to the differential equation is:

y(x) = c1e2x+c2xe2x+1

2e2x

Question 8

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −5y′+ 6y= 20e2x

Solution

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

termined coefficients, we first need to find the complementary function (the

solution to the associated homogeneous equation) and the particular solution.

Step 1: Find the complementary function

The associated homogeneous equation is:

y′′ −5y′+ 6y= 0

The characteristic equation is:

r2−5r+ 6 = 0

Factoring the characteristic equation gives:

(r−2)(r−3) = 0

So, the complementary function is:

yc(x) = c1e2x+c2e3x

Step 2: Find the particular solution

We assume the particular solution has the form:

yp(x) = Ae2x

where Ais a constant.

Substitute yp(x) and its derivatives into the original equation:

y′′

p−5y′

p+ 6yp= 20e2x

(4A−10A+ 6A)e2x= 20e2x

Simplifying gives:

A=−5

So, the particular solution is:

yp(x) = −5e2x

Step 3: Find the general solution

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

particular solution:

y(x) = yc(x) + yp(x)

y(x) = c1e2x+c2e3x−5e2x

y(x)=(c1−5)e2x+c2e3x

Therefore, the general solution to the differential equation is:

y(x)=(c1−5)e2x+c2e3x

Question 9

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 2e2x+ 3x

Solution

We first find the complementary function by solving the associated homogeneous

equation:

y′′

h−4y′

h+ 4yh= 0

The characteristic equation is r2−4r+ 4 = 0, which gives a repeated root of

r= 2. Therefore, the complementary function is given by:

yc= (c1+c2x)e2x

Next, we find the particular solution through the method of undetermined

coefficients as follows:

yp=A1e2x+A2x+A3

where A1e2xaccounts for the term 2e2xand A2x+A3accounts for the term 3x.

Step 1: Calculate the first and second derivatives of yp.

y′

p= 2A1e2x+A2, y′′

p= 4A1e2x

Step 2: Substitute yp,y′

p, and y′′

pback into the original differential equation

and simplify.

4A1e2x−4(2A1e2x+A2) + 4(A1e2x+A2x+A3)=2e2x+ 3x

Step 3: Equate coefficients of e2x,x, and the constant term on both sides

of the equation to find A1,A2, and A3.

A1=1

4, A2=−1

8, A3=3

Therefore, the particular solution is:

yp=1

4e2x−1

8x+3

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

the particular solution.

y=yc+yp= (c1+c2x)e2x+1

4e2x−1

8x+3

Question 10

Question

Solve the differential equation y′′ −4y=e2x+ 3x2using the method of unde-

termined coefficients.

Solution

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

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

(r−2)(r+ 2) = 0. So the solutions are yc=c1e2x+c2e−2x.

Step 2: Assume the particular solution has the form yp=Ae2x+Bx2+

Cx +D. Then, calculate y′

pand y′′

p:y′

p= 2Ae2x+ 2Bx +C y′′

p= 4Ae2x+ 2B

Step 3: Substitute yp,y′

p, and y′′

pinto the original differential equation and

simplify: (4Ae2x+ 2B)−4(Ae2x+Bx2+Cx +D) = e2x+ 3x2

Step 4: Equate coefficients to solve for A,B,C, and D: For e2xterms:

4A−4A= 1 =⇒0 = 1 (Not possible, so adjust yp) For x2terms: 0 −4B=

3 =⇒B=−3

4For xterms: 0 −4C= 0 =⇒C= 0 For constant terms:

0−4D= 0 =⇒D= 0

Step 5: Adjust ypusing the new values of B,C, and D:yp=Ae2x−3

4x2

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

solutions: y=yc+yp=c1e2x+c2e−2x+Ae2x−3

4x2

Question 11

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 12e2x.

Solution

The characteristic equation is r2−4r+ 4 = 0, which has a repeated root of

r= 2. Therefore, the complementary function is yc(x) = c1e2x+c2xe2x.

To find the particular solution, we assume a particular solution of the form

yp(x) = Ae2x. Since 2 is a repeated root of the characteristic equation, we

multiply by xto avoid overlap. Thus, we assume yp(x) = Ax2e2x.

Step 1: Calculate the derivatives of yp(x).

y′

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

y′′

p(x) = (2A+ 4Ax + 2Ax2)e2x

Step 2: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation.

(2A+ 4Ax + 2Ax2)e2x−4(2Ax +Ax2)e2x+ 4Ax2e2x= 12e2x

Step 3: Simplify the equation and solve for A.

2A+ 4Ax + 2Ax2−8Ax −4Ax2+ 4Ax2= 12

2A= 12

A= 6

Step 4: The particular solution is yp(x)=6x2e2x.

Step 5: The general solution is the sum of the complementary function and

the particular solution.

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x+ 6x2e2x

Question 12

Question

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

undetermined coefficients:

y′′ + 4y′+ 4y= 2e−2x+xcos x

Solution

We first find the complementary solution yc(x) by solving the homogeneous part

of the differential equation, which is y′′ + 4y′+ 4y= 0.

Step 1: Find the characteristic equation The characteristic equation

is given by r2+ 4r+ 4 = 0, which simplifies to (r+ 2)2= 0. Thus, we have a

repeated root r=−2.

Step 2: Write down the complementary solution Since we have a

repeated root, the complementary solution will be in the form:

yc(x) = c1e−2x+c2xe−2x

Now, we apply the method of undetermined coefficients to find the particular

solution yp(x).

Step 3: Assume a form for the particular solution We assume the

particular solution has the form:

yp(x) = Ae−2x+Bx cos x+Cx sin x

Step 4: Find the derivatives of yp(x)

y′

p(x) = −2Ae−2x+Bcos x−Bx sin x+Csin x+Cx cos x

y′′

p(x) = 4Ae−2x+ 4Bcos x−2Bsin x−2Csin x+ 2Cx cos x

Step 5: Substitute yp(x)and its derivatives into the differential

equation

4Ae−2x+4Bcos x−2Bsin x−2Csin x+2Cx cos x+4(−2Ae−2x+Bcos x−Bx sin x+Csin x+Cx cos x)+4(Ae−2x+Bx cos x+Cx sin x)=2e−2x+xcos x

Step 6: Solve for unknown coefficients Matching coefficients:

4Ae−2x−4Ae−2x+ 4Ae−2x= 2e−2x=⇒A=1

4Bcos x−2Bsin x+ 4Bcos x=xcos x=⇒B=1

−2Csin x+ 2Cx cos x+ 4Csin x= 0 =⇒C= 0

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

yp(x) = 1

4e−2x+1

4xcos x

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

sum of the complementary and particular solutions:

y(x) = yc(x) + yp(x) = c1e−2x+c2xe−2x+1

4e−2x+1

4xcos x

Question 13

Question

Find the particular solution to the differential equation y′′ −6y′+ 9y= 6x2e3x

using the method of undetermined coefficients.

Solution

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

we first find the complementary solution by solving the associated homogeneous

equation y′′ −6y′+ 9y= 0.

Step 1: Find the complementary solution The characteristic equation

is r2−6r+ 9 = 0, which factors as (r−3)2= 0. So, the complementary solution

is of the form yc=c1e3x+c2xe3x, where c1and c2are constants.

Step 2: Find the particular solution Assume the particular solution

has the form yp=Ax2e3x. Then calculate its first and second derivatives:

y′

p= (2Ax+3Ax2)e3xy′′

p= (2A+6Ax+6Ax+9Ax2)e3x= (9Ax2+12Ax+2A)e3x

Substitute ypand its derivatives into the differential equation: (9Ax2+

12Ax + 2A)e3x−6(2Ax + 3Ax2)e3x+ 9(Ax2)e3x= 6x2e3x

Equating coefficients of like terms gives: (9A−6·2A+ 9A)x2e3x+ (12A−

18A)x+ 2A= 6x2Simplify this equation to solve for A.

Step 3: Solve for ASolving the equation obtained in Step 2, we get:

12A−18A= 0 =⇒ −6A= 0 =⇒A= 0

Step 4: Write down the particular solution Since A= 0, the particular

solution is yp= 0.

Step 5: Write down the general solution The general solution to the

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

y=yc+yp=c1e3x+c2xe3x

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

c1e3x+c2xe3x.

Question 14

Question

Find the particular solution to the following differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 5e2x

Solution

To find the particular solution to the differential equation, we will assume the

particular solution has the form yp=Ae2x, where Ais a constant to be deter-

mined.

Step 1: Find y′

pand y′′

y′

p= 2Ae2x

y′′

p= 4Ae2x

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation

y′′ −4y′+ 4y= 5e2x

4Ae2x−4(2Ae2x) + 4(Ae2x)=5e2x

4Ae2x−8Ae2x+ 4Ae2x= 5e2x

0=5e2x

Step 3: Determine the value of AThe equation 0 = 5e2xis clearly not

true for all values of x, which means our assumption was incorrect. To find the

correct particular solution, we need to include a term of the form Bxe2xwhere

Bis a constant to be determined.

Step 4: Modify our assumption for ypWe will now assume the partic-

ular solution has the form yp=Ae2x+Bxe2x, where Aand Bare constants to

be determined.

Step 5: Find y′

pand y′′

y′

p= 2Ae2x+ 2Be2x+Bxe2x

y′′

p= 4Ae2x+ 4Be2x+ 4Be2x+ 2Be2x

Step 6: Substitute yp,y′

p, and y′′

pinto the differential equation

y′′ −4y′+ 4y= 5e2x

(4Ae2x+ 4Be2x+ 4Be2x+ 2Be2x)−4(2Ae2x+ 2Be2x+Bxe2x) + 4(Ae2x+Bxe2x)=5e2x

Step 7: Solve for Aand BCompare coefficients of e2xand xe2xto

determine the values of Aand B.

Question 15

Question

Find the particular solution to the differential equation: y′′ −4y′+ 4y= 3e2x+

5xe2x, where y(0) = 1 and y′(0) = 2.

Solution

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

method of undetermined coefficients.

Step 1: Guess the form of the particular solution based on the form of the

non-homogeneous terms. We will assume the particular solution has the form:

yp=Ae2x+Bxe2x

where Aand Bare undetermined coefficients to be found.

Step 2: Calculate the first and second derivatives of yp.

y′

p= 2Ae2x+ (B+ 2Bx)e2x

y′′

p= 4Ae2x+ 4Be2x+ 2Be2x+ 2Bxe2x

Step 3: Substitute yp,y′

p, and y′′

pinto the differential equation y′′ −4y′+4y=

3e2x+ 5xe2x.

4Ae2x+ 4Be2x+ 2Be2x+ 2Bxe2x−4(2Ae2x+ 2Be2x+B(1 + 2x)e2x) + 4(Ae2x+Bxe2x)

= 3e2x+ 5xe2x

Step 4: Simplify and equate coefficients.

(2B−8B+ 4B)e2x+ (2B−8B)xe2x= 3e2x+ 5xe2x

Equate coefficients of e2xand xe2xseparately:

−2B= 3 for e2x

−6B= 5 for xe2x

Solving the above two equations gives B=−3

Step 5: Substitute Bback into the equation −2B= 3 to find A.

−2(−3

2)=3

3=3

This is true for all A, so Ais arbitrary.

Step 6: Write the particular solution using the values of Aand B.

yp=Ae2x−3

2xe2x

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

where ycis the complementary function and ypis the particular solution.

Question 16

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 4e2x+ 5x2.

Solution

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

determined coefficients, we need to find the particular solution yp(x) to the

equation.

Step 1: Find the complementary function yc(x) by solving the correspond-

ing homogeneous equation:

y′′ −3y′+ 2y= 0.

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

Therefore, the complementary function is

yc(x) = C1ex+C2e2x,

where C1and C2are constants.

Step 2: Find the particular solution yp(x) by assuming the form of the

solution based on the non-homogeneous terms:

yp(x) = Ae2x+Bx2+Cx +D.

Here, A,B,C, and Dare coefficients to be determined.

Step 3: Substitute the particular solution yp(x) back into the original dif-

ferential equation and solve for the coefficients:

y′′ −3y′+ 2y= 4e2x+ 5x2,

⇒y′′

p−3y′

p+ 2yp= 4e2x+ 5x2,

⇒4A−3(2A)+2Ae2x+ 2B−3(2B+C) + 2(Bx2+Cx +D)=4e2x+ 5x2.

Step 4: Equate like terms and solve the system of equations to find the

values of A,B,C, and D:

2A−6A= 4 ⇒ −4A= 4 ⇒A=−1,

2B−6B= 5 ⇒ −4B= 5 ⇒B=−5

−3(2B+C)=0 ⇒ −3−5

2+C= 0 ⇒C=−5

2D−3C= 0 ⇒2D−3−5

2= 0 ⇒D=15

Step 5: The particular solution is then

yp(x) = −e2x−5

4x2−5

2x+15

Step 6: The general solution to the non-homogeneous differential equation

y(x) = yc(x) + yp(x) = C1ex+C2e2x−e2x−5

4x2−5

2x+15

Question 17

Question

Solve the following nonhomogeneous linear differential equation using the method

of undetermined coefficients:

y′′ −2y′+y= 3ex+ 2x2.

Solution

Step 1: Find the complementary solution ycby solving the associated homoge-

neous equation y′′ −2y′+y= 0. The characteristic equation is r2−2r+ 1 = 0,

which factors as (r−1)2= 0. So, we have a repeated root r= 1. Hence, the

complementary solution is yc=c1ex+c2xex.

Step 2: Find the particular solution ypby guessing a form based on the

nonhomogeneous terms. Since the right-hand side contains terms of the form ex

and x2, we can guess the particular solution of the form yp=Aex+Bx2+Cx+D.

Step 3: Substitute ypinto the differential equation and solve for the coeffi-

cients A,B,C, and D.

yp=Aex+Bx2+Cx +D

y′

p=Aex+ 2Bx +C

y′′

p=Aex+ 2B

Substitute these into the differential equation:

(Aex+ 2B)−2(Aex+ 2Bx +C)+(Aex+Bx2+Cx +D)=3ex+ 2x2.

Step 4: Collect like terms and solve for A,B,C, and D. Matching coeffi-

cients of ex,x2, and constant terms on both sides, we get:

A−2A+A= 3 ⇒A= 3

−2B−2B= 2 ⇒B=−1

−2C= 0 ⇒C= 0

2D= 0 ⇒D= 0

Thus, the particular solution is yp= 3ex−1

2x2.

Step 5: The general solution is the sum of the complementary and particular

solutions. Therefore, the general solution to the given differential equation is:

y=yc+yp=c1ex+c2xex+ 3ex−1

2x2.

Question 18

Question

Find the general solution to the following non-homogeneous differential equa-

tion:

y′′ −4y′+ 4y= 2e2x+ 3x

where y(0) = 1 and y′(0) = 0.

Solution

Step 1: First, we find the general solution to the corresponding homogeneous

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

gives a repeated root r= 2. So the homogeneous solution is yh= (c1+c2x)e2x.

Step 2: Next, we find a particular solution to the non-homogeneous equation

using the method of undetermined coefficients. Assume a particular solution of

the form yp=Ae2x+Bx +C. Substitute this into the differential equation and

solve for A,B, and C.

Step 3: Differentiate yptwice to find y′

pand y′′

y′

p= 2Ae2x+Band y′′

p= 4Ae2x.

Step 4: Substitute yp,y′

p, and y′′

pback into the non-homogeneous equation.

(4A−4B+ 4(Ae2x+Bx +C)) −4(2Ae2x+B) + 4(Ae2x+Bx +C) = 2e2x+ 3x.

Step 5: Simplify the equation and solve for A,B, and Cto get the particular

solution.

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

solutions.

y=yh+yp

Question 19

Question

Consider the differential equation y′′ −4y′+ 4y= 2e2x. Solve the differential

equation using the method of undetermined coefficients.

Solution

Step 1: 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. This gives us a repeated root r= 2, so the complementary

solution is yc=C1e2x+C2xe2x.

Step 2: Assume the particular solution has the form yp=Ae2x.

Step 3: Calculate the derivatives of ypand substitute into the differential

equation to find the values of A.

Substitute ypand its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 2e2x

4Ae2x−8Ae2x+ 4Ae2x= 2e2x

−4Ae2x= 2e2x

Step 4: Solve for A.

A=−1

Step 5: The particular solution is yp=−1

2e2x.

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

solutions.

y=yc+yp=C1e2x+C2xe2x−1

2e2x

Therefore, the solution to the differential equation y′′ −4y′+4y= 2e2xusing

the method of undetermined coefficients is y=C1e2x+C2xe2x−1

2e2x.

Question 20

Question

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

using the method of undetermined coefficients.

Solution

Step 1: First, we find the characteristic equation of the homogeneous equation.

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

So, the homogeneous solution is yh(x)=(c1+c2x)ex.

Step 2: Now, we guess a particular solution of the form yp(x) = Aexcos x+

Bexsin xfor the non-homogeneous equation.

Step 3: Find y′

p(x) and y′′

p(x):

y′

p(x) = Aexcos x−Aexsin x+Bexsin x+Bexcos x

y′′

p(x) = Aexcos x−2Aexsin x−Aexsin x+Aexcos x

+Bexsin x+ 2Bexcos x−Bexcos x+Bexsin x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation

and simplify:

−2Aexsin x+ 2Bexcos x=excos x

By comparing coefficients, we find A=−1

2and B= 0.

Step 5: The particular solution is yp(x) = −1

2excos x.

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

solutions:

y(x)=(c1+c2x)ex−1

2excos x

Therefore, the particular solution to the given differential equation is y(x) =

(c1+c2x)ex−1

2excos x.

Question 21

Question

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

the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary function by solving the homogeneous

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

factors as (r−2)2= 0, giving a repeated root r= 2. Hence, the complementary

function is of the form yc=c1e2x+c2xe2x.

Step 2: Now, we guess the form of the particular solution for the nonho-

mogeneous equation y′′ −4y′+ 4y=e2x. Since the right side of the equation is

e2x, we try a particular solution of the form yp=Ae2x, where Ais a constant

to be determined.

Step 3: Compute the derivatives of yp:

y′

p= 2Ae2x, y′′

p= 4Ae2x

Step 4: Substitute ypand its derivatives back into the differential equation:

4Ae2x−4(2Ae2x) + 4(Ae2x) = e2x

Step 5: Simplify the equation:

4Ae2x−8Ae2x+ 4Ae2x=e2x⇒0 = e2x

Step 6: Since 0 =e2x, there is no solution for A. To resolve this, we

multiply the guess by x:yp=Axe2x.

Step 7: Compute the derivatives of the new guess:

y′

p= (Ax +A)e2x, y′′

p= (Ax + 2A)e2x

Step 8: Substitute the new guess and its derivatives back into the differential

equation:

(Ax + 2A)e2x−4(Ax +A)e2x+ 4Axe2x=e2x

Step 9: Simplify the equation:

Ae2x=e2x

Step 10: Comparing coefficients, we find that A= 1. Therefore, the par-

ticular solution is yp=xe2x.

Step 11: The general solution is the sum of the complementary function

and the particular solution:

y=yc+yp=c1e2x+c2xe2x+xe2x

Question 22

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −3y′+ 2y= 8e4x.

Solution

To solve the given nonhomogeneous differential equation, we will first find

the complementary function, then determine the particular solution using the

method of undetermined coefficients, and finally combine both solutions.

Step 1: Find the complementary function The complementary func-

tion yc(x) satisfies the associated homogeneous differential equation, y′′ −3y′+

2y= 0. We assume the solution has the form yc(x) = emx and substitute into

the homogeneous equation:

Plugging yc(x) = emx into y′′ −3y′+ 2y= 0 gives:

(m2−3m+ 2)emx = 0.

The characteristic equation is m2−3m+2 = 0, which factors to (m−1)(m−

2) = 0. Thus, m1= 1 and m2= 2.

Therefore, the complementary function is:

yc(x) = C1ex+C2e2x,

where C1and C2are constants.

Step 2: Find the particular solution We seek a particular solution in the

form yp(x) = Ae4x, since 8e4xmatches the form of the forcing term. Substitute

yp(x) = Ae4xinto the differential equation y′′ −3y′+ 2y= 8e4x:

Differentiating yp(x) twice and substituting back into the differential equa-

tion gives:

(16A−12A+ 2A)e4x= 8e4x.

Solving 6A= 8 ⇒A=4

Therefore, the particular solution is:

yp(x) = 4

3e4x.

Step 3: Combine solutions The general solution is the sum of the com-

plementary function and the particular solution:

y(x) = yc(x) + yp(x) = C1ex+C2e2x+4

3e4x,

where C1and C2are constants that can be determined from initial conditions.

Question 23

Question

Find the particular solution to the differential equation y′′ −4y′+ 4y=xe2x

using the method of undetermined coefficients.

Solution

To find the particular solution to the differential equation y′′ −4y′+ 4y=xe2x,

we will first find the complementary solution and then the particular solution.

Step 1: Find the complementary solution The characteristic equation

for the given homogeneous differential equation y′′ −4y′+4y= 0 is r2−4r+4 = 0.

This equation has a repeated root of r= 2.

Thus, the complementary solution is yc(x)=(c1+c2x)e2x, where c1and c2

are arbitrary constants.

Step 2: Find the particular solution Assume the particular solution

has the form yp(x) = Ax2e2x. We then find the first and second derivatives:

y′

p(x) = (2Ax + 2A)e2xand y′′

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

Substitute these into the differential equation to find A: (2A+ 4A)x+

4A)e2x−4((2Ax+2A)e2x)+4(Ax2e2x) = xe2x. This simplifies to 4Ae2x=xe2x,

so solving for Agives A=1

Therefore, the particular solution is yp(x) = 1

4x2e2x.

Step 3: Find the general solution The general solution to the differential

equation is the sum of the complementary solution and the particular solution:

y(x) = yc(x) + yp(x) = (c1+c2x)e2x+1

4x2e2x, where c1and c2are constants

to be determined by initial conditions, if provided.

Question 24

Question

Find a particular solution to the differential equation y′′ −4y′+ 4y= 5e2x−

4x2+ 3x−2 using the method of undetermined coefficients.

Solution

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

assume the form of the particular solution based on the terms on the right-hand

side of the equation.

Step 1: Find the particular solution for ypto the homogeneous equation

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

repeated root r= 2. Therefore, the solution to yhis yh=C1e2x+C2xe2x.

Step 2: Evaluate the terms on the right-hand side: 5e2x,−4x2, 3x, and −2.

We assume the form of ypto be:

yp=Ae2x+Bx2+Cx +D

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

solve for the coefficients A,B,C, and D.

y′′

p−4y′

p+ 4yp= 5e2x−4x2+ 3x−2

Step 4: Calculate the derivatives:

y′

p= 2Ae2x+ 2Bx +C

y′′

p= 4Ae2x+ 2B

Step 5: Substitute yp,y′

p, and y′′

pinto the differential equation and simplify.

(4Ae2x+ 2B)−4(2Ae2x+ 2Bx +C) + 4(Ae2x+Bx2+Cx +D)

= 5e2x−4x2+ 3x−2

Step 6: Match coefficients of like terms and solve the resulting system of

equations to find A,B,C, and D.

Step 7: Substitute the values of A,B,C, and Dback into ypto obtain the

particular solution.

Therefore, the particular solution to the given differential equation is

yp=5

2e2x−4x2+ 3x−9

Question 25

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x+ 5x2.

Solution

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

efficients, we first find the complementary function by solving the associated

homogeneous equation:

y′′ −3y′+ 2y= 0.

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

0, giving us roots r1= 1 and r2= 2. Therefore, the complementary function is

yc(x) = c1ex+c2e2x,

where c1and c2are arbitrary constants.

Next, we find the particular solution for the nonhomogeneous equation. We

guess the form of the particular solution to be:

yp(x) = Ae2x+Bx2+Cx +D.

We then find the derivatives needed to substitute into the original differential

equation:

y′

p(x)=2Ae2x+ 2Bx +C,

y′′

p(x)=4Ae2x+ 2B.

Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential equation:

(4Ae2x+ 2B)−3(2Ae2x+ 2Bx +C) + 2(Ae2x+Bx2+Cx +D)=6e2x+ 5x2.

This simplifies to the following equation:

2Ae2x+Bx2+ (2A−2B+C)x+ (4A−3C+ 2D)=6e2x+ 5x2.

Matching coefficients, we get the system of equations:

2A= 6,

B= 5,

2A−2B+C= 0,

4A−3C+ 2D= 0.

Solving this system of equations gives A= 3, B= 5, C= 7, and D= 9.

Therefore, the particular solution is:

yp(x)=3e2x+ 5x2+ 7x+ 9.

Thus, the general solution to the nonhomogeneous differential equation is:

y(x) = yc(x) + yp(x) = c1ex+c2e2x+ 3e2x+ 5x2+ 7x+ 9.

Question 26

Question

Find the particular solution of the nonhomogeneous differential equation: y′′ −

y= sin(x) + ex, given that the complementary solution is yc=c1cos(x) +

c2sin(x).

Solution

To find the particular solution of the nonhomogeneous differential equation using

the method of undetermined coefficients, we assume the particular solution has

the form yp=Asin(x) + Bex.

Step 1: Calculate the first and second derivatives:

yp=Asin(x) + Bex

y′

p=Acos(x) + Bex

y′′

p=−Asin(x) + Bex

Step 2: Substitute into the differential equation:

−Asin(x) + Bex−(Asin(x) + Bex) = sin(x) + ex

Step 3: Equate coefficients and solve for A and B:

−2A= sin(x) (coefficient of sin(x) on both sides)

2B=ex(coefficient of exon both sides)

⇒A=−1

2sin(x), B =1

2ex

Step 4: Write the particular solution:

yp=−1

2sin(x) + 1

2ex

Step 5: Write the general solution:

y=yc+yp=c1cos(x) + c2sin(x)−1

2sin(x) + 1

2ex

Therefore, the particular solution of the nonhomogeneous differential equa-

tion is y=c1cos(x) + c2sin(x)−1

2sin(x) + 1

2ex.

Question 27

Question

Use the method of undetermined coefficients to find a particular solution to the

nonhomogeneous differential equation:

y′′ −4y′+ 4y= 12e2x.

Solution

To find a particular solution to the nonhomogeneous differential equation, we

assume a particular solution of the form yp=Ae2x, where Ais a constant to

be determined.

Step 1: Find ypand its derivatives

yp=Ae2x, y′

p= 2Ae2x, y′′

p= 4Ae2x

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

tion Substitute ypand its derivatives into the differential equation:

y′′ −4y′+ 4y= 12e2x

4Ae2x−4(2Ae2x) + 4(Ae2x) = 12e2x

Step 3: Simplify the equation and solve for A

4Ae2x−8Ae2x+ 4Ae2x= 12e2x

0 = 12e2x

Since there is no solution to this equation, we must try a different form for

the particular solution.

Step 4: Try a different form for ypSince our assumed form Ae2xis a

solution to the homogeneous equation, we need to multiply by xto get linearly

independent solutions. Let’s try yp=Axe2x.

Step 5: Find ypand its derivatives

yp=Axe2x, y′

p= (2Axe2x+Ae2x), y′′

p= (4Axe2x+ 4Ae2x+ 2Ae2x)

Step 6: Substitute ypand its derivatives into the differential equa-

tion Substitute ypand its derivatives into the differential equation:

(4Axe2x+ 4Ae2x+ 2Ae2x)−4(2Axe2x+Ae2x)+4Axe2x= 12e2x

Step 7: Simplify the equation and solve for A

4Axe2x+ 4Ae2x+ 2Ae2x−8Axe2x−4Ae2x+ 4Axe2x= 12e2x

2Ae2x= 12e2x

A= 6

Therefore, a particular solution to the differential equation is yp= 6xe2x.

Question 28

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 3e2x+ 2x2

Solution

Step 1: First, we find the general solution to the corresponding homogeneous

equation:

y′′ −4y′+ 4y= 0

The characteristic equation is r2−4r+ 4 = 0, which factors as (r−2)2= 0.

Thus, we have a repeated root r= 2, and the solution to the homogeneous

equation is of the form:

yh(x) = c1e2x+c2xe2x

Step 2: Now, we find a particular solution yp(x) to the non-homogeneous

equation. For the term 3e2x, we try a particular solution of the form Ae2x, and

for the term 2x2, we try a particular solution of the form Bx2+Cx +D.

Step 3: Substitute yp(x) = Ae2x+Bx2+Cx +Dback into the original

differential equation:

(Ae2x)′′ −4(Ae2x)′+ 4(Ae2x)+2x2= 3e2x+ 2x2

Step 4: Compute the derivatives and plug them back into the equation:

4Ae2x−4Ae2x+ 4Ae2x+ 2x2= 3e2x+ 2x2

Step 5: Equate like terms:

2x2= 2x2=⇒there are no restrictions on the values of A, B, C, and D

Step 6: Therefore, a particular solution is yp(x) = Ae2x+Bx2+Cx +D.

Combining the particular and homogeneous solutions, the general solution to

the non-homogeneous equation is:

y(x) = yh(x) + yp(x) = c1e2x+c2xe2x+Ae2x+Bx2+Cx +D

Question 29

Question

Consider the differential equation y′′ −5y′+ 6y=e2x. Find the particular

solution using the method of undetermined coefficients.

Solution

To find the particular solution to the given differential equation using the

method of undetermined coefficients, we first need to find the complementary

solution by solving the associated homogeneous equation y′′ −5y′+ 6y= 0.

Step 1: Find the complementary solution. The characteristic equation

is r2−5r+ 6 = (r−2)(r−3) = 0. Thus, the complementary solution is given

by yc=c1e2x+c2e3x.

Step 2: Assume the form of the particular solution. Since the right-

hand side is e2x, and 2 is the root of the characteristic equation with multiplicity

1, we assume the particular solution has the form yp=Ae2x.

Step 3: Find the derivatives of yp.We have yp=Ae2x,y′

p= 2Ae2x,

and y′′

p= 4Ae2x.

Step 4: Substitute back into the differential equation. Substitute

yp,y′

p, and y′′

pback into the differential equation y′′ −5y′+ 6y=e2x:

4Ae2x−5(2Ae2x) + 6(Ae2x) = e2x

Step 5: Solve for the unknown coefficient. Solving the equation above,

we get:

A(4 −10 + 6)e2x=e2x=⇒A=1

Therefore, the particular solution is yp=1

2e2x.

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=c1e2x+c2e3x+1

2e2x

where c1and c2are arbitrary constants.

Question 30

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 6xe2x

Solution

Step 1: First, find the general solution to the associated homogeneous equation:

y′′

h−4y′

h+ 4yh= 0

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

Thus, the characteristic root is r= 2 with multiplicity 2. Therefore, the general

solution to the associated homogeneous equation is:

yh(x) = c1e2x+c2xe2x

Step 2: Next, find a particular solution for the non-homogeneous equation

by assuming:

yp(x)=(ax +b)e2x

Then find y′

pand y′′

y′

p=ae2x+ 2axe2x+be2x

y′′

p= 2ae2x+ 2ae2x+ 4axe2x+ 2be2x

Substitute yp,y′

p, and y′′

pinto the original differential equation:

(2ae2x+2ae2x+4axe2x+2be2x)−4(ae2x+2axe2x+be2x)+4(ax+b)e2x= 6xe2x

This simplifies to:

2ae2x+ 4axe2x+ 2be2x= 6xe2x

Step 3: Equate coefficients to solve for aand b:

2a= 0

4a= 6

2b= 0

Solving these equations gives a= 3 and b= 0.

Step 4: Find the particular solution:

yp(x) = (3x)e2x

Step 5: The general solution to the non-homogeneous equation is the sum of

the general solution to the associated homogeneous equation and the particular

solution:

y(x) = yh(x) + yp(x) = c1e2x+c2xe2x+ 3xe2x

So, the solution to the differential equation y′′ −4y′+ 4y= 6xe2xis y(x) =

c1e2x+c2xe2x+ 3xe2x.

Question 31

Question

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

undetermined coefficients:

y′′ −6y′+ 9y= 2e3x.

Solution

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

termined coefficients, we will first find the general solution to the corresponding

homogeneous equation, then determine the particular solution.

Step 1: Find the general solution to the homogeneous equation

The characteristic equation for the homogeneous equation y′′ −6y′+ 9y= 0 is

r2−6r+ 9 = 0, which factors to (r−3)2= 0. Thus, the characteristic roots

are r= 3 (a repeated root).

Therefore, the general solution to the homogeneous equation is

yh(x) = (c1+c2x)e3x,

where c1and c2are arbitrary constants.

Step 2: Guess the particular solution Since the non-homogeneous term

is of the form aebx, we guess the particular solution to be of the form yp(x) =

Ae3x, where Ais a coefficient to be determined.

Step 3: Find the derivatives of the particular solution We have

yp(x) = Ae3x, so the first and second derivatives are

y′

p(x) = 3Ae3xand y′′

p(x)=9Ae3x.

Step 4: Substitute the particular solution into the differential

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

9Ae3x−6(3Ae3x) + 9(Ae3x)=2e3x.

Step 5: Solve for the unknown coefficient Solving the equation 9Ae3x−

18Ae3x+ 9Ae3x= 2e3xgives A=1

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

is yp(x) = 1

9e3x.

Step 7: Write the general solution The general solution to the non-

homogeneous differential equation is

y(x) = yh(x) + yp(x)=(c1+c2x)e3x+1

9e3x,

where c1and c2are arbitrary constants.

Question 32

Question

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

undetermined coefficients:

y′′ −4y′+ 4y= 6e2x.

Solution

1. First, we solve the homogeneous equation y′′ −4y′+ 4y= 0 to find the

complementary function. The characteristic equation is r2−4r+ 4 = 0,

which factors as (r−2)2= 0. So, the complementary function is yc=

c1e2x+c2xe2x.

2. Next, we need to find a particular solution to the non-homogeneous equa-

tion. We propose a particular solution of the form yp=Ae2x, where Ais

a constant to be determined.

3. Substitute ypand its derivatives back into the differential equation:

y′

p= 2Ae2x,

y′′

p= 4Ae2x.

Substituting into the original equation gives:

4Ae2x−4(2Ae2x) + 4(Ae2x)=6e2x.

4. Simplifying the equation gives 4Ae2x−8Ae2x+4Ae2x= 6e2x. This reduces

to 0 = 6e2x. This equation has no solution, so we need to adjust our

particular solution.

5. Since e2xis already present in the homogeneous solution, we try a new

particular solution of the form yp=Ax2e2x, where Ais a new constant

to be determined.

6. Calculate the first and second derivatives of ypand substitute them back

into the original differential equation.

7. Solving for the value of Aby substituting ypinto the original differential

equation gives A= 3/2. Therefore, the particular solution is yp=3

2x2e2x.

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

sum of the complementary function and the particular solution, i.e.,

y=yc+yp=c1e2x+c2xe2x+3

2x2e2x.

Question 33

Question

Find the particular solution to the differential equation y′′ −4y′+ 4y= 2xe2x

using the method of undetermined coefficients.

Solution

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

we will assume that the particular solution has the form:

yp(x) = (Ax2+Bx +C)e2x

Step 1: Find the derivatives

y′

p(x) = (2Ax +B)e2x+ (Ax2+Bx +C)2e2x

y′′

p(x)=2Ae2x+ 2Axe2x+ 2Be2x+ 2Ax2e2x+ 2Be2x+ 2Ce2x

= (2A+ 2B)e2x+ (4A+ 2B)xe2x+ 2Cx2e2x

Step 2: Substitute derivatives into the differential equation Substi-

tute yp(x), y′

p(x), and y′′

p(x) into the differential equation:

y′′

p−4y′

p+4yp= (2A+2B)e2x+(4A+2B)xe2x+2Cx2e2x−4((2A+2B)e2x+(4A+2B)xe2x+2Cx2e2x)+4((Ax2+Bx+C)e2x)

Step 3: Solve for coefficients Match the terms on both sides of the

equation and solve for the coefficients A,B, and C. From the constant terms,

we get C=1

From the coefficient of e2x, we get 2A+ 2B−4(2A+ 2B) + 4A= 0 =⇒

−6A−6B= 0 =⇒A=−B.

From the coefficient of xe2x, we get 4A+ 2B−4(4A+ 2B) + 4B= 2 =⇒

−14A−6B= 2 =⇒A=−1

14 .

Therefore, A=−1

14 ,B=1

14 , and C=1

Step 4: Write the particular solution The particular solution is:

yp(x) = −1

14x2+1

14x+1

4e2x

Question 34

Question

Solve the following second order linear nonhomogeneous differential equation

using the method of undetermined coefficients:

y′′ −3y′+ 2y= 3ex+ 2x.

Solution

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

ficients, we will first find the general solution to the corresponding homogeneous

equation, and then find a particular solution to the nonhomogeneous equation.

Step 1: Find the general solution to the homogeneous equation

The homogeneous equation corresponding to the given differential equation

is:

y′′ −3y′+ 2y= 0.

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

0. Thus, the roots are r= 1 and r= 2.

Therefore, the general solution to the homogeneous equation is:

yh(x) = c1ex+c2e2x,

where c1and c2are arbitrary constants.

Step 2: Find a particular solution to the nonhomogeneous equation

For the nonhomogeneous part 3ex+2x, we make an educated guess based on

the form of this term. We hypothesize that a particular solution has the form:

yp(x) = Aex+Bx +C.

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

y′

p(x) = Aex+B,

y′′

p(x) = Aex.

Substitute yp(x), y′

p(x), and y′′

p(x) back into the differential equation:

−Aex+ 3(Aex+B) + 2(Aex+Bx +C)=3ex+ 2x.

By comparing coefficients, we can find A= 1, B=−2, and C= 0.

Therefore, a particular solution to the nonhomogeneous equation is:

yp(x) = ex−2x.

Step 3: Find the general solution

The general solution to the nonhomogeneous equation is the sum of the

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

nonhomogeneous equation:

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

y(x) = c1ex+c2e2x+ex−2x,

where c1and c2are arbitrary constants.

Question 35

Question

Find the particular solution of the nonhomogeneous differential equation

y′′ −3y′+ 2y= 6xex.

Solution

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

we assume the particular solution has the form yp=Ax2ex+Bxex.

yp=Ax2ex+Bxex,

y′

p= (2Ax +Aex+B)ex,

y′′

p= (2A+ 2Aex+ 3Bex).

(1)

Step 1: Substitute the assumed form of ypand its derivatives into the dif-

ferential equation.

(2A+ 2Aex+ 3Bex)−3((2Ax +Aex+B)ex) + 2(Ax2ex+Bxex)

= 6xex.(2)

Step 2: Simplify the equation.

2A−3(2Ax +Aex+B)+2Ax2+ 2Bx = 6x. (3)

Step 3: Equate coefficients of like terms.

2A−6A= 0,

−3A+ 2A−3B−3B= 0,

2B= 6.

(4)

Step 4: Solve for Aand B.

A= 0,

−A−6B= 0 =⇒B=−1.(5)

Therefore, the particular solution is yp=−xex.

MATH 334 - DIFFERENTIAL EQUATIONS Method of undetermined coefficients Question Bank Set 2

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