MATH 334 - DIFFERENTIAL EQUATIONS Method of undetermined coefficients Question Bank Set 4

MATH 334 - DIFFERENTIAL

EQUATIONS - Method of

undetermined coefficients

Question Bank - Set 4

Liberty University

Question 1

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 4e2x.

Solution

Step 1: First, find the general solution yh(x) to the associated homogeneous

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

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

the general solution to the homogeneous equation is yh(x) = c1ex+c2e2x.

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

equation y′′ −3y′+ 2y= 4e2x. Since e2xis a solution to the homogeneous

equation, we try the particular solution in the form yp(x) = Axe2x. Now, we

differentiate yp(x) to find y′

p(x) and y′′

p(x).

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) back into the original differential

equation and simplify to solve for A.

y′′

p−3y′

p+ 2yp= 4e2x

(4A−6A+ 4Ax)e2x= 4e2x

(−2A+ 4Ax)e2x=e2x

Step 4: Equate coefficients to find the value of A. From the previous step,

we have −2A+ 4A= 1, so 2A= 1 and A=1

Step 5: Write down the particular solution with the value of A. Therefore,

the particular solution is yp(x) = 1

2xe2x.

Step 6: Combine the general solution yh(x) and the particular solution

yp(x) to get the final solution. The general solution is y(x) = yh(x) + yp(x) =

c1ex+c2e2x+1

2xe2x.

Question 2

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −4y′+ 4y= 3e2x.

Solution

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

termined coefficients, we first find the complementary solution by solving the

corresponding homogeneous equation:

y′′ −4y′+ 4y= 0.

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

Thus, the complementary solution is yc=c1e2x+c2xe2x.

Next, we guess the form of the particular solution yp. Since the nonhomo-

geneous term is 3e2x, and the characteristic roots are 2 with multiplicity 2, we

try yp=Ax2e2x.

Step 1: Calculate the derivatives needed for the differential equation.

y′

p= (2Ax + 2A)e2x,

y′′

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

Step 2: Substitute the particular solution and its derivatives into the dif-

ferential equation.

(4A+ 4Ax)e2x−4(2Ax + 2A)e2x+ 4Ax2e2x= 3e2x

(4A+ 4Ax −8Ax −8A+ 4Ax2)e2x= 3e2x

(−4A+ 4Ax2)e2x= 3e2x.

Step 3: Equate coefficients of like terms.

−4A+ 4Ax2= 3,

4Ax2= 3 (coefficient of e2xterms must be equal).

Step 4: Solve for A.

4A= 3 ⇒A=3

Therefore, the particular solution is yp=3

4x2e2x.

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

solutions:

y=yc+yp=c1e2x+c2xe2x+3

4x2e2x.

Question 3

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 complementary solution by solving the homogeneous

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

which simplifies to (r−2)2= 0. Thus, the complementary solution is yc=

c1e2x+c2xe2x.

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

tion. Since the right-hand side contains e2xand x2, we assume the particular

solution has the form yp=Ae2x+Bx2+Cx +D.

Step 3: We substitute ypinto the differential equation and its derivatives to

find A,B,C, and D.

y′′

p−4y′

p+ 4yp= 3e2x+ 2x2

(4A−8Be2x)−4(2A+ 2Bx) + 4(Ae2x+Bx2+Cx +D)=3e2x+ 2x2

Step 4: Equating coefficients, we get the following system of equations:

−4A+ 4B= 0

10A−12B= 2

−8A+ 8B−4C= 0

20A−24B+ 8C−4D= 3

Step 5: Solve the system of equations to find A,B,C, and D. After solving,

we get A=3

5,B=3

10 ,C=−3

5, and D=−7

10 .

Step 6: Finally, the general solution to the non-homogeneous equation is the

sum of the complementary and particular solutions:

y=yc+yp=c1e2x+c2xe2x+3

5e2x+3

10x2−3

5x−7

Question 4

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −2y′+y= 3ex.

Solution

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

undetermined coefficients, we will first find the general solution to the asso-

ciated homogeneous equation and then find a particular solution to the non-

homogeneous equation.

Step 1: Find the general solution to the homogeneous equation

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

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

root is r= 1 with multiplicity 2.

Therefore, the general solution to the homogeneous equation is yh(x) =

c1ex+c2xex.

Step 2: Find a particular solution to the non-homogeneous equa-

tion For the particular solution to the non-homogeneous equation, we look for

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

Substitute yp(x) = Aexinto the non-homogeneous equation:

(Aex)′′ −2(Aex)′+Aex= 3ex.

Aex−2Aex+Aex= 3ex.

Aex= 3ex.

A= 3.

So, a particular solution is yp(x)=3ex.

Step 3: Determine the general solution of the non-homogeneous

equation The general solution of the non-homogeneous equation is the sum of

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

the non-homogeneous equation:

y(x) = yh(x) + yp(x) = c1ex+c2xex+ 3ex.

Therefore, the general solution to the given non-homogeneous differential

equation is y(x) = c1ex+c2xex+ 3ex.

Question 5

Question

Consider the following second-order nonhomogeneous differential equation:

y′′ −3y′+ 2y= 2ex+ 4 sin(x)

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

this differential equation.

Solution

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

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

ycto the corresponding homogeneous equation y′′ −3y′+2y= 0, and then guess

the form of the particular solution ypbased on the form of the nonhomogeneous

terms.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous equation y′′ −3y′+ 2y= 0 is given by:

r2−3r+ 2 = 0

This equation factors as (r−1)(r−2) = 0, so the roots are r= 1 and r= 2.

Therefore, the complementary solution is:

yc=c1ex+c2e2x

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

as:

yp=Aex+B

where Aand Bare constants to be determined.

Step 3: Find derivatives of ypFirst, we find the first and second deriva-

tives of yp:

y′

p=Aexand y′′

p=Aex

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

tion Substitute yp,y′

p, and y′′

pinto the differential equation y′′ −3y′+ 2y=

2ex+ 4 sin(x) and simplify:

(Aex)−3(Aex) + 2(Aex+B)=2ex+ 4 sin(x)

This simplifies to:

2Aex+ 2B= 2ex+ 4 sin(x)

Step 5: Solve for Aand BEquating coefficients of exon both sides, we

get:

2A= 2 =⇒A= 1

Equating constants on both sides, we get:

2B= 0 =⇒B= 0

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

to the differential equation is:

yp=ex

Step 7: Find the general solution The general solution to the nonho-

mogeneous differential equation is the sum of the complementary solution and

the particular solution:

y=yc+yp=c1ex+c2e2x+ex

Question 6

Question

Find the particular solution to the following nonhomogeneous differential equa-

tion using the method of undetermined coefficients:

y′′ −2y′+y= 4e3x−2 sin(x)

Solution

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

equation m2−2m+ 1 = 0. This gives us the characteristic roots m= 1 (with

multiplicity 2).

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

Step 2: Next, we seek a particular solution in the form of yp=ae3x+

bsin(x) + ccos(x), where a,b, and care constants to be determined.

Step 3: Calculate the derivatives of yp:

y′

p= 3ae3x+bcos(x)−csin(x)

y′′

p= 9ae3x−bsin(x)−ccos(x)

Step 4: Substitute yp,y′

p, and y′′

pback into the original differential equation:

9ae3x−bsin(x)−ccos(x)−2(3ae3x+bcos(x)−csin(x))+ae3x+bsin(x)+ccos(x)=4e3x−2 sin(x)

Step 5: Simplify the equation and collect like terms:

(10a−2b+c)e3x+ (b+ 2c) sin(x)+(−2b+c) cos(x)=4e3x−2 sin(x)

Step 6: Equate the coefficients of like terms on both sides of the equation:

10a−2b+c= 4 (1)

b+ 2c=−2 (2)

−2b+c= 0 (3)

Step 7: Solve the system of equations (1), (2), and (3) to find the values of

a,b, and c.

Step 8: Once you have found the values of a,b, and c, substitute them back

into the particular solution ypto get the final particular solution.

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

5e3x−2

5sin(x).

Question 7

Question

Use the Method of Undetermined Coefficients to solve the following nonhomo-

geneous differential equation:

y′′ −3y′+ 2y=e2x+ 4 cos(x)

Assume yphas the form Ae2x+Bcos(x) + Csin(x).

Solution

We will first find the complementary solution ycto the associated homogeneous

equation y′′ −3y′+ 2y= 0, before finding the particular solution ypto the

nonhomogeneous equation.

Step 1: Find the complementary solution. The characteristic equation

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

(r−1)(r−2) = 0. This gives us characteristic roots r1= 1 and r2= 2.

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

Step 2: Find the particular solution. Given y′′ −3y′+ 2y=e2x+

4 cos(x), we assume a particular solution of the form yp=Ae2x+Bcos(x) +

Csin(x).

Substitute ypinto the differential equation:

(Ae2x+Bcos(x)+Csin(x))′′ −3(Ae2x+Bcos(x)+Csin(x))′+2(Ae2x+Bcos(x)+Csin(x)) = e2x+4 cos(x)

Simplify before solving for coefficients:

A(e2x)′′ +Bcos(x)′′ +Csin(x)′′ −3(Ae2x)′−3Bsin(x)−3Ccos(x)+2Ae2x+2Bcos(x)+2Csin(x) = e2x+4 cos(x)

Differentiate and simplify further:

4Ae2x−Bcos(x)−Csin(x)−6Ae2x−3Bsin(x)−3Ccos(x)+2Ae2x+2Bcos(x)+2Csin(x) = e2x+4 cos(x)

Grouping terms with the same trigonometric functions:

(4A−6A+2A)e2x+(−B−3B+2B) cos(x)+(−C−3C+2C) sin(x) = e2x+4 cos(x)

This simplifies to:

0e2x−2Bcos(x) + −2Csin(x) = e2x+ 4 cos(x)

Equating coefficients gives us the system of equations:

0=1

−2B= 4

−2C= 0

Solving this system, we find B=−2 and C= 0. There is no solution to

0 = 1, which implies that Acan be any value. Let’s set A= 1 for simplicity.

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

Step 3: Write the general solution. The general solution to the nonho-

mogeneous differential equation is:

y=yc+yp=c1ex+c2e2x+e2x−2 cos(x)

where c1and c2are constants determined by initial conditions or boundary

conditions.

Question 8

Question

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

undetermined coefficients:

y′′ −4y′+ 4y=e2x+ sin x

Solution

To solve the given non-homogeneous differential equation using the method

of undetermined coefficients, we will first find the complementary solution by

solving the associated homogeneous equation. Then, we will guess the particular

solution based on the form of the right-hand side.

Step 1: Find the complementary solution: The associated homoge-

neous equation is:

y′′ −4y′+ 4y= 0

Characteristics equation:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2

Therefore, the complementary solution is:

yc=c1e2x+c2xe2x

Step 2: Find the particular solution:

For the particular solution, we guess the form:

yp=Ae2x+Bx cos x+Cx sin x

Step 3: Find y′

pand y′′

y′

p= 2Ae2x−Bsin x+Bx cos x+Ccos x+Cx sin x

y′′

p= 4Ae2x−2Bcos x−2Bsin x−2Cx cos x+Csin x+ 2Csin x

Step 4: Substitute into the original equation:

4Ae2x−2Bcos x−2Bsin x−2Cx cos x+Csin x+2Csin x−4(2Ae2x−Bsin x+Bx cos x+Ccos x+Cx sin x)+4(Ae2x+Bx cos x+Cx sin x) = e2x+sin x

Step 5: Solve for coefficients: Matching coefficients of terms on both

sides of the equation, we get the system of equations:

4A−8A+ 4A= 1

−2B+ 4B= 0

−2B−2C+ 4A+ 4C= 1

Solving the system of equations gives A=1

6,B=1

4, and C=−1

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

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

ular solutions:

y=yc+yp

y=c1e2x+c2xe2x+1

6e2x+1

4xcos x−1

8xsin x

Question 9

Question

Consider the differential equation

y′′ −4y′+ 4y= 2e2x+ sin(2x)

Find the particular solution using the method of undetermined coefficients.

Solution

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

we first need to determine the complementary solution, yc, by solving the ho-

mogeneous version of the differential equation:

y′′ −4y′+ 4y= 0.

Step 1: Find the complementary solution The characteristic equation

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

solution is given by:

yc=c1e2x+c2xe2x.

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

side of the differential equation includes e2xand sin(2x), we guess the particular

solution to have the form:

yp=ae2x+bsin(2x) + ccos(2x).

Step 3: Find the first and second derivatives of yp

y′

p= 2ae2x+ 2bcos(2x)−2csin(2x),

y′′

p= 4ae2x−4bsin(2x)−4ccos(2x).

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

tion Substitute yp,y′

p, and y′′

pinto the differential equation and simplify to

get:

4ae2x−4bsin(2x)−4ccos(2x)−8ae2x+8bcos(2x)−8csin(2x)+4ae2x+4bsin(2x)+4ccos(2x)=2e2x+sin(2x).

Step 5: Solve for the coefficients By comparing coefficients on both

sides of the equation, we find that:

a=−1

2, b =1

8, c = 0.

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

is:

yp=−1

2e2x+1

8sin(2x).

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

the complementary and particular solutions:

y=yc+yp=c1e2x+c2xe2x−1

2e2x+1

8sin(2x).

Question 10

Question

Find the particular solution to the differential equation y′′ −5y′+ 6y= 4e2x

using the method of undetermined coefficients.

Solution

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

we will assume the form of the particular solution. Since e2xis a solution to the

homogenous equation y′′ −5y′+ 6y= 0, we need to multiply our assumed form

by xto prevent overlap. Therefore, we will assume the particular solution has

the form yp(x) = Ae2x.

Step 1: Find y′

pand y′′



y′

p= 2Ae2x



y′′

p= 4Ae2x

Step 2: Substitute into the differential equation Substitute yp,y′

and y′′

pback into the differential equation:

4Ae2x−5(2Ae2x) + 6(Ae2x)=4e2x

Step 3: Solve for A

4Ae2x−10Ae2x+ 6Ae2x= 4e2x

(4 −10 + 6)Ae2x= 4e2x

A=4

Since the right side is a constant, we can conclude that the solution does

not exist. This means that our assumed form for the particular solution is not

correct. We need to modify our assumed form and try again.

Question 11

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −3y′+ 2y= 4et+t

Solution

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

mined coefficients, we will first find the complementary function yc(t) by solving

the associated homogenous equation y′′ −3y′+ 2y= 0. Then, we will find a

particular solution yp(t) that satisfies the nonhomogeneous equation, and the

general solution will be given by y(t) = yc(t) + yp(t).

Step 1: Find the complementary function yc(t) The characteristic

equation of the homogeneous equation is:

r2−3r+ 2 = 0

Solving for r, we get r= 1 and r= 2. Therefore, the complementary function

is:

yc(t) = c1et+c2e2t

Step 2: Find the particular solution yp(t) Let’s guess a particular

solution of the form yp(t) = Aet+Bt +C, where A,B, and Care constants to

be determined.

Substitute yp(t) into the nonhomogeneous equation:

y′′

p−3y′

p+ 2yp= 4et+t

(Aet+Bt +C)′′ −3(Aet+Bt +C)′+ 2(Aet+Bt +C) = 4et+t

Differentiate and substitute the above expression into the nonhomogeneous

equation, we get:

−Aet−3(Aet+B) + 2(Aet+Bt +C)=4et+t

Rearranging terms, we find:

(A−3A+ 2A)et+ (−3B+ 2B)t+ 2C= 4et+t

This gives us the system of equations:

A−3A+ 2A= 4

−3B+ 2B= 1

2C= 0

Solving the system of equations, we find A=−2, B=−1, C= 0. Thus,

the particular solution is:

yp(t) = −2et−t

Step 3: Find the general solution y(t) The general solution is the sum

of the complementary function and the particular solution:

y(t) = yc(t) + yp(t) = c1et+c2e2t−2et−t= (c1−2)et+c2e2t−t

Therefore, the general solution of the nonhomogeneous differential equation

is:

y(t) = (c1−2)et+c2e2t−t

Question 12

Question

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

3y′+ 2y= 3exsin x, given that the homogeneous solution is yh=c1e2x+c2ex.

Solution

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

undetermined coefficients, we must first guess the form of the particular solution.

Step 1: Guess the form of the particular solution

Since the right-hand side of the non-homogeneous differential equation is

3exsin x, we can choose the particular solution to have the form yp= (Ax2+

Bx +C)exsin x+ (Dx +E)excos x.

Step 2: Find the derivatives of yp

The first and second derivatives of ypare:

y′

p= [(2Ax +B)exsin x+ (Ax2+Bx +C)excos x

+ (D+Ex)excos x−(Dx +E)exsin x],

y′′

p= [(2A+ 2Bx)exsin x+ (2Ax +B)excos x+ (2A+ 2Bx)excos x

−(2Ax +B)exsin x+ (D+E+Ex)excos x−(D+E)exsin x].

Step 3: Substitute yp,y′

p, and y′′

pinto the differential equation

Substitute yp,y′

p, and y′′

pback into the non-homogeneous differential equa-

tion to find the values of the coefficients A,B,C,D, and E.

Step 4: Solve for the coefficients

Solve for the coefficients A,B,C,D, and Eby equating coefficients of the

terms involving exsin xand excos xon both sides of the differential equation.

Step 5: Calculate the particular solution

Once the coefficients are determined, substitute them back into the assumed

form of ypto obtain the particular solution yp.

Question 13

Question

Solve the following second order linear homogeneous differential equation using

the method of undetermined coefficients:

y′′ −3y′+ 2y= 4e2x+ 3x2

Solution

1. First, solve the associated homogeneous equation:

y′′ −3y′+ 2y= 0

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

2) = 0. So the roots are r= 1 and r= 2. Thus, the general solution to

the homogeneous equation is:

yh(x) = c1ex+c2e2x

2. Next, find the particular solution for the given non-homogeneous equation.

Assume the particular solution has the form:

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

Compute the first and second derivatives:

y′

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

y′′

p(x) = 4Ae2x+ 2B

3. Substituting yp(x), y′

p(x), and y′′

p(x) into the original differential equation,

we get:

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

4. Simplifying the above equation gives us:

(4A−6A+2A)e2x+(2B−6B+2B)x+(2B−3C+2C)x+2C= 4e2x+3x2

5. Equating the coefficients of the exponential and polynomial terms, we get

the following system of equations:

−Ae2x= 4e2x

−4Bx = 3x2

2B−C= 0

2C= 0

6. Solve the system of equations to find A,B,C, and D. Then, substitute

these values back into the particular solution.

7. The particular solution is:

yp(x) = −4e2x+3

4x2+3

8. Therefore, the general solution to the non-homogeneous equation is:

y(x) = c1ex+c2e2x−4e2x+3

4x2+3

Question 14

Question

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

undetermined coefficients:

y′′ −3y′+ 2y= 6e4x

Solution

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

undetermined coefficients, we first need to find the complementary function

solution yc(x) by solving the corresponding homogeneous equation:

y′′ −3y′+ 2y= 0

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

0. Therefore, the complementary function is given by:

yc(x) = Aex+Be2x

Next, we assume a particular solution of the form yp(x) = Ce4xfor the

non-homogeneous equation.

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

yp(x) = Ce4x

y′

p(x) = 4Ce4x

y′′

p(x) = 16Ce4x

Step 2: Substitute yp(x) and its derivatives into the non-homogeneous equa-

tion to solve for C:

16Ce4x−3(4Ce4x) + 2(Ce4x)=6e4x

(16C−12C+ 2C)e4x= 6e4x

6C= 6

C= 1

So, the particular solution is yp(x) = e4x.

Step 3: The general solution to the non-homogeneous differential equation

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

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

y(x) = Aex+Be2x+e4x

Therefore, the general solution to the non-homogeneous differential equation

y′′ −3y′+ 2y= 6e4xis y(x) = Aex+Be2x+e4x.

Question 15

Question

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

undetermined coefficients:

y′′ −3y′+ 2y= 6e2x

Solution

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

termined coefficients, we will assume the particular solution has the form:

yp=Ae2x

where Ais a constant to be determined.

Step 1: Find the first and second derivatives of yp

yp=Ae2x

y′

p= 2Ae2x

y′′

p= 4Ae2x

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation

4Ae2x−3(2Ae2x) + 2(Ae2x)=6e2x

4Ae2x−6Ae2x+ 2Ae2x= 6e2x

(4A−6A+ 2A)e2x= 6e2x

0=6e2x

Step 3: Solve for the constant ASince the right side is not equal to

zero, there is a contradiction. To resolve this, we introduce a factor of xinto

the assumed particular solution:

yp=Axe2x

Step 4: Find the first and second derivatives of the new yp

yp=Axe2x

y′

p=Ae2x+ 2Axe2x

y′′

p= 2Ae2x+ 2Ae2x+ 2Axe2x

Step 5: Substitute yp,y′

p, and y′′

pinto the differential equation

2Ae2x+ 2Ae2x+ 2Axe2x−3(Ae2x+ 2Axe2x) + 2(Axe2x)=6e2x

(4A−6A+ 2A)e2x= 6e2x

0=6e2x

Step 6: Solve for the constant ASince the right side is not equal to

zero, there is still a contradiction. To resolve this, we introduce another factor

of xinto the assumed particular solution:

yp=Ax2e2x

Continue this process until a valid particular solution is found.

Question 16

Question

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

efficients:

y′′ −y′−2y= 2e3x.

Solution

Step 1: Find the complementary solution to the homogeneous equation y′′ −y′−

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

1) = 0. Thus, the complementary solution is given by yc=c1e2x+c2e−x.Step

2: Assume the particular solution has the form yp=Ae3x.Step 3: Calculate

the derivatives of ypand substitute them into the differential equation to find

A. Substitute ypand its derivatives into the differential equation:

y′′

p−y′

p−2yp= 2e3x.

This gives (9A−3A)e3x−(3A)e3x−2Ae3x= 2e3x. Simplify to get 3Ae3x= 2e3x,

which implies A=2

3.Step 4: The particular solution is yp=2

3e3x.Step 5:

The general solution is the sum of the complementary and particular solutions:

y=yc+yp=c1e2x+c2e−x+2

3e3x.

Question 17

Question

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

undetermined coefficients:

y′′ −4y′+ 4y= 3e2x.

Solution

We begin by finding the general solution to the corresponding homogeneous

equation y′′ −4y′+ 4y= 0:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2

So the general solution to the homogeneous equation is yh(x) = c1e2x+c2xe2x.

Next, we determine a particular solution to the non-homogeneous equation

y′′ −4y′+ 4y= 3e2x. Since the right-hand side is 3e2x, we try a particular

solution of the form yp(x) = Ae2x, where Ais a constant to be determined.

Step 1: Calculate the first and second derivatives of yp(x):

yp(x) = Ae2x

y′

p(x)=2Ae2x

y′′

p(x)=4Ae2x

Step 2: Substitute yp(x), y′

p(x), and y′′

p(x) into the non-homogeneous equa-

tion, and simplify:

4Ae2x−4(2Ae2x) + 4(Ae2x)=3e2x

4Ae2x−8Ae2x+ 4Ae2x= 3e2x

0=3e2x

Step 3: Since 0 = 3e2x, we conclude that our initial assumption for yp(x)

was incorrect. We need to modify our guess. Since e2xis a solution to the

homogeneous equation, we multiply our guess by xto avoid overlap:

yp(x) = Axe2x

Step 4: Calculate the first and second derivatives of the revised yp(x):

yp(x) = Axe2x

y′

p(x) = (2Axe2x) + Ae2x

y′′

p(x) = 2Ae2x+ 4Axe2x

Step 5: Substitute the revised yp(x), y′

p(x), and y′′

p(x) into the non-homogeneous

equation and simplify:

(2Ae2x+ 4Axe2x)−4((2Axe2x) + Ae2x) + 4(Axe2x) = 3e2x

2Ae2x+ 4Axe2x−8Axe2x−4Ae2x+ 4Ae2x= 3e2x

2Ae2x= 3e2x

Step 6: Equating coefficients gives 2A= 3, so A=3

Therefore, the particular solution is yp(x) = 3

2xe2x.

Step 7: The general solution to the non-homogeneous equation is the sum

of the homogeneous solution and the particular solution:

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

2xe2x.

Question 18

Question

Find the particular solution to the differential equation y′′ −5y′+ 6y= 3e2x

using the method of undetermined coefficients.

Solution

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

we assume the particular solution has the form yp=Ae2x, where Ais a constant

to be determined.

Step 1: Find y′

pand y′′



y′

p= 2Ae2x



y′′

p= 4Ae2x

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation.

(4Ae2x)−5(2Ae2x) + 6(Ae2x)=3e2x

4Ae2x−10Ae2x+ 6Ae2x= 3e2x

A(4 −10 + 6)e2x= 3e2x

A(0)e2x= 3e2x

Step 3: Solve for A.Since A(0)e2x= 3e2xmust hold true for all x, we

have 0 = 3, which is a contradiction. This means our assumption was incorrect.

Step 4: Revise the assumption. Since our initial assumption did not

work, we need to revise our guess. Since 3e2xis a solution to the homogeneous

equation (D2−5D+ 6)y= 0, we need to multiply our guess by xto get linearly

independent solutions. We will try yp=Ax1e2x.

Step 5: Find y′

pand y′′



y′

p=A(2x+ 1)e2x



y′′

p=A(2x+ 1)e2x+ 2Ae2x

Step 6: Substitute yp,y′

p, and y′′

pinto the differential equation.

(A(2x+ 1)e2x+ 2Ae2x)−5(A(2x+ 1)e2x) + 6(Ax1e2x)=3e2x

A(2x+ 1)e2x+ 2Ae2x−5A(2x+ 1)e2x+ 6Ax1e2x= 3e2x

A(2x+ 1)e2x+ 2Ae2x−10Axe2x−5Ae2x+ 6Ax1e2x= 3e2x

2Axe2x+Ae2x+ 2Ae2x−10Axe2x−5Ae2x+ 6Axe2x= 3e2x

(Ae2x)=3e2x

Step 7: Solve for A.Comparing coefficients, we get A= 3.

Step 8: Write the particular solution. The particular solution is yp=

3x1e2x.

Question 19

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 12e2x

Solution

Step 1: First, find the solutions to the associated homogeneous equation y′′ −

4y′+4y= 0. This equation has characteristic polynomial r2−4r+4 = (r−2)2=

0, so we have a repeated root r1=r2= 2. Therefore, the homogeneous solution

is yh(x) = c1e2x+c2xe2x.

Step 2: Now, let’s find a particular solution to the non-homogeneous equa-

tion in the form of yp(x) = Ae2x, where Ais a constant to be determined.

Step 3: Compute the derivatives needed for the substitution into the non-

homogeneous equation:

y′

p= 2Ae2xand y′′

p= 4Ae2x

Step 4: Substitute yp,y′

p, and y′′

pback into the non-homogeneous equation

and solve for A:

4Ae2x−4(2Ae2x)+4Ae2x= 12e2x

4Ae2x−8Ae2x+ 4Ae2x= 12e2x

0 = 12e2x

Step 5: The above equation is not possible, so we need to try a different

form of the particular solution. Since e2xis also a solution of the homogeneous

equation, multiply the initial part of the particular solution by x:

yp(x) = Ax2e2x

Step 6: Compute the derivatives needed for the substitution:

y′

p= (2Ax +A2x)e2x= (2A+ 2Ax)e2x

y′′

p= (2A+ 2Ax)e2x+ 2Ae2x= (4A+ 2A+ 2Ax)e2x= (6A+ 2Ax)e2x

Step 7: Substitute yp,y′

p, and y′′

pback into the non-homogeneous equation

and solve for A:

(6A+ 2Ax)e2x−4(2A+ 2Ax)e2x+ 4(Ax2)e2x= 12e2x

(6A+ 2Ax −8A−8Ax + 4Ax2)e2x= 12e2x

(6A−2A+ 2Ax −8Ax + 4Ax2)e2x= 12e2x

(4A−6Ax + 4Ax2)e2x= 12e2x

Step 8: Equate coefficients of e2xon both sides:

4A−6Ax + 4Ax2= 12

Solve for Aby matching coefficients.

Question 20

Question

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

undetermined coefficients:

y′′ −3y′+ 2y=x2+ex

Solution

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

determined coefficients, we first find the general solution to the corresponding

homogeneous equation:

y′′

h−3y′

h+ 2yh= 0

Characteristics equation: r2−3r+ 2 = 0 Solving the characteristics equation,

we find:

r1= 2 and r2= 1

Therefore, the solution to the homogeneous equation is:

yh=c1e2x+c2ex

Next, we find a particular solution to the non-homogeneous equation in the

form of yp=Ax2+Bex:

y′′

p−3y′

p+ 2yp=x2+ex

Substitute ypinto the differential equation and solve for Aand B:

(4A−6Ax + 2Ax2)−3(2Ax + 2B) + 2(Ax2+Bex) = x2+ex

Step 1: Find A

2A= 1 =⇒A=1

Step 2: Find B

2B−3(2A) = 1 =⇒2B−3 = 1 =⇒B= 2

Thus, the particular solution is:

yp=1

2x2+ 2ex

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

of the homogeneous and particular solutions:

y=yh+yp=c1e2x+c2ex+1

2x2+ 2ex

Therefore, the solution to the non-homogeneous differential equation y′′ −

3y′+ 2y=x2+exis y=c1e2x+c2ex+1

2x2+ 2ex.

Question 21

Question

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

the method of undetermined coefficients:

y′′ −5y′+ 6y= 12e3x.

Solution

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

undetermined coefficients, we follow these steps:

Step 1: Find the complementary solution First, find the complemen-

tary solution ycby solving the corresponding homogeneous equation:

y′′ −5y′+ 6y= 0.

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

0. So, the complementary solution is

yc=c1e2x+c2e3x.

Step 2: Find the particular solution Assume the particular solution has

the form yp=Ae3x, where Ais a constant to be determined. Substitute ypand

its derivatives into the differential equation:

yp=Ae3x,

y′

p= 3Ae3x,

y′′

p= 9Ae3x.

Substitute these expressions into the differential equation to get:

9Ae3x−5(3Ae3x) + 6(Ae3x) = 12e3x.

Step 3: Solve for the constant Simplify the equation from Step 2 to

solve for A:

9Ae3x−15Ae3x+ 6Ae3x= 12e3x.

This simplifies to:

A=12

9−15 + 6 =12

which is undefined. Therefore, we need to modify our assumption for yp.

Step 4: Modify the assumption Since A=12

0is undefined, we need to

modify our assumption by multiplying ypby x:

yp=Axe3x.

Step 5: Find the new particular solution Substitute ypand its deriva-

tives into the differential equation and solve for A:

9Axe3x−5(3Axe3x) + 6(Axe3x) = 12e3x.

Solving this equation gives A=−2.

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

homogeneous differential equation is:

y=yc+yp=c1e2x+c2e3x−2xe3x.

Question 22

Question

Find a particular solution to the non-homogeneous differential equation:

y′′ −3y′+ 2y= 2ex+ 5 cos(x)−4 sin(x)

Solution

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

we will use the method of undetermined coefficients.

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

homogeneous differential equation:

y′′ −3y′+ 2y= 0

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

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

yc(x) = c1ex+c2e2x

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

equation. Since the right-hand side contains terms of the form ex, cos(x), and

sin(x), we guess a particular solution of the form:

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

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

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

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

y′

p(x) = Aex−Bsin(x) + Ccos(x)

y′′

p(x) = Aex−Bcos(x)−Csin(x)

Substitute these into the differential equation:

(A−3A+2A)ex+(−B−3B−2B) sin(x)+(C−3C+2C) cos(x) = 2ex+5 cos(x)−4 sin(x)

Step 4: Equate the coefficients of ex, cos(x), and sin(x) on both sides to

determine the values of A,B, and C. From the equation above, we get the

following system of equations:

A−3A+ 2A= 2

−3B−2B= 5

[C - 3C + 2C = -4] Solving this system gives A=−4, B=−5

5, and C=−4

Step 5: Therefore, the particular solution to the non-homogeneous differ-

ential equation is:

yp(x) = −4ex−5

5cos(x)−4

2sin(x)

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

equation is the sum of the complementary and particular solutions:

y(x) = yc(x) + yp(x) = c1ex+c2e2x−4ex−cos(x)−2 sin(x)

Question 23

Question

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

3y′+ 2y= 5ex+ 4 sin xusing the method of undetermined coefficients.

Solution

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

we assume the particular solution has the form:

yp=Aex+Bsin x+Ccos x

where A,B, and Care constants to be determined. We also need to find

y′′

p, y′

Step 1: Find y′′

pand y′

yp=Aex+Bsin x+Ccos x

y′

p=Aex+Bcos x−Csin x

y′′

p=Aex−Bsin x−Ccos x

Step 2: Substitute yp, y′

p, y′′

pinto the differential equation.

y′′ −3y′+ 2y= 5ex+ 4 sin x

Aex−Bsin x−Ccos x−3(Aex+Bcos x−Csin x)+2(Aex+Bsin x+Ccos x)=5ex+4 sin x

Step 3: Simplify the equation.

(A−3A+ 2A)ex+ (−B+ 3B+ 2B) sin x+ (−C+ 3C+ 2C) cos x= 5ex+ 4 sin x

Matching the coefficients:

0·ex+ 2B·sin x+ 2C·cos x= 5ex+ 4 sin x

Step 4: Compare coefficients and solve for the constants. From the

previous equation, we have:

2B= 4 =⇒B= 2

2C= 5 =⇒C=5

Therefore, the particular solution to the nonhomogeneous differential equa-

tion is:

yp=Aex+ 2 sin x+5

2cos x

Question 24

Question

Consider the differential equation:

y′′ −4y′+ 4y=x2e2x+ 3 sin(2x)

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

the differential equation.

Solution

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

determine the form of the particular solution based on the functions on the

right side of the equation. Let’s denote the particular solution as yp(x).

Step 1: Determine the form of the particular solution

The functions x2e2xand 3 sin(2x) are of the form P(x)eQx and P(x) sin(Qx),

respectively. So, we will assume the particular solution has the form:

yp(x) = (Ax2+Bx +C)e2x+Dsin(2x)

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

Step 2: Find the derivatives of yp(x)

We will need the first and second derivatives of yp(x) to substitute them

back into the differential equation.

y′

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

y′′

p(x) = (2A)e2x+ 4(Ax +B)e2x+ 4(Ax2+Bx +C)e2x−4Dsin(2x)

Step 3: Substitute yp(x),y′

p(x), and y′′

p(x)into the differential equa-

tion

Substitute yp(x), y′

p(x), and y′′

p(x) back into the original differential equation

and simplify.

(2A−4A+4A−4D)e2x+(4B−4B+4B)e2x+4Ce2x−4Dsin(2x) = x2e2x+3 sin(2x)

This will give us a system of equations to solve for constants A,B,C, and

D. This is quite a complex process and may require further steps to simplify

and solve the system.

Question 25

Question

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

undetermined coefficients:

y′′ −2y′+y= 3ex+ 2 sin x

Solution

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

mined coefficients, we will first find the complementary function and particular

solution separately. Then, we will combine them to find the general solution.

Step 1: Find the complementary function

The complementary function yc(x) is the solution to the homogeneous equa-

tion:

y′′ −2y′+y= 0

The characteristic equation is:

r2−2r+ 1 = 0

(r−1)2= 0

r= 1

Therefore, the complementary function is:

yc(x) = c1ex+c2xex

Step 2: Find the particular solution

We assume the particular solution has the form:

yp(x) = Aex+Bxex+Cex+Dsin x+Ecos x

where A,B,C,D, and Eare constants to be determined.

Substitute yp(x) into the original differential equation:

(A−2A+A)ex+ (B−2A+B)xex+Cex+Dsin x+Ecos x= 3ex+ 2 sin x

Comparing coefficients, we get the following equations:

A−2A+A= 3

B−2A+B= 0

C= 0

D= 2

E= 0

Solving these equations, we find A= 1, B= 1, C= 0, D= 2, and E= 0.

Therefore, the particular solution is:

yp(x) = ex+xex+ 2 sin x

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) = c1ex+c2xex+ex+xex+ 2 sin x

Question 26

Question

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

using the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary solution to the 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. The complementary

solution is yc=C1e2x+C2xe2x.

Step 2: Next, we guess the particular solution to the given non-homogeneous

equation. Since the right-hand side is a polynomial times an exponential func-

tion, we make the particular solution of the form yp= (Ax +B)xe2x.

Step 3: Now, we find the first and second derivatives of yp:y′

p= (2Ax +

A+B)e2x+ 2(Ax +B)xe2xy′′

p= (4Ax + 2A+ 2B)e2x+ 2(2Ax +A+B)e2x+

2(Ax +B)xe2x

Step 4: Substituting yp,y′

p, and y′′

pinto the original differential equation

and simplifying, we get: (4Ax + 2A+ 2B)e2x+ 2(2Ax +A+B)e2x+ 2(Ax +

B)xe2x−4((2Ax +A+B)e2x+ 2(Ax +B)xe2x) + 4(Ax +B)xe2x= 4xe2x

Step 5: Equating coefficients of like terms, we find: For terms with e2x:

4Ax + 2A+ 2B−4(2Ax +A+B) = 0 =⇒A=−1 For terms with xe2x:

2(2Ax +A+B)−4(2Ax +A+B) + 4(Ax +B) = 4 =⇒B= 1

Step 6: Therefore, the particular solution is yp=−(x+1)xe2x. Finally, the

general solution to the differential equation is y=yc+yp=C1e2x+C2xe2x−

(x+ 1)xe2x.

Question 27

Question

Find the particular solution to the differential equation y′′ −2y′−3y= 15 sin(2x)

using the method of undetermined coefficients.

Solution

To use the method of undetermined coefficients, we need to propose a particular

solution that matches the form of the non-homogeneous term. In this case, since

the non-homogeneous term is 15 sin(2x), the particular solution should be of the

form:

yp=Asin(2x) + Bcos(2x)

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

y′

p= 2Acos(2x)−2Bsin(2x)

y′′

p=−4Asin(2x)−4Bcos(2x)

Step 2: Substitute yp,y′

p, and y′′

pinto the original differential equation:

(−4Asin(2x)−4Bcos(2x))−2(2Acos(2x)−2Bsin(2x))−3(Asin(2x)+Bcos(2x)) = 15 sin(2x)

Step 3: Simplify the equation and equate coefficients of like terms:

(−4A+ 4B−3A) sin(2x)+(−4B−4A−3B) cos(2x) = 15 sin(2x)

Equating coefficients:

−7A+ 4B= 15

−7B−4A= 0

Solving these equations simultaneously, we find A=−60

65 =−12

13 and B=

−84

65 =−12

Step 4: Therefore, the particular solution is:

yp=−12

13 sin(2x)−12

5cos(2x)

So, the particular solution to the differential equation is yp=−12

13 sin(2x)−

5cos(2x).

Question 28

Question

Consider the following differential equation:

y′′ −3y′+ 2y= 2e4x

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

differential equation.

Solution

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

undetermined coefficients, we will first find the complementary solution, then

determine the form of the particular solution.

Step 1: Find the Complementary Solution The characteristic equation

for the corresponding homogeneous differential equation y′′ −3y′+ 2y= 0 is:

r2−3r+ 2 = 0

This equation factors as (r−1)(r−2) = 0, so the complementary solution is:

yc=c1ex+c2e2x

Step 2: Determine the Form of the Particular Solution Since the

right-hand side of the differential equation is 2e4x, we guess a particular solution

of the form:

yp=Ae4x

Step 3: Find the Derivatives of the Particular Solution

y′

p= 4Ae4xand y′′

p= 16Ae4x

Step 4: Substitute the Particular Solution into the Differential

Equation Substitute yp,y′

p, and y′′

pinto the differential equation:

16Ae4x−3(4Ae4x)+2Ae4x= 2e4x

Step 5: Solve for A Simplify the left side of the equation:

(16A−12A+ 2A)e4x= 2e4x

6Ae4x= 2e4x

A=1

Step 6: Write Down the Particular Solution The particular solution

is:

yp=1

3e4x

Therefore, the general solution to the differential equation is:

y=yc+yp=c1ex+c2e2x+1

3e4x

Question 29

Question

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

4y′+ 4y= 4e2xusing the method of undetermined coefficients.

Solution

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

homogeneous equation:

y′′ −4y′+ 4y= 0

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

So, the complementary solution is:

yc(x) = c1e2x+c2xe2x

Step 2: Next, find the particular solution yp(x) for the nonhomogeneous

equation y′′ −4y′+ 4y= 4e2x. Since 4e2xhas the same form as the complemen-

tary solution, we try the particular solution in the form:

yp(x) = Bx2e2x

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

y′

p(x) = (2Bx +Bx2)e2x

y′′

p(x) = (2B+ 2Bx + 2Bx + 2Bx2)e2x= (2B+ 4Bx + 2Bx2)e2x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation:

(2B+ 4Bx + 2Bx2)e2x−4(2Bx +Bx2)e2x+ 4Bx2e2x= 4e2x

Step 5: Equate coefficients of similar terms to find B:

2B−8Bx + 6Bx2= 4

Step 6: Solve for Bby matching coefficients:

6B= 0 ⇒B= 0

Step 7: The particular solution is yp(x) = 0, so the general solution to the

nonhomogeneous equation is:

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

Therefore, the particular solution to the given nonhomogeneous differential

equation is y(x) = c1e2x+c2xe2x.

Question 30

Question

Consider the second-order linear non-homogeneous differential equation

y′′ −3y′+ 2y= 6e2x+ 5x.

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

this differential equation.

Solution

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

method of undetermined coefficients, we will first find the complementary solu-

tion and then the particular solution.

Step 1: Find the complementary solution The complementary solution

solves the corresponding homogeneous differential equation y′′ −3y′+ 2y= 0.

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

Thus, the complementary solution is

yc(x) = c1ex+c2e2x.

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

has the form

yp(x) = Ae2x+Bx +C.

Taking the derivatives,

y′

p(x)=2Ae2x+B, y′′

p(x)=4Ae2x.

Substitute yp,y′

p, and y′′

pinto the differential equation:

4Ae2x−3(2Ae2x+B) + 2(Ae2x+Bx +C)=6e2x+ 5x.

Simplify and equate coefficients:

(−6A+ 2B)e2x+ (2A+ 2B) = 6e2x+ 5x.

Equating coefficients of like terms, we get the system of equations:

−6A+ 2B= 6,

2A+ 2B= 5.

Solving the system of equations gives A=−1 and B= 3. Therefore, the

particular solution is

yp(x) = −e2x+ 3x+C.

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

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

ular solutions:

y(x) = yc(x) + yp(x) = c1ex+c2e2x−e2x+ 3x+C).

Question 31

Question

Solve the following second-order homogeneous differential equation using the

method of undetermined coefficients: y′′ −3y′+ 2y= 3e2x+ 5 sin(x).

Solution

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

method of undetermined coefficients, we need to follow these steps:

Step 1: Find the complementary solution ycby solving the homogeneous

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

factors to (r−1)(r−2) = 0. So, the complementary solution is yc=c1ex+c2e2x.

Step 2: Guess the form of the particular solution ypbased on the form

of the nonhomogeneous terms. Since the nonhomogeneous terms are 3e2xand

5 sin(x), we guess yp=Ae2x+Bsin(x) + Ccos(x).

Step 3: Find y′

pand y′′

p:y′

p= 2Ae2x+Bcos(x)−Csin(x), y′′

p= 4Ae2x−

Bsin(x)−Ccos(x).

Step 4: Substitute yp,y′

p, and y′′

pinto the original differential equation and

solve for the coefficients A,B, and C: 4Ae2x−Bsin(x)−Ccos(x)−3(2Ae2x+

Bcos(x)−Csin(x)) + 2(Ae2x+Bsin(x) + Ccos(x)) = 3e2x+ 5 sin(x).

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

and solve for A,B, and C. Solving for A,B, and Cwe get A=3

4,B=−4

and C=3

Step 6: Write down the general solution by adding the complementary

solution and the particular solution: y=yc+yp=c1ex+c2e2x+3

4e2x−

5sin(x) + 3

5cos(x).

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

3e2x+ 5 sin(x) is y=c1ex+c2e2x+3

4e2x−4

5sin(x) + 3

5cos(x).

Question 32

Question

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

nonhomogeneous differential equation

y′′ −3y′+ 2y= 6ex.

Solution

To find a particular solution to the differential equation, we assume the par-

ticular solution has the form yp=Aex.We then substitute this form into the

differential equation and solve for A.

Step 1: Find the derivatives of yp:Since yp=Aex,we have:

y′

p=Aex

y′′

p=Aex

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

tion: Substitute yp, y′

p,and y′′

pinto the differential equation:

(Aex)′′ −3(Aex)′+ 2(Aex)=6ex

Aex−3Aex+ 2Aex= 6ex

Step 3: Solve for A:Simplify the equation obtained in Step 2 and solve

for A:

Aex= 6ex

A= 6

Thus, the particular solution to the given differential equation is yp= 6ex.

Question 33

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −y′−2y= 2et+ 3 sin(t)

Solution

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

mined coefficients, we first need to find the general solution to the corresponding

homogeneous equation and then find a particular solution to the nonhomoge-

neous equation.

Step 1: Find the homogeneous solution The corresponding homoge-

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

which factors to (r−2)(r+ 1) = 0, giving us roots r1= 2 and r2=−1.

Therefore, the homogeneous solution is yh=c1e2t+c2e−t.

Step 2: Find the particular solution For the particular solution, we

consider the right-hand side of the nonhomogeneous equation as a sum of the

individual functions: 2etand 3 sin(t).

Let’s assume the particular solution has the form yp=Aet+Bsin(t) +

Ccos(t), where A,B, and Care constants to be determined.

Substitute ypand its derivatives into the differential equation to determine

the values of A,B, and C.

We have:

y′′

p−y′

p−2yp= 2et+ 3 sin(t)

Substitute ypinto the equation and solve for A,B, and C.

Step 3: Solve for the unknown coefficients First, find the derivatives

of yp:

y′

p=Aet+Bcos(t)−Csin(t)

y′′

p=Aet−Bsin(t)−Ccos(t)

Now, substitute yp,y′

p, and y′′

pinto the differential equation:

(Aet−Bsin(t)−Ccos(t))−(Aet+Bcos(t)−Csin(t))−2(Aet+Bsin(t)+Ccos(t)) = 2et+3 sin(t)

Solve for A,B, and Cby comparing coefficients of et, sin(t), and cos(t) on

both sides of the equation.

Step 4: Write the general solution Once we have found the values of

A,B, and C, the particular solution ypis determined. Therefore, the general

solution to the nonhomogeneous differential equation is:

y=yh+yp=c1e2t+c2e−t+Aet+Bsin(t) + Ccos(t)

Where c1,c2,A,B, and Care constants to be determined based on initial

conditions or boundary conditions.

Question 34

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −3y′+ 2y= 4ex−3 sin x

Solution

Step 1: First, find the complementary function by solving the associated homo-

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

0, which factors to (r−1)(r−2) = 0. So the complementary function is

yc=c1ex+c2e2x.

Step 2: Next, find the particular solution yp. Since the right-hand side

contains exand sin x, we assume a form for yp:

yp=Aex+Bsin x+Ccos x

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

y′

p=Aex+Bcos x−Csin x

y′′

p=Aex−Bsin x−Ccos x

Step 4: Substitute yp, y′

p, y′′

pinto the original differential equation and solve

for the coefficients A, B, C. Plugging into the equation y′′ −3y′+ 2y= 4ex−

3 sin xgives:

Aex−Bsin x−Ccos x−3(Aex+Bcos x−Csin x)+2(Aex+Bsin x+Ccos x)=4ex−3 sin x

Step 5: Equate the terms with similar functions on both sides and solve for

A, B, C. For the terms with ex:

A−3A+ 2A= 4 =⇒0=4

This shows that there is no solution for Awhich means the form for ypis

incorrect.

Step 6: Since the first guess was not successful, let’s try a new form for yp

involving a polynomial of degree 1 times ex:

yp= (Ax +B)ex

Step 7: Calculate the first two derivatives of the revised yp:

y′

p= (A+Ax +B)ex

y′′

p= (2A+Ax +B)ex

Step 8: Substitute yp, y′

p, y′′

pinto the original differential equation and solve

for the coefficients A, B. Plugging into the equation y′′ −3y′+2y= 4ex−3 sin x

gives:

(2A+Ax +B)ex−3(A+Ax +B)ex+ 2(Ax +B)ex= 4ex−3 sin x

Step 9: Equate the terms with exon both sides and solve for A, B. This will

give you the values for Aand Bneeded to find yp.

Step 10: Finally, write down the general solution by combining the comple-

mentary function and the particular solution:

y=yc+yp=c1ex+c2e2x+ (Ax +B)ex

Question 35

Question

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

64 cos(x) using the method of undetermined coefficients.

Solution

We first find the complementary solution to the homogeneous equation: y′′ −

4y′+ 4y= 0.

Step 5: Write down the particular solution with the value of A. Therefore,

the particular solution is yp(x) = 1

2xe2x.

Step 6: Combine the general solution yh(x) and the particular solution

yp(x) to get the final solution. The general solution is y(x) = yh(x) + yp(x) =

c1ex+c2e2x+1

2xe2x.

Question 2

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −4y′+ 4y= 3e2x.

Solution

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

termined coefficients, we first find the complementary solution by solving the

corresponding homogeneous equation:

y′′ −4y′+ 4y= 0.

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

Thus, the complementary solution is yc=c1e2x+c2xe2x.

Next, we guess the form of the particular solution yp. Since the nonhomo-

geneous term is 3e2x, and the characteristic roots are 2 with multiplicity 2, we

try yp=Ax2e2x.

Step 1: Calculate the derivatives needed for the differential equation.

y′

p= (2Ax + 2A)e2x,

y′′

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

Step 2: Substitute the particular solution and its derivatives into the dif-

ferential equation.

(4A+ 4Ax)e2x−4(2Ax + 2A)e2x+ 4Ax2e2x= 3e2x

(4A+ 4Ax −8Ax −8A+ 4Ax2)e2x= 3e2x

(−4A+ 4Ax2)e2x= 3e2x.

Step 3: Equate coefficients of like terms.

−4A+ 4Ax2= 3,

4Ax2= 3 (coefficient of e2xterms must be equal).

Step 4: Solve for A.

4A= 3 ⇒A=3

Therefore, the particular solution is yp=3

4x2e2x.

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

solutions:

y=yc+yp=c1e2x+c2xe2x+3

4x2e2x.

Question 3

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 complementary solution by solving the homogeneous

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

which simplifies to (r−2)2= 0. Thus, the complementary solution is yc=

c1e2x+c2xe2x.

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

tion. Since the right-hand side contains e2xand x2, we assume the particular

solution has the form yp=Ae2x+Bx2+Cx +D.

Step 3: We substitute ypinto the differential equation and its derivatives to

find A,B,C, and D.

y′′

p−4y′

p+ 4yp= 3e2x+ 2x2

(4A−8Be2x)−4(2A+ 2Bx) + 4(Ae2x+Bx2+Cx +D)=3e2x+ 2x2

Step 4: Equating coefficients, we get the following system of equations:

−4A+ 4B= 0

10A−12B= 2

−8A+ 8B−4C= 0

20A−24B+ 8C−4D= 3

Step 5: Solve the system of equations to find A,B,C, and D. After solving,

we get A=3

5,B=3

10 ,C=−3

5, and D=−7

10 .

Step 6: Finally, the general solution to the non-homogeneous equation is the

sum of the complementary and particular solutions:

y=yc+yp=c1e2x+c2xe2x+3

5e2x+3

10x2−3

5x−7

Question 4

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −2y′+y= 3ex.

Solution

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

undetermined coefficients, we will first find the general solution to the asso-

ciated homogeneous equation and then find a particular solution to the non-

homogeneous equation.

Step 1: Find the general solution to the homogeneous equation

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

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

root is r= 1 with multiplicity 2.

Therefore, the general solution to the homogeneous equation is yh(x) =

c1ex+c2xex.

Step 2: Find a particular solution to the non-homogeneous equa-

tion For the particular solution to the non-homogeneous equation, we look for

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

Substitute yp(x) = Aexinto the non-homogeneous equation:

(Aex)′′ −2(Aex)′+Aex= 3ex.

Aex−2Aex+Aex= 3ex.

Aex= 3ex.

A= 3.

So, a particular solution is yp(x)=3ex.

Step 3: Determine the general solution of the non-homogeneous

equation The general solution of the non-homogeneous equation is the sum of

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

the non-homogeneous equation:

y(x) = yh(x) + yp(x) = c1ex+c2xex+ 3ex.

Therefore, the general solution to the given non-homogeneous differential

equation is y(x) = c1ex+c2xex+ 3ex.

Question 5

Question

Consider the following second-order nonhomogeneous differential equation:

y′′ −3y′+ 2y= 2ex+ 4 sin(x)

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

this differential equation.

Solution

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

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

ycto the corresponding homogeneous equation y′′ −3y′+2y= 0, and then guess

the form of the particular solution ypbased on the form of the nonhomogeneous

terms.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous equation y′′ −3y′+ 2y= 0 is given by:

r2−3r+ 2 = 0

This equation factors as (r−1)(r−2) = 0, so the roots are r= 1 and r= 2.

Therefore, the complementary solution is:

yc=c1ex+c2e2x

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

as:

yp=Aex+B

where Aand Bare constants to be determined.

Step 3: Find derivatives of ypFirst, we find the first and second deriva-

tives of yp:

y′

p=Aexand y′′

p=Aex

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

tion Substitute yp,y′

p, and y′′

pinto the differential equation y′′ −3y′+ 2y=

2ex+ 4 sin(x) and simplify:

(Aex)−3(Aex) + 2(Aex+B)=2ex+ 4 sin(x)

This simplifies to:

2Aex+ 2B= 2ex+ 4 sin(x)

Step 5: Solve for Aand BEquating coefficients of exon both sides, we

get:

2A= 2 =⇒A= 1

Equating constants on both sides, we get:

2B= 0 =⇒B= 0

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

to the differential equation is:

yp=ex

Step 7: Find the general solution The general solution to the nonho-

mogeneous differential equation is the sum of the complementary solution and

the particular solution:

y=yc+yp=c1ex+c2e2x+ex

Question 6

Question

Find the particular solution to the following nonhomogeneous differential equa-

tion using the method of undetermined coefficients:

y′′ −2y′+y= 4e3x−2 sin(x)

Solution

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

equation m2−2m+ 1 = 0. This gives us the characteristic roots m= 1 (with

multiplicity 2).

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

Step 2: Next, we seek a particular solution in the form of yp=ae3x+

bsin(x) + ccos(x), where a,b, and care constants to be determined.

Step 3: Calculate the derivatives of yp:

y′

p= 3ae3x+bcos(x)−csin(x)

y′′

p= 9ae3x−bsin(x)−ccos(x)

Step 4: Substitute yp,y′

p, and y′′

pback into the original differential equation:

9ae3x−bsin(x)−ccos(x)−2(3ae3x+bcos(x)−csin(x))+ae3x+bsin(x)+ccos(x)=4e3x−2 sin(x)

Step 5: Simplify the equation and collect like terms:

(10a−2b+c)e3x+ (b+ 2c) sin(x)+(−2b+c) cos(x)=4e3x−2 sin(x)

Step 6: Equate the coefficients of like terms on both sides of the equation:

10a−2b+c= 4 (1)

b+ 2c=−2 (2)

−2b+c= 0 (3)

Step 7: Solve the system of equations (1), (2), and (3) to find the values of

a,b, and c.

Step 8: Once you have found the values of a,b, and c, substitute them back

into the particular solution ypto get the final particular solution.

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

5e3x−2

5sin(x).

Question 7

Question

Use the Method of Undetermined Coefficients to solve the following nonhomo-

geneous differential equation:

y′′ −3y′+ 2y=e2x+ 4 cos(x)

Assume yphas the form Ae2x+Bcos(x) + Csin(x).

Solution

We will first find the complementary solution ycto the associated homogeneous

equation y′′ −3y′+ 2y= 0, before finding the particular solution ypto the

nonhomogeneous equation.

Step 1: Find the complementary solution. The characteristic equation

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

(r−1)(r−2) = 0. This gives us characteristic roots r1= 1 and r2= 2.

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

Step 2: Find the particular solution. Given y′′ −3y′+ 2y=e2x+

4 cos(x), we assume a particular solution of the form yp=Ae2x+Bcos(x) +

Csin(x).

Substitute ypinto the differential equation:

(Ae2x+Bcos(x)+Csin(x))′′ −3(Ae2x+Bcos(x)+Csin(x))′+2(Ae2x+Bcos(x)+Csin(x)) = e2x+4 cos(x)

Simplify before solving for coefficients:

A(e2x)′′ +Bcos(x)′′ +Csin(x)′′ −3(Ae2x)′−3Bsin(x)−3Ccos(x)+2Ae2x+2Bcos(x)+2Csin(x) = e2x+4 cos(x)

Differentiate and simplify further:

4Ae2x−Bcos(x)−Csin(x)−6Ae2x−3Bsin(x)−3Ccos(x)+2Ae2x+2Bcos(x)+2Csin(x) = e2x+4 cos(x)

Grouping terms with the same trigonometric functions:

(4A−6A+2A)e2x+(−B−3B+2B) cos(x)+(−C−3C+2C) sin(x) = e2x+4 cos(x)

This simplifies to:

0e2x−2Bcos(x) + −2Csin(x) = e2x+ 4 cos(x)

Equating coefficients gives us the system of equations:

0=1

−2B= 4

−2C= 0

Solving this system, we find B=−2 and C= 0. There is no solution to

0 = 1, which implies that Acan be any value. Let’s set A= 1 for simplicity.

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

Step 3: Write the general solution. The general solution to the nonho-

mogeneous differential equation is:

y=yc+yp=c1ex+c2e2x+e2x−2 cos(x)

where c1and c2are constants determined by initial conditions or boundary

conditions.

Question 8

Question

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

undetermined coefficients:

y′′ −4y′+ 4y=e2x+ sin x

Solution

To solve the given non-homogeneous differential equation using the method

of undetermined coefficients, we will first find the complementary solution by

solving the associated homogeneous equation. Then, we will guess the particular

solution based on the form of the right-hand side.

Step 1: Find the complementary solution: The associated homoge-

neous equation is:

y′′ −4y′+ 4y= 0

Characteristics equation:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2

Therefore, the complementary solution is:

yc=c1e2x+c2xe2x

Step 2: Find the particular solution:

For the particular solution, we guess the form:

yp=Ae2x+Bx cos x+Cx sin x

Step 3: Find y′

pand y′′

y′

p= 2Ae2x−Bsin x+Bx cos x+Ccos x+Cx sin x

y′′

p= 4Ae2x−2Bcos x−2Bsin x−2Cx cos x+Csin x+ 2Csin x

Step 4: Substitute into the original equation:

4Ae2x−2Bcos x−2Bsin x−2Cx cos x+Csin x+2Csin x−4(2Ae2x−Bsin x+Bx cos x+Ccos x+Cx sin x)+4(Ae2x+Bx cos x+Cx sin x) = e2x+sin x

Step 5: Solve for coefficients: Matching coefficients of terms on both

sides of the equation, we get the system of equations:

4A−8A+ 4A= 1

−2B+ 4B= 0

−2B−2C+ 4A+ 4C= 1

Solving the system of equations gives A=1

6,B=1

4, and C=−1

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

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

ular solutions:

y=yc+yp

y=c1e2x+c2xe2x+1

6e2x+1

4xcos x−1

8xsin x

Question 9

Question

Consider the differential equation

y′′ −4y′+ 4y= 2e2x+ sin(2x)

Find the particular solution using the method of undetermined coefficients.

Solution

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

we first need to determine the complementary solution, yc, by solving the ho-

mogeneous version of the differential equation:

y′′ −4y′+ 4y= 0.

Step 1: Find the complementary solution The characteristic equation

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

solution is given by:

yc=c1e2x+c2xe2x.

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

side of the differential equation includes e2xand sin(2x), we guess the particular

solution to have the form:

yp=ae2x+bsin(2x) + ccos(2x).

Step 3: Find the first and second derivatives of yp

y′

p= 2ae2x+ 2bcos(2x)−2csin(2x),

y′′

p= 4ae2x−4bsin(2x)−4ccos(2x).

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

tion Substitute yp,y′

p, and y′′

pinto the differential equation and simplify to

get:

4ae2x−4bsin(2x)−4ccos(2x)−8ae2x+8bcos(2x)−8csin(2x)+4ae2x+4bsin(2x)+4ccos(2x)=2e2x+sin(2x).

Step 5: Solve for the coefficients By comparing coefficients on both

sides of the equation, we find that:

a=−1

2, b =1

8, c = 0.

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

is:

yp=−1

2e2x+1

8sin(2x).

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

the complementary and particular solutions:

y=yc+yp=c1e2x+c2xe2x−1

2e2x+1

8sin(2x).

Question 10

Question

Find the particular solution to the differential equation y′′ −5y′+ 6y= 4e2x

using the method of undetermined coefficients.

Solution

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

we will assume the form of the particular solution. Since e2xis a solution to the

homogenous equation y′′ −5y′+ 6y= 0, we need to multiply our assumed form

by xto prevent overlap. Therefore, we will assume the particular solution has

the form yp(x) = Ae2x.

Step 1: Find y′

pand y′′



y′

p= 2Ae2x



y′′

p= 4Ae2x

Step 2: Substitute into the differential equation Substitute yp,y′

and y′′

pback into the differential equation:

4Ae2x−5(2Ae2x) + 6(Ae2x)=4e2x

Step 3: Solve for A

4Ae2x−10Ae2x+ 6Ae2x= 4e2x

(4 −10 + 6)Ae2x= 4e2x

A=4

Since the right side is a constant, we can conclude that the solution does

not exist. This means that our assumed form for the particular solution is not

correct. We need to modify our assumed form and try again.

Question 11

Question

Solve the following nonhomogeneous differential equation using the method of

undetermined coefficients:

y′′ −3y′+ 2y= 4et+t

Solution

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

mined coefficients, we will first find the complementary function yc(t) by solving

the associated homogenous equation y′′ −3y′+ 2y= 0. Then, we will find a

particular solution yp(t) that satisfies the nonhomogeneous equation, and the

general solution will be given by y(t) = yc(t) + yp(t).

Step 1: Find the complementary function yc(t) The characteristic

equation of the homogeneous equation is:

r2−3r+ 2 = 0

Solving for r, we get r= 1 and r= 2. Therefore, the complementary function

is:

yc(t) = c1et+c2e2t

Step 2: Find the particular solution yp(t) Let’s guess a particular

solution of the form yp(t) = Aet+Bt +C, where A,B, and Care constants to

be determined.

Substitute yp(t) into the nonhomogeneous equation:

y′′

p−3y′

p+ 2yp= 4et+t

(Aet+Bt +C)′′ −3(Aet+Bt +C)′+ 2(Aet+Bt +C) = 4et+t

Differentiate and substitute the above expression into the nonhomogeneous

equation, we get:

−Aet−3(Aet+B) + 2(Aet+Bt +C)=4et+t

Rearranging terms, we find:

(A−3A+ 2A)et+ (−3B+ 2B)t+ 2C= 4et+t

This gives us the system of equations:

A−3A+ 2A= 4

−3B+ 2B= 1

2C= 0

Solving the system of equations, we find A=−2, B=−1, C= 0. Thus,

the particular solution is:

yp(t) = −2et−t

Step 3: Find the general solution y(t) The general solution is the sum

of the complementary function and the particular solution:

y(t) = yc(t) + yp(t) = c1et+c2e2t−2et−t= (c1−2)et+c2e2t−t

Therefore, the general solution of the nonhomogeneous differential equation

is:

y(t) = (c1−2)et+c2e2t−t

Question 12

Question

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

3y′+ 2y= 3exsin x, given that the homogeneous solution is yh=c1e2x+c2ex.

Solution

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

undetermined coefficients, we must first guess the form of the particular solution.

Step 1: Guess the form of the particular solution

Since the right-hand side of the non-homogeneous differential equation is

3exsin x, we can choose the particular solution to have the form yp= (Ax2+

Bx +C)exsin x+ (Dx +E)excos x.

Step 2: Find the derivatives of yp

The first and second derivatives of ypare:

y′

p= [(2Ax +B)exsin x+ (Ax2+Bx +C)excos x

+ (D+Ex)excos x−(Dx +E)exsin x],

y′′

p= [(2A+ 2Bx)exsin x+ (2Ax +B)excos x+ (2A+ 2Bx)excos x

−(2Ax +B)exsin x+ (D+E+Ex)excos x−(D+E)exsin x].

Step 3: Substitute yp,y′

p, and y′′

pinto the differential equation

Substitute yp,y′

p, and y′′

pback into the non-homogeneous differential equa-

tion to find the values of the coefficients A,B,C,D, and E.

Step 4: Solve for the coefficients

Solve for the coefficients A,B,C,D, and Eby equating coefficients of the

terms involving exsin xand excos xon both sides of the differential equation.

Step 5: Calculate the particular solution

Once the coefficients are determined, substitute them back into the assumed

form of ypto obtain the particular solution yp.

Question 13

Question

Solve the following second order linear homogeneous differential equation using

the method of undetermined coefficients:

y′′ −3y′+ 2y= 4e2x+ 3x2

Solution

1. First, solve the associated homogeneous equation:

y′′ −3y′+ 2y= 0

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

2) = 0. So the roots are r= 1 and r= 2. Thus, the general solution to

the homogeneous equation is:

yh(x) = c1ex+c2e2x

2. Next, find the particular solution for the given non-homogeneous equation.

Assume the particular solution has the form:

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

Compute the first and second derivatives:

y′

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

y′′

p(x) = 4Ae2x+ 2B

3. Substituting yp(x), y′

p(x), and y′′

p(x) into the original differential equation,

we get:

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

4. Simplifying the above equation gives us:

(4A−6A+2A)e2x+(2B−6B+2B)x+(2B−3C+2C)x+2C= 4e2x+3x2

5. Equating the coefficients of the exponential and polynomial terms, we get

the following system of equations:

−Ae2x= 4e2x

−4Bx = 3x2

2B−C= 0

2C= 0

6. Solve the system of equations to find A,B,C, and D. Then, substitute

these values back into the particular solution.

7. The particular solution is:

yp(x) = −4e2x+3

4x2+3

8. Therefore, the general solution to the non-homogeneous equation is:

y(x) = c1ex+c2e2x−4e2x+3

4x2+3

Question 14

Question

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

undetermined coefficients:

y′′ −3y′+ 2y= 6e4x

Solution

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

undetermined coefficients, we first need to find the complementary function

solution yc(x) by solving the corresponding homogeneous equation:

y′′ −3y′+ 2y= 0

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

0. Therefore, the complementary function is given by:

yc(x) = Aex+Be2x

Next, we assume a particular solution of the form yp(x) = Ce4xfor the

non-homogeneous equation.

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

yp(x) = Ce4x

y′

p(x) = 4Ce4x

y′′

p(x) = 16Ce4x

Step 2: Substitute yp(x) and its derivatives into the non-homogeneous equa-

tion to solve for C:

16Ce4x−3(4Ce4x) + 2(Ce4x)=6e4x

(16C−12C+ 2C)e4x= 6e4x

6C= 6

C= 1

So, the particular solution is yp(x) = e4x.

Step 3: The general solution to the non-homogeneous differential equation

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

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

y(x) = Aex+Be2x+e4x

Therefore, the general solution to the non-homogeneous differential equation

y′′ −3y′+ 2y= 6e4xis y(x) = Aex+Be2x+e4x.

Question 15

Question

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

undetermined coefficients:

y′′ −3y′+ 2y= 6e2x

Solution

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

termined coefficients, we will assume the particular solution has the form:

yp=Ae2x

where Ais a constant to be determined.

Step 1: Find the first and second derivatives of yp

yp=Ae2x

y′

p= 2Ae2x

y′′

p= 4Ae2x

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation

4Ae2x−3(2Ae2x) + 2(Ae2x)=6e2x

4Ae2x−6Ae2x+ 2Ae2x= 6e2x

(4A−6A+ 2A)e2x= 6e2x

0=6e2x

Step 3: Solve for the constant ASince the right side is not equal to

zero, there is a contradiction. To resolve this, we introduce a factor of xinto

the assumed particular solution:

yp=Axe2x

Step 4: Find the first and second derivatives of the new yp

yp=Axe2x

y′

p=Ae2x+ 2Axe2x

y′′

p= 2Ae2x+ 2Ae2x+ 2Axe2x

Step 5: Substitute yp,y′

p, and y′′

pinto the differential equation

2Ae2x+ 2Ae2x+ 2Axe2x−3(Ae2x+ 2Axe2x) + 2(Axe2x)=6e2x

(4A−6A+ 2A)e2x= 6e2x

0=6e2x

Step 6: Solve for the constant ASince the right side is not equal to

zero, there is still a contradiction. To resolve this, we introduce another factor

of xinto the assumed particular solution:

yp=Ax2e2x

Continue this process until a valid particular solution is found.

Question 16

Question

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

efficients:

y′′ −y′−2y= 2e3x.

Solution

Step 1: Find the complementary solution to the homogeneous equation y′′ −y′−

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

1) = 0. Thus, the complementary solution is given by yc=c1e2x+c2e−x.Step

2: Assume the particular solution has the form yp=Ae3x.Step 3: Calculate

the derivatives of ypand substitute them into the differential equation to find

A. Substitute ypand its derivatives into the differential equation:

y′′

p−y′

p−2yp= 2e3x.

This gives (9A−3A)e3x−(3A)e3x−2Ae3x= 2e3x. Simplify to get 3Ae3x= 2e3x,

which implies A=2

3.Step 4: The particular solution is yp=2

3e3x.Step 5:

The general solution is the sum of the complementary and particular solutions:

y=yc+yp=c1e2x+c2e−x+2

3e3x.

Question 17

Question

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

undetermined coefficients:

y′′ −4y′+ 4y= 3e2x.

Solution

We begin by finding the general solution to the corresponding homogeneous

equation y′′ −4y′+ 4y= 0:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2

So the general solution to the homogeneous equation is yh(x) = c1e2x+c2xe2x.

Next, we determine a particular solution to the non-homogeneous equation

y′′ −4y′+ 4y= 3e2x. Since the right-hand side is 3e2x, we try a particular

solution of the form yp(x) = Ae2x, where Ais a constant to be determined.

Step 1: Calculate the first and second derivatives of yp(x):

yp(x) = Ae2x

y′

p(x)=2Ae2x

y′′

p(x)=4Ae2x

Step 2: Substitute yp(x), y′

p(x), and y′′

p(x) into the non-homogeneous equa-

tion, and simplify:

4Ae2x−4(2Ae2x) + 4(Ae2x)=3e2x

4Ae2x−8Ae2x+ 4Ae2x= 3e2x

0=3e2x

Step 3: Since 0 = 3e2x, we conclude that our initial assumption for yp(x)

was incorrect. We need to modify our guess. Since e2xis a solution to the

homogeneous equation, we multiply our guess by xto avoid overlap:

yp(x) = Axe2x

Step 4: Calculate the first and second derivatives of the revised yp(x):

yp(x) = Axe2x

y′

p(x) = (2Axe2x) + Ae2x

y′′

p(x) = 2Ae2x+ 4Axe2x

Step 5: Substitute the revised yp(x), y′

p(x), and y′′

p(x) into the non-homogeneous

equation and simplify:

(2Ae2x+ 4Axe2x)−4((2Axe2x) + Ae2x) + 4(Axe2x) = 3e2x

2Ae2x+ 4Axe2x−8Axe2x−4Ae2x+ 4Ae2x= 3e2x

2Ae2x= 3e2x

Step 6: Equating coefficients gives 2A= 3, so A=3

Therefore, the particular solution is yp(x) = 3

2xe2x.

Step 7: The general solution to the non-homogeneous equation is the sum

of the homogeneous solution and the particular solution:

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

2xe2x.

Question 18

Question

Find the particular solution to the differential equation y′′ −5y′+ 6y= 3e2x

using the method of undetermined coefficients.

Solution

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

we assume the particular solution has the form yp=Ae2x, where Ais a constant

to be determined.

Step 1: Find y′

pand y′′



y′

p= 2Ae2x



y′′

p= 4Ae2x

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation.

(4Ae2x)−5(2Ae2x) + 6(Ae2x)=3e2x

4Ae2x−10Ae2x+ 6Ae2x= 3e2x

A(4 −10 + 6)e2x= 3e2x

A(0)e2x= 3e2x

Step 3: Solve for A.Since A(0)e2x= 3e2xmust hold true for all x, we

have 0 = 3, which is a contradiction. This means our assumption was incorrect.

Step 4: Revise the assumption. Since our initial assumption did not

work, we need to revise our guess. Since 3e2xis a solution to the homogeneous

equation (D2−5D+ 6)y= 0, we need to multiply our guess by xto get linearly

independent solutions. We will try yp=Ax1e2x.

Step 5: Find y′

pand y′′



y′

p=A(2x+ 1)e2x



y′′

p=A(2x+ 1)e2x+ 2Ae2x

Step 6: Substitute yp,y′

p, and y′′

pinto the differential equation.

(A(2x+ 1)e2x+ 2Ae2x)−5(A(2x+ 1)e2x) + 6(Ax1e2x)=3e2x

A(2x+ 1)e2x+ 2Ae2x−5A(2x+ 1)e2x+ 6Ax1e2x= 3e2x

A(2x+ 1)e2x+ 2Ae2x−10Axe2x−5Ae2x+ 6Ax1e2x= 3e2x

2Axe2x+Ae2x+ 2Ae2x−10Axe2x−5Ae2x+ 6Axe2x= 3e2x

(Ae2x)=3e2x

Step 7: Solve for A.Comparing coefficients, we get A= 3.

Step 8: Write the particular solution. The particular solution is yp=

3x1e2x.

Question 19

Question

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

the method of undetermined coefficients:

y′′ −4y′+ 4y= 12e2x

Solution

Step 1: First, find the solutions to the associated homogeneous equation y′′ −

4y′+4y= 0. This equation has characteristic polynomial r2−4r+4 = (r−2)2=

0, so we have a repeated root r1=r2= 2. Therefore, the homogeneous solution

is yh(x) = c1e2x+c2xe2x.

Step 2: Now, let’s find a particular solution to the non-homogeneous equa-

tion in the form of yp(x) = Ae2x, where Ais a constant to be determined.

Step 3: Compute the derivatives needed for the substitution into the non-

homogeneous equation:

y′

p= 2Ae2xand y′′

p= 4Ae2x

Step 4: Substitute yp,y′

p, and y′′

pback into the non-homogeneous equation

and solve for A:

4Ae2x−4(2Ae2x)+4Ae2x= 12e2x

4Ae2x−8Ae2x+ 4Ae2x= 12e2x

0 = 12e2x

Step 5: The above equation is not possible, so we need to try a different

form of the particular solution. Since e2xis also a solution of the homogeneous

equation, multiply the initial part of the particular solution by x:

yp(x) = Ax2e2x

Step 6: Compute the derivatives needed for the substitution:

y′

p= (2Ax +A2x)e2x= (2A+ 2Ax)e2x

y′′

p= (2A+ 2Ax)e2x+ 2Ae2x= (4A+ 2A+ 2Ax)e2x= (6A+ 2Ax)e2x

Step 7: Substitute yp,y′

p, and y′′

pback into the non-homogeneous equation

and solve for A:

(6A+ 2Ax)e2x−4(2A+ 2Ax)e2x+ 4(Ax2)e2x= 12e2x

(6A+ 2Ax −8A−8Ax + 4Ax2)e2x= 12e2x

(6A−2A+ 2Ax −8Ax + 4Ax2)e2x= 12e2x

(4A−6Ax + 4Ax2)e2x= 12e2x

Step 8: Equate coefficients of e2xon both sides:

4A−6Ax + 4Ax2= 12

Solve for Aby matching coefficients.

Question 20

Question

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

undetermined coefficients:

y′′ −3y′+ 2y=x2+ex

Solution

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

determined coefficients, we first find the general solution to the corresponding

homogeneous equation:

y′′

h−3y′

h+ 2yh= 0

Characteristics equation: r2−3r+ 2 = 0 Solving the characteristics equation,

we find:

r1= 2 and r2= 1

Therefore, the solution to the homogeneous equation is:

yh=c1e2x+c2ex

Next, we find a particular solution to the non-homogeneous equation in the

form of yp=Ax2+Bex:

y′′

p−3y′

p+ 2yp=x2+ex

Substitute ypinto the differential equation and solve for Aand B:

(4A−6Ax + 2Ax2)−3(2Ax + 2B) + 2(Ax2+Bex) = x2+ex

Step 1: Find A

2A= 1 =⇒A=1

Step 2: Find B

2B−3(2A) = 1 =⇒2B−3 = 1 =⇒B= 2

Thus, the particular solution is:

yp=1

2x2+ 2ex

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

of the homogeneous and particular solutions:

y=yh+yp=c1e2x+c2ex+1

2x2+ 2ex

Therefore, the solution to the non-homogeneous differential equation y′′ −

3y′+ 2y=x2+exis y=c1e2x+c2ex+1

2x2+ 2ex.

Question 21

Question

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

the method of undetermined coefficients:

y′′ −5y′+ 6y= 12e3x.

Solution

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

undetermined coefficients, we follow these steps:

Step 1: Find the complementary solution First, find the complemen-

tary solution ycby solving the corresponding homogeneous equation:

y′′ −5y′+ 6y= 0.

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

0. So, the complementary solution is

yc=c1e2x+c2e3x.

Step 2: Find the particular solution Assume the particular solution has

the form yp=Ae3x, where Ais a constant to be determined. Substitute ypand

its derivatives into the differential equation:

yp=Ae3x,

y′

p= 3Ae3x,

y′′

p= 9Ae3x.

Substitute these expressions into the differential equation to get:

9Ae3x−5(3Ae3x) + 6(Ae3x) = 12e3x.

Step 3: Solve for the constant Simplify the equation from Step 2 to

solve for A:

9Ae3x−15Ae3x+ 6Ae3x= 12e3x.

This simplifies to:

A=12

9−15 + 6 =12

which is undefined. Therefore, we need to modify our assumption for yp.

Step 4: Modify the assumption Since A=12

0is undefined, we need to

modify our assumption by multiplying ypby x:

yp=Axe3x.

Step 5: Find the new particular solution Substitute ypand its deriva-

tives into the differential equation and solve for A:

9Axe3x−5(3Axe3x) + 6(Axe3x) = 12e3x.

Solving this equation gives A=−2.

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

homogeneous differential equation is:

y=yc+yp=c1e2x+c2e3x−2xe3x.

Question 22

Question

Find a particular solution to the non-homogeneous differential equation:

y′′ −3y′+ 2y= 2ex+ 5 cos(x)−4 sin(x)

Solution

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

we will use the method of undetermined coefficients.

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

homogeneous differential equation:

y′′ −3y′+ 2y= 0

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

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

yc(x) = c1ex+c2e2x

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

equation. Since the right-hand side contains terms of the form ex, cos(x), and

sin(x), we guess a particular solution of the form:

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

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

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

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

y′

p(x) = Aex−Bsin(x) + Ccos(x)

y′′

p(x) = Aex−Bcos(x)−Csin(x)

Substitute these into the differential equation:

(A−3A+2A)ex+(−B−3B−2B) sin(x)+(C−3C+2C) cos(x) = 2ex+5 cos(x)−4 sin(x)

Step 4: Equate the coefficients of ex, cos(x), and sin(x) on both sides to

determine the values of A,B, and C. From the equation above, we get the

following system of equations:

A−3A+ 2A= 2

−3B−2B= 5

[C - 3C + 2C = -4] Solving this system gives A=−4, B=−5

5, and C=−4

Step 5: Therefore, the particular solution to the non-homogeneous differ-

ential equation is:

yp(x) = −4ex−5

5cos(x)−4

2sin(x)

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

equation is the sum of the complementary and particular solutions:

y(x) = yc(x) + yp(x) = c1ex+c2e2x−4ex−cos(x)−2 sin(x)

Question 23

Question

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

3y′+ 2y= 5ex+ 4 sin xusing the method of undetermined coefficients.

Solution

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

we assume the particular solution has the form:

yp=Aex+Bsin x+Ccos x

where A,B, and Care constants to be determined. We also need to find

y′′

p, y′

Step 1: Find y′′

pand y′

yp=Aex+Bsin x+Ccos x

y′

p=Aex+Bcos x−Csin x

y′′

p=Aex−Bsin x−Ccos x

Step 2: Substitute yp, y′

p, y′′

pinto the differential equation.

y′′ −3y′+ 2y= 5ex+ 4 sin x

Aex−Bsin x−Ccos x−3(Aex+Bcos x−Csin x)+2(Aex+Bsin x+Ccos x)=5ex+4 sin x

Step 3: Simplify the equation.

(A−3A+ 2A)ex+ (−B+ 3B+ 2B) sin x+ (−C+ 3C+ 2C) cos x= 5ex+ 4 sin x

Matching the coefficients:

0·ex+ 2B·sin x+ 2C·cos x= 5ex+ 4 sin x

Step 4: Compare coefficients and solve for the constants. From the

previous equation, we have:

2B= 4 =⇒B= 2

2C= 5 =⇒C=5

Therefore, the particular solution to the nonhomogeneous differential equa-

tion is:

yp=Aex+ 2 sin x+5

2cos x

Question 24

Question

Consider the differential equation:

y′′ −4y′+ 4y=x2e2x+ 3 sin(2x)

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

the differential equation.

Solution

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

determine the form of the particular solution based on the functions on the

right side of the equation. Let’s denote the particular solution as yp(x).

Step 1: Determine the form of the particular solution

The functions x2e2xand 3 sin(2x) are of the form P(x)eQx and P(x) sin(Qx),

respectively. So, we will assume the particular solution has the form:

yp(x) = (Ax2+Bx +C)e2x+Dsin(2x)

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

Step 2: Find the derivatives of yp(x)

We will need the first and second derivatives of yp(x) to substitute them

back into the differential equation.

y′

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

y′′

p(x) = (2A)e2x+ 4(Ax +B)e2x+ 4(Ax2+Bx +C)e2x−4Dsin(2x)

Step 3: Substitute yp(x),y′

p(x), and y′′

p(x)into the differential equa-

tion

Substitute yp(x), y′

p(x), and y′′

p(x) back into the original differential equation

and simplify.

(2A−4A+4A−4D)e2x+(4B−4B+4B)e2x+4Ce2x−4Dsin(2x) = x2e2x+3 sin(2x)

This will give us a system of equations to solve for constants A,B,C, and

D. This is quite a complex process and may require further steps to simplify

and solve the system.

Question 25

Question

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

undetermined coefficients:

y′′ −2y′+y= 3ex+ 2 sin x

Solution

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

mined coefficients, we will first find the complementary function and particular

solution separately. Then, we will combine them to find the general solution.

Step 1: Find the complementary function

The complementary function yc(x) is the solution to the homogeneous equa-

tion:

y′′ −2y′+y= 0

The characteristic equation is:

r2−2r+ 1 = 0

(r−1)2= 0

r= 1

Therefore, the complementary function is:

yc(x) = c1ex+c2xex

Step 2: Find the particular solution

We assume the particular solution has the form:

yp(x) = Aex+Bxex+Cex+Dsin x+Ecos x

where A,B,C,D, and Eare constants to be determined.

Substitute yp(x) into the original differential equation:

(A−2A+A)ex+ (B−2A+B)xex+Cex+Dsin x+Ecos x= 3ex+ 2 sin x

Comparing coefficients, we get the following equations:

A−2A+A= 3

B−2A+B= 0

C= 0

D= 2

E= 0

Solving these equations, we find A= 1, B= 1, C= 0, D= 2, and E= 0.

Therefore, the particular solution is:

yp(x) = ex+xex+ 2 sin x

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) = c1ex+c2xex+ex+xex+ 2 sin x

Question 26

Question

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

using the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary solution to the 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. The complementary

solution is yc=C1e2x+C2xe2x.

Step 2: Next, we guess the particular solution to the given non-homogeneous

equation. Since the right-hand side is a polynomial times an exponential func-

tion, we make the particular solution of the form yp= (Ax +B)xe2x.

Step 3: Now, we find the first and second derivatives of yp:y′

p= (2Ax +

A+B)e2x+ 2(Ax +B)xe2xy′′

p= (4Ax + 2A+ 2B)e2x+ 2(2Ax +A+B)e2x+

2(Ax +B)xe2x

Step 4: Substituting yp,y′

p, and y′′

pinto the original differential equation

and simplifying, we get: (4Ax + 2A+ 2B)e2x+ 2(2Ax +A+B)e2x+ 2(Ax +

B)xe2x−4((2Ax +A+B)e2x+ 2(Ax +B)xe2x) + 4(Ax +B)xe2x= 4xe2x

Step 5: Equating coefficients of like terms, we find: For terms with e2x:

4Ax + 2A+ 2B−4(2Ax +A+B) = 0 =⇒A=−1 For terms with xe2x:

2(2Ax +A+B)−4(2Ax +A+B) + 4(Ax +B) = 4 =⇒B= 1

Step 6: Therefore, the particular solution is yp=−(x+1)xe2x. Finally, the

general solution to the differential equation is y=yc+yp=C1e2x+C2xe2x−

(x+ 1)xe2x.

Question 27

Question

Find the particular solution to the differential equation y′′ −2y′−3y= 15 sin(2x)

using the method of undetermined coefficients.

Solution

To use the method of undetermined coefficients, we need to propose a particular

solution that matches the form of the non-homogeneous term. In this case, since

the non-homogeneous term is 15 sin(2x), the particular solution should be of the

form:

yp=Asin(2x) + Bcos(2x)

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

y′

p= 2Acos(2x)−2Bsin(2x)

y′′

p=−4Asin(2x)−4Bcos(2x)

Step 2: Substitute yp,y′

p, and y′′

pinto the original differential equation:

(−4Asin(2x)−4Bcos(2x))−2(2Acos(2x)−2Bsin(2x))−3(Asin(2x)+Bcos(2x)) = 15 sin(2x)

Step 3: Simplify the equation and equate coefficients of like terms:

(−4A+ 4B−3A) sin(2x)+(−4B−4A−3B) cos(2x) = 15 sin(2x)

Equating coefficients:

−7A+ 4B= 15

−7B−4A= 0

Solving these equations simultaneously, we find A=−60

65 =−12

13 and B=

−84

65 =−12

Step 4: Therefore, the particular solution is:

yp=−12

13 sin(2x)−12

5cos(2x)

So, the particular solution to the differential equation is yp=−12

13 sin(2x)−

5cos(2x).

Question 28

Question

Consider the following differential equation:

y′′ −3y′+ 2y= 2e4x

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

differential equation.

Solution

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

undetermined coefficients, we will first find the complementary solution, then

determine the form of the particular solution.

Step 1: Find the Complementary Solution The characteristic equation

for the corresponding homogeneous differential equation y′′ −3y′+ 2y= 0 is:

r2−3r+ 2 = 0

This equation factors as (r−1)(r−2) = 0, so the complementary solution is:

yc=c1ex+c2e2x

Step 2: Determine the Form of the Particular Solution Since the

right-hand side of the differential equation is 2e4x, we guess a particular solution

of the form:

yp=Ae4x

Step 3: Find the Derivatives of the Particular Solution

y′

p= 4Ae4xand y′′

p= 16Ae4x

Step 4: Substitute the Particular Solution into the Differential

Equation Substitute yp,y′

p, and y′′

pinto the differential equation:

16Ae4x−3(4Ae4x)+2Ae4x= 2e4x

Step 5: Solve for A Simplify the left side of the equation:

(16A−12A+ 2A)e4x= 2e4x

6Ae4x= 2e4x

A=1

Step 6: Write Down the Particular Solution The particular solution

is:

yp=1

3e4x

Therefore, the general solution to the differential equation is:

y=yc+yp=c1ex+c2e2x+1

3e4x

Question 29

Question

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

4y′+ 4y= 4e2xusing the method of undetermined coefficients.

Solution

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

homogeneous equation:

y′′ −4y′+ 4y= 0

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

So, the complementary solution is:

yc(x) = c1e2x+c2xe2x

Step 2: Next, find the particular solution yp(x) for the nonhomogeneous

equation y′′ −4y′+ 4y= 4e2x. Since 4e2xhas the same form as the complemen-

tary solution, we try the particular solution in the form:

yp(x) = Bx2e2x

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

y′

p(x) = (2Bx +Bx2)e2x

y′′

p(x) = (2B+ 2Bx + 2Bx + 2Bx2)e2x= (2B+ 4Bx + 2Bx2)e2x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation:

(2B+ 4Bx + 2Bx2)e2x−4(2Bx +Bx2)e2x+ 4Bx2e2x= 4e2x

Step 5: Equate coefficients of similar terms to find B:

2B−8Bx + 6Bx2= 4

Step 6: Solve for Bby matching coefficients:

6B= 0 ⇒B= 0

Step 7: The particular solution is yp(x) = 0, so the general solution to the

nonhomogeneous equation is:

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

Therefore, the particular solution to the given nonhomogeneous differential

equation is y(x) = c1e2x+c2xe2x.

Question 30

Question

Consider the second-order linear non-homogeneous differential equation

y′′ −3y′+ 2y= 6e2x+ 5x.

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

this differential equation.

Solution

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

method of undetermined coefficients, we will first find the complementary solu-

tion and then the particular solution.

Step 1: Find the complementary solution The complementary solution

solves the corresponding homogeneous differential equation y′′ −3y′+ 2y= 0.

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

Thus, the complementary solution is

yc(x) = c1ex+c2e2x.

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

has the form

yp(x) = Ae2x+Bx +C.

Taking the derivatives,

y′

p(x)=2Ae2x+B, y′′

p(x)=4Ae2x.

Substitute yp,y′

p, and y′′

pinto the differential equation:

4Ae2x−3(2Ae2x+B) + 2(Ae2x+Bx +C)=6e2x+ 5x.

Simplify and equate coefficients:

(−6A+ 2B)e2x+ (2A+ 2B) = 6e2x+ 5x.

Equating coefficients of like terms, we get the system of equations:

−6A+ 2B= 6,

2A+ 2B= 5.

Solving the system of equations gives A=−1 and B= 3. Therefore, the

particular solution is

yp(x) = −e2x+ 3x+C.

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

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

ular solutions:

y(x) = yc(x) + yp(x) = c1ex+c2e2x−e2x+ 3x+C).

Question 31

Question

Solve the following second-order homogeneous differential equation using the

method of undetermined coefficients: y′′ −3y′+ 2y= 3e2x+ 5 sin(x).

Solution

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

method of undetermined coefficients, we need to follow these steps:

Step 1: Find the complementary solution ycby solving the homogeneous

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

factors to (r−1)(r−2) = 0. So, the complementary solution is yc=c1ex+c2e2x.

Step 2: Guess the form of the particular solution ypbased on the form

of the nonhomogeneous terms. Since the nonhomogeneous terms are 3e2xand

5 sin(x), we guess yp=Ae2x+Bsin(x) + Ccos(x).

Step 3: Find y′

pand y′′

p:y′

p= 2Ae2x+Bcos(x)−Csin(x), y′′

p= 4Ae2x−

Bsin(x)−Ccos(x).

Step 4: Substitute yp,y′

p, and y′′

pinto the original differential equation and

solve for the coefficients A,B, and C: 4Ae2x−Bsin(x)−Ccos(x)−3(2Ae2x+

Bcos(x)−Csin(x)) + 2(Ae2x+Bsin(x) + Ccos(x)) = 3e2x+ 5 sin(x).

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

and solve for A,B, and C. Solving for A,B, and Cwe get A=3

4,B=−4

and C=3

Step 6: Write down the general solution by adding the complementary

solution and the particular solution: y=yc+yp=c1ex+c2e2x+3

4e2x−

5sin(x) + 3

5cos(x).

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

3e2x+ 5 sin(x) is y=c1ex+c2e2x+3

4e2x−4

5sin(x) + 3

5cos(x).

Question 32

Question

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

nonhomogeneous differential equation

y′′ −3y′+ 2y= 6ex.

Solution

To find a particular solution to the differential equation, we assume the par-

ticular solution has the form yp=Aex.We then substitute this form into the

differential equation and solve for A.

Step 1: Find the derivatives of yp:Since yp=Aex,we have:

y′

p=Aex

y′′

p=Aex

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

tion: Substitute yp, y′

p,and y′′

pinto the differential equation:

(Aex)′′ −3(Aex)′+ 2(Aex)=6ex

Aex−3Aex+ 2Aex= 6ex

Step 3: Solve for A:Simplify the equation obtained in Step 2 and solve

for A:

Aex= 6ex

A= 6

Thus, the particular solution to the given differential equation is yp= 6ex.

Question 33

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −y′−2y= 2et+ 3 sin(t)

Solution

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

mined coefficients, we first need to find the general solution to the corresponding

homogeneous equation and then find a particular solution to the nonhomoge-

neous equation.

Step 1: Find the homogeneous solution The corresponding homoge-

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

which factors to (r−2)(r+ 1) = 0, giving us roots r1= 2 and r2=−1.

Therefore, the homogeneous solution is yh=c1e2t+c2e−t.

Step 2: Find the particular solution For the particular solution, we

consider the right-hand side of the nonhomogeneous equation as a sum of the

individual functions: 2etand 3 sin(t).

Let’s assume the particular solution has the form yp=Aet+Bsin(t) +

Ccos(t), where A,B, and Care constants to be determined.

Substitute ypand its derivatives into the differential equation to determine

the values of A,B, and C.

We have:

y′′

p−y′

p−2yp= 2et+ 3 sin(t)

Substitute ypinto the equation and solve for A,B, and C.

Step 3: Solve for the unknown coefficients First, find the derivatives

of yp:

y′

p=Aet+Bcos(t)−Csin(t)

y′′

p=Aet−Bsin(t)−Ccos(t)

Now, substitute yp,y′

p, and y′′

pinto the differential equation:

(Aet−Bsin(t)−Ccos(t))−(Aet+Bcos(t)−Csin(t))−2(Aet+Bsin(t)+Ccos(t)) = 2et+3 sin(t)

Solve for A,B, and Cby comparing coefficients of et, sin(t), and cos(t) on

both sides of the equation.

Step 4: Write the general solution Once we have found the values of

A,B, and C, the particular solution ypis determined. Therefore, the general

solution to the nonhomogeneous differential equation is:

y=yh+yp=c1e2t+c2e−t+Aet+Bsin(t) + Ccos(t)

Where c1,c2,A,B, and Care constants to be determined based on initial

conditions or boundary conditions.

Question 34

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −3y′+ 2y= 4ex−3 sin x

Solution

Step 1: First, find the complementary function by solving the associated homo-

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

0, which factors to (r−1)(r−2) = 0. So the complementary function is

yc=c1ex+c2e2x.

Step 2: Next, find the particular solution yp. Since the right-hand side

contains exand sin x, we assume a form for yp:

yp=Aex+Bsin x+Ccos x

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

y′

p=Aex+Bcos x−Csin x

y′′

p=Aex−Bsin x−Ccos x

Step 4: Substitute yp, y′

p, y′′

pinto the original differential equation and solve

for the coefficients A, B, C. Plugging into the equation y′′ −3y′+ 2y= 4ex−

3 sin xgives:

Aex−Bsin x−Ccos x−3(Aex+Bcos x−Csin x)+2(Aex+Bsin x+Ccos x)=4ex−3 sin x

Step 5: Equate the terms with similar functions on both sides and solve for

A, B, C. For the terms with ex:

A−3A+ 2A= 4 =⇒0=4

This shows that there is no solution for Awhich means the form for ypis

incorrect.

Step 6: Since the first guess was not successful, let’s try a new form for yp

involving a polynomial of degree 1 times ex:

yp= (Ax +B)ex

Step 7: Calculate the first two derivatives of the revised yp:

y′

p= (A+Ax +B)ex

y′′

p= (2A+Ax +B)ex

Step 8: Substitute yp, y′

p, y′′

pinto the original differential equation and solve

for the coefficients A, B. Plugging into the equation y′′ −3y′+2y= 4ex−3 sin x

gives:

(2A+Ax +B)ex−3(A+Ax +B)ex+ 2(Ax +B)ex= 4ex−3 sin x

Step 9: Equate the terms with exon both sides and solve for A, B. This will

give you the values for Aand Bneeded to find yp.

Step 10: Finally, write down the general solution by combining the comple-

mentary function and the particular solution:

y=yc+yp=c1ex+c2e2x+ (Ax +B)ex

Question 35

Question

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

64 cos(x) using the method of undetermined coefficients.

Solution

We first find the complementary solution to the homogeneous equation: y′′ −

4y′+ 4y= 0.



Characteristic equation: r2−4r+ 4 = 0



Roots: r= 2 (with multiplicity 2)



Complementary solution: yc=c1e2x+c2xe2x

Next, we proceed to find a particular solution to the non-homogeneous equa-

tion: y′′ −4y′+ 4y= 10e2x+ 64 cos(x).



Assume the particular solution has the form yp=Ae2x+Bcos(x).



Calculate the first and second derivatives of yp:

y′

p= 2Ae2x−Bsin(x)

y′′

p= 4Ae2x−Bcos(x)



Substitute these derivatives back into the differential equation:

4Ae2x−Bcos(x)−4(2Ae2x−Bsin(x)) + 4(Ae2x+Bcos(x)) = 10e2x+ 64 cos(x)

−4Ae2x+ 4Bsin(x)+4Ae2x−4Bsin(x)+4Ae2x+ 4Bcos(x) = 10e2x+ 64 cos(x)

4Ae2x+ 4Bcos(x) = 10e2x+ 64 cos(x)



Solve for Aand Bby equating coefficients:

4A= 10 4B= 64

A=5

2B= 16



Therefore, our particular solution is yp=5

2e2x+ 16 cos(x).

The general solution to the original non-homogeneous equation is the sum

of the complementary and particular solutions: y=yc+yp=c1e2x+c2xe2x+

2e2x+ 16 cos(x).

MATH 334 - DIFFERENTIAL EQUATIONS Method of undetermined coefficients Question Bank Set 4

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