MATH 334 - DIFFERENTIAL EQUATIONS Method of undetermined coefficients Question Bank Set 3

MATH 334 - DIFFERENTIAL

EQUATIONS - Method of

undetermined coefficients

Question Bank - Set 3

Liberty University

Question 1

Question

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

undetermined coefficients:

y′′ −3y′+ 2y= 6e2x

Solution

Step 1: Find the complementary solution First, find the solutions to the

associated 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(x) = c1ex+c2e2x

Step 2: Guess the form of the particular solution We guess that the

particular solution has the form:

yp(x) = Ae2x

Step 3: Find the first and second derivatives of the guess

yp(x) = Ae2x

y′

p(x)=2Ae2x

y′′

p(x)=4Ae2x

Step 4: Substitute into the original equation and solve for ASub-

stitute yp(x), y′

p(x), and y′′

p(x) into the original equation:

4Ae2x−3(2Ae2x) + 2(Ae2x)=6e2x

Simplify to get:

3Ae2x= 6e2x

Therefore, A= 2.

Step 5: Write the particular solution The particular solution is:

yp(x)=2e2x

Step 6: Write the general solution The general solution is the sum of

the complementary and particular solutions:

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

Question 2

Question

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

efficients: y′′ −2y′+y=x2ex.

Solution

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

efficients, we will first find the complementary function yc(x) by solving the

characteristic equation. Then, we will find a particular solution yp(x) by as-

suming a suitable form based on the non-homogeneous term of the differential

equation.

Step 1: Find the complementary function The characteristic equation

for y′′ −2y′+y= 0 is r2−2r+ 1 = 0. Solving this, we get (r−1)2= 0 which

gives us a repeated root r= 1. Therefore, the complementary function is of the

form yc(x) = c1ex+c2xex.

Step 2: Assume the particular solution For y′′ −2y′+y=x2ex, we

will assume a particular solution of the form yp(x)=(Ax2+Bx +C)ex, where

A,B, and Care constants to be determined.

Step 3: Find the derivative of yp(x)

y′

p(x) = (2Ax +B+A)x2ex

y′′

p(x) = (2A+ 2)x2ex+ 4Ax2ex

Step 4: 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) into the differential equation y′′ −2y′+

y=x2ex:

(2A+ 2)x2ex+ 4Ax2ex−2(2Ax +B+A)x2ex+ (Ax2+Bx +C)ex=x2ex

Step 5: Simplify and solve for constants Equating coefficients of like

terms on both sides of the equation, we get the following system of equations:

2A−2A+ 2 = 0 (coefficients of x2ex)

4A−2B+A= 1 (coefficients of xex)

−2B+B+C= 0 (coefficients of ex)

Solving this system of equations, we find A= 1, B= 2, and C= 2.

Step 6: Form the general solution Therefore, the general solution to

the differential equation y′′ −2y′+y=x2exis:

y(x) = yc(x) + yp(x) = c1ex+c2xex+ (x2+ 2x+ 2)ex

Question 3

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x+ 5x2−1

Solution

Step 1: First, we find the complementary solution ycby solving the corre-

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

is r2−3r+ 2 = (r−1)(r−2) = 0, so the solutions are yc=c1ex+c2e2x.

Step 2: Next, we determine the particular solution ypfor the nonho-

mogeneous equation. Since the right-hand side is a sum of exponential and

polynomial functions, we assume the particular solution has the form yp=

Ae2x+Bx2+Cx +D. We then substitute ypback into the differential equation

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

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

y′

p= 2Ae2x+ 2Bx +C

y′′

p= 4Ae2x+ 2B

Substitute yp,y′

p, and y′′

pinto the differential equation:

4Ae2x+ 2B−3(2Ae2x+ 2Bx +C) + 2(Ae2x+Bx2+Cx +D)=6e2x+ 5x2−1

Step 4: Simplify the equation and match coefficients of like terms. This

will give us a system of equations to solve for A,B,C, and D.

Let’s simplify and equate coefficients:

A−2B+ 2C= 0

4A−3(2B)+2C= 6

2B+ 2D=−1

4A−3(2B)+2C= 0

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

substitute these values back into ypto get the particular solution.

After solving the system of equations, we find A= 2, B=−1, C= 1, and

D=−1

2. Thus, the particular solution is yp= 2e2x−x2+x−1

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

solutions: y=yc+yp. Therefore, the final solution to the differential equation

is:

y=c1ex+c2e2x+ 2e2x−x2+x−1

Question 4

Question

Find the particular solution to the differential equation y′′ −2y′−3y= 2e3x+x

using the method of undetermined coefficients.

Solution

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

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

factors as (r−3)(r+ 1) = 0. So, the roots are r= 3 and r=−1, making the

complementary solution yc=c1e3x+c2e−x.

Step 2: Next, we consider the form of the particular solution. Since the right-

hand side of the given equation includes terms 2e3xand xwhich are similar to

the terms in the complementary solution, we need to use a modification of the

form of the complementary solution for the particular solution. Therefore, we

assume the particular solution has the form:

yp=Ax2e3x+Bx +C

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

y′

p= (2Ax +B)e3x+ 3Ax2e3x

y′′

p= (6Ax2+ 6Ax + 2B)e3x+ 6Ax2e3x

Step 4: Substitute yp,y′

p, and y′′

pinto the original differential equation and

simplify:

(6Ax2+6Ax+2B−2(2Ax+B)−3(Ax2e3x+Bx+C))e3x−3(Ax2e3x+Bx+C)=2e3x+x

Step 5: Collecting like terms, we get the following equations:

(−6A+ 4B−3A)x2e3x= 0

(6A−6A−6B−3B)x= 1

(6A+ 6A+ 2B−3C)=0

Step 6: Solve the system of equations to find the values of A,B, and C.

Step 7: After finding the values of A,B, and C, the particular solution yp

can be written as:

yp= (value of A)x2e3x+ (value of B)x+ (value of C)

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

(value of A)x2e3x+ (value of B)x+ (value of C).

Question 5

Question

Find the particular solution to the following non-homogeneous second-order

differential equation using the method of undetermined coefficients:

y′′ −4y′+ 4y= 3e2x+ 10x2

Solution

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

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

which factors to (r−2)2= 0. So the complementary solution is of the form

yc(x) = c1e2x+c2xe2x.

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

equation. Since 3e2xis of the same form as the complementary solution, we

need to use xtimes a function of xfor that term. Since the repeated root has

multiplicity 2, we take yp(x) = Ax2e2x.

Step 3: Now, we find the derivative terms needed for the equation: y′

p(x) =

(2Ax + 2A)e2xand y′′

p(x) = (4A+ 4A)xe2x+ 4Ae2x.

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the original non-homogeneous

equation:

y′′

p−4y′

p+ 4yp= 3e2x+ 10x2

(4A+ 4A)xe2x+ 4Ae2x−4(2Ax + 2A)e2x+ 4Ax2e2x= 3e2x+ 10x2

Step 5: Simplify the equation and match coefficients on both sides:

(4A+ 4A)xe2x+ 4Ae2x−8Axe2x−8Ae2x+ 4Ax2e2x= 3e2x+ 10x2

Step 6: Equating the coefficients of like terms:

For e2x:

4A−8A= 3 =⇒ −4A= 3 =⇒A=−3

For x2e2x:

4A= 10 =⇒4(−3

4) = 10 =⇒ −3 = 10

There is no solution for A, so we correct the form of ypby multiplying by x:

yp(x) = −3

4x2e2x

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

ticular solutions:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x−3

4x2e2x

Question 6

Question

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

3y′+ 2y= 3e2xusing the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary solution of the corresponding homoge-

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

which factors as (r−1)(r−2) = 0. So the complementary solution is yc(x) =

c1ex+c2e2x.

Step 2: Next, we assume the particular solution has the form yp(x) = Ae2x,

where Ais a constant to be determined.

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

y′′ −3y′+ 2y= (A·4e2x)−3(A·2e2x) + 2(Ae2x)

= 4Ae2x−6Ae2x+ 2Ae2x= 3e2x

Step 4: Simplify the equation 3Ae2x= 3e2xand solve for A. This gives

A= 1.

Step 5: Therefore, the particular solution is yp(x) = e2x.

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

the particular solution: y(x) = yc(x) + yp(x) = c1ex+c2e2x+e2x, where c1and

c2are arbitrary constants.

Question 7

Question

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

3y′+ 2y= 2e−t+ 3 sin(2t) using the method of undetermined coefficients.

Solution

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

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

which factors to (r+ 1)(r+ 2) = 0. Therefore, the complementary solution is

yc(t) = c1e−t+c2e−2t.

Step 2: Next, we assume a particular solution of the form yp(t) = Ae−t+

Bsin(2t) + Ccos(2t). Then calculate the first and second derivatives: y′

p(t) =

−Ae−t+ 2Bcos(2t)−2Csin(2t), y′′

p(t) = Ae−t−4Bsin(2t)−4Ccos(2t).

Step 3: Substitute yp(t), y′

p(t), and y′′

p(t) into the differential equation:

(Ae−t−4Bsin(2t)−4Ccos(2t))+3(−Ae−t+2Bcos(2t)−2Csin(2t))+2(Ae−t+

Bsin(2t) + Ccos(2t)) = 2e−t+ 3 sin(2t).

Step 4: Group terms with the same functions together and solve for A,B,

and C: (A+3A+2A)e−t+(−4B+6B+2B) sin(2t)+(−4C−6C+2C) cos(2t) =

2e−t+ 3 sin(2t). Simplify to get: 6Ae−t+ 4Bsin(2t)−8Ccos(2t) = 2e−t+

3 sin(2t).

Step 5: Equate coefficients to find A,B, and C: 6A= 2 =⇒A=1

4B= 3 =⇒B=3

4,−8C= 0 =⇒C= 0.

Step 6: Finally, the particular solution is yp(t) = 1

3e−t+3

4sin(2t). Therefore,

the general solution is y(t) = yc(t) + yp(t) = c1e−t+c2e−2t+1

3e−t+3

4sin(2t).

Question 8

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 first need to find the complementary solution.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous part of the differential equation is r2−5r+ 6 = 0. This

factors as (r−2)(r−3) = 0, so the characteristic roots are r1= 2 and r2= 3.

Therefore, the complementary solution is

yc(x) = c1e2x+c2e3x

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

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

Step 3: Calculate the derivatives We have y′

p(x) = 2Ae2xand y′′

p(x) =

4Ae2x.

Step 4: Substitute into the differential equation Substitute yp(x),

y′

p(x), and y′′

p(x) into the differential equation y′′ −5y′+ 6y= 3e2x:

4Ae2x−5(2Ae2x) + 6(Ae2x)=3e2x

Step 5: Solve for A Solving the above equation for Agives:

A=3

6−4−5=−1

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

solution is

yp(x) = −e2x

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

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

y(x) = yc(x) + yp(x) = c1e2x+c2e3x−e2x

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

c1e2x+c2e3x−e2x.

Question 9

Question

Solve the following second order linear differential equation using the method

of undetermined coefficients:

y′′ −y= 2 sin(3x)+3ex

Solution

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

efficients, we first find the complementary function yc(x) by solving the corre-

sponding homogeneous equation y′′ −y= 0.

The characteristic equation is r2−1 = 0, which has roots r=±1. Therefore,

the complementary function is given by:

yc(x) = C1ex+C2e−x

Next, we look for the particular solution yp(x) based on the form of the

forcing term. Since the right-hand side includes 2 sin(3x) and 3ex, we assume

the particular solution to have the form:

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

Now, we find the derivatives of yp(x) needed to substitute back into the

original differential equation:

y′

p(x) = 3Acos(3x) + Bex

y′′

p(x) = −9Asin(3x) + Bex

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

tion:

−9Asin(3x) + Bex−Asin(3x)−Bex= 2 sin(3x)+3ex

Matching coefficients of like terms, we get the following system of equations:

−9A−A= 2

B−B= 3

Solving this system, we find A=−2

10 =−1

5and B= 3.

Therefore, the particular solution is:

yp(x) = −1

5sin(3x)+3ex

The general solution to the original differential equation is the sum of the

complementary function and the particular solution:

y(x) = yc(x) + yp(x) = C1ex+C2e−x−1

5sin(3x)+3ex

Thus, the solution to the differential equation is:

y(x) = C1ex+C2e−x−1

5sin(3x)+3ex

Question 10

Question

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

3y′+ 2y= 4exusing the method of undetermined coefficients.

Solution

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

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

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

is yc=c1ex+c2e2x.

Step 2: Now, we guess the form of the particular solution ypto be of the

form Aex. Substitute yp=Aexback into the differential equation: (Aex)′′ −

3(Aex)′+ 2(Aex)=4ex.

Step 3: Calculate the derivatives and substitute them back into the dif-

ferential equation. We have Aex−3Aex+ 2Aex= 4ex. Simplifying, we get

Aex= 4exwhich implies A= 4.

Step 4: Therefore, the particular solution is yp= 4ex.

Step 5: The general solution is given by adding the complementary and

particular solutions: y=yc+yp. Hence, y=c1ex+c2e2x+ 4exis the general

solution to the non-homogeneous differential equation.

Question 11

Question

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

undetermined coefficients:

y′′ −4y′+ 4y= 2e2x.

Solution

Step 1: Find the complementary solution yc(x) by solving the associated homo-

geneous 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 then given by yc(x) = c1e2x+c2xe2x.

Step 2: Assume a particular solution of the form yp(x) = Ae2x. Now, we

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

y′

p(x)=2Ae2xand y′′

p(x) = 4Ae2x.

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential

equation:

4Ae2x−4(2Ae2x)+4Ae2x= 2e2x.

Simplify this equation to get 4Ae2x−8Ae2x+4Ae2x= 2e2x, which simplifies

to 0 = 2e2x, which is not true.

Step 4: Since Ae2xdoes not satisfy the original differential equation, we

need to modify our guess to include a factor of xto account for the repeated

root:

yp(x) = Ax2e2x.

Step 5: Now, find the first and second derivatives of yp(x):

y′

p(x) = (2Ax + 2A)e2xand y′′

p(x) = (4Ax2+ 8Ax + 2A)e2x.

Step 6: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential

equation:

(4Ax2+ 8Ax + 2A)e2x−4(2Ax + 2A)e2x+ 4Ax2e2x= 2e2x.

Step 7: Simplify this equation to get (4Ax2+ 8Ax + 2A−8Ax −8A+

4Ax2)e2x= 2e2x, which simplifies to (8Ax2−6A)e2x= 2e2x.

Step 8: Equate the coefficients of like terms:

8Ax2−6A= 2.

Step 9: Solve for A:

8A= 8 ⇒A= 1.

Step 10: The particular solution is yp(x) = x2e2x. Thus, the general

solution to the original differential equation is:

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

Question 12

Question

Find the particular solution to the nonhomogeneous differential equation:

y′′ −4y′+ 4y= 2e2x+ 5x

using the method of undetermined coefficients.

Solution

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

we first need to find the complementary solution. Let’s assume yp(x) is the

particular solution we are trying to find.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous part of the differential equation is:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2

Thus, the complementary solution is:

yc(x) = c1e2x+c2xe2x

Step 2: Find the particular solution Our particular solution yp(x)

should have the same form as the nonhomogeneous terms, so we can guess

that yp(x) = Ae2x+Bx +C.

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

Ae2x+Bx +Cand its derivatives into the differential equation:

y′′ −4y′+ 4y= 2e2x+ 5x

(4A−4B)e2x+ 5 = 2e2x+ 5x

Step 4: Solve for the coefficients Equating coefficients, we get:

4A−4B= 2

5=5

Solving the equations, we find A=1

2and B= 0. We can disregard the constants

here.

Step 5: Assemble the particular solution Therefore, the particular

solution is:

yp(x) = 1

2e2x

Step 6: Find the general solution The general solution is the sum of

the complementary solution and the particular solution:

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

2e2x

So, the particular solution to the nonhomogeneous differential equation is

y(x) = c1e2x+c2xe2x+1

2e2x.

Question 13

Question

Find the particular solution to the differential equation x′′ −4x′+4x=e2tcos(t)

using the method of undetermined coefficients.

Solution

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

we need to guess a form for the particular solution based on the form of the

non-homogeneous term. Since e2tcos(t) is a product of an exponential and

trigonometric function, we guess the form of the particular solution to be xp(t) =

(At2+Bt +C)e2tcos(t) + (Dt2+Et +F)e2tsin(t), where A,B,C,D,E, and

Fare constants to be determined.

Step 1: Calculate the derivatives of xp(t).

x′

p(t) = (2At+B)e2tcos(t)−(At2+Bt+C)e2tsin(t)+(2Dt+E)e2tsin(t)+(Dt2+Et+F)e2tcos(t)

x′′

p(t) = (2A)e2tcos(t)+(4At+2B)e2tsin(t)−(4At+2B+2C)e2tcos(t)+(2Dt+E)e2tcos(t)−(2Dt+E)e2tsin(t)+(2D+2E)e2tsin(t)−(Dt2+Et+F)e2tsin(t)−(2Dt+E)e2tcos(t)

Step 2: Substitute xp(t), x′

p(t), and x′′

p(t) into the differential equation

x′′ −4x′+ 4x=e2tcos(t) and simplify.

LHS = (2A)e2tcos(t)+(4At+2B)e2tsin(t)−(4At+2B+2C)e2tcos(t)+(2Dt+E)e2tcos(t)−(2Dt+E)e2tsin(t)+(2D+2E)e2tsin(t)−(Dt2+Et+F)e2tsin(t)−(2Dt+E)e2tcos(t)

RHS = e2tcos(t)

Step 3: Equate the coefficients of like terms on both sides to determine the

values of A,B,C,D,E, and F.

2A−4C= 1 (e2tcos(t) terms)

4A+ 2B−4C+ 2E= 0 (e2tsin(t) terms)

−4A+ 2B+ 2E= 0 (e2tcos(t) terms)

2D+ 2E= 0 (e2tsin(t) terms)

−2D−E= 0 (e2tcos(t) terms)

D= 0 (e2tsin(t) terms)

Step 4: Solve the system of equations to find the values of A,B,C,D,E,

and F, then substitute back into xp(t) to get the particular solution.

Step 5: Finally, the particular solution is

xp(t) = 3

10t2+1

5t+1

10e2tcos(t)

Question 14

Question

Find the particular solution of the nonhomogeneous differential equation:

y′′ −4y′+ 4y= 6e2x

Solution

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

we assume the particular solution has the form yp(x) = Ae2x.



Step 1: Find the derivatives of yp.

y′

p= 2Ae2x

y′′

p= 4Ae2x



Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation.

4Ae2x−4(2Ae2x) + 4(Ae2x)=6e2x



Step 3: Simplify the equation and solve for A.

4Ae2x−8Ae2x+ 4Ae2x= 6e2x

0=6e2x

This equation has no solutions, so we need to add a multiplying factor of

xto yp(x).



Step 4: Update the form of yp(x) to include the additional factor. Let’s

try yp(x) = Axe2x.

yp=Axe2x

y′

p=Ae2x+ 2Axe2x

y′′

p= 2Ae2x+ 2Ae2x+ 4Axe2x



Step 5: Substitute yp,y′

p, and y′′

pinto the differential equation.

2Ae2x+ 2Ae2x+ 4Axe2x−4(Ae2x+ 2Axe2x) + 4(Axe2x)=6e2x



Step 6: Simplify the equation and solve for A.

6Ae2x= 6e2x

A= 1



Step 7: The particular solution is yp(x) = xe2x.

Question 15

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6ex−5x2+ 2

Solution

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

0 to find the complementary function yc(x). The characteristic equation is

r2−3r+2 = 0, which factors as (r−1)(r−2) = 0. Thus, the roots are r1= 1 and

r2= 2. Therefore, the complementary function is given by yc(x) = C1ex+C2e2x,

where C1and C2are arbitrary constants.

Step 2: Next, we assume the particular solution has the form yp(x) =

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

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

y′

p(x) = Aex+ 2Bx +C

y′′

p(x) = Aex+ 2B

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential

equation and simplify to solve for A,B,C, and D.

Aex+ 2B−3(Aex+ 2Bx +C) + 2(Aex+Bx2+Cx +D)=6ex−5x2+ 2

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

to find the values of A,B,C, and D. Solving these simultaneous equations will

give the particular solution yp(x).

Step 6: Finally, the general solution is the sum of the complementary func-

tion and particular solution:

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

y(x) = C1ex+C2e2x+ (Aex+Bx2+Cx +D)

Question 16

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −3y′+ 2y= 5e2x+ 4 sin x

Solution

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

y′′ −3y′+ 2y= 0

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

Thus, the characteristic roots are r1= 1 and r2= 2. The complementary

solution is:

yc(x) = c1ex+c2e2x

Step 2: Determine the form of the particular solution based on the form

of the nonhomogeneous terms. Since the right-hand side contains e2x(which

is also a solution to the homogeneous equation), we multiply by xto make it

linearly independent:

yp(x) = Axe2x+Bsin x+Ccos x

Now, find the derivatives of yp(x):

y′

p(x) = Ae2x+ 2Axe2x+Bcos x−Csin x

y′′

p(x)=3Ae2x+ 4Axe2x−Bsin x−Ccos x

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential equa-

tion to determine the coefficients A,B, and C. The equation becomes:

3Ae2x+4Axe2x−Bsin x−Ccos x−3(Ae2x+2Axe2x+Bcos x−Csin x)+2(Axe2x+Bsin x+Ccos x)=5e2x+4 sin x

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

to solve for A,B, and C. This will lead to a system of equations which can be

solved to find the values of the constants.

Step 5: Once you have the values of A,B, and C, the particular solution

yp(x) can be written. Add the complementary and particular solutions to get

the general solution:

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

Question 17

Question

Solve the following second order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 6e2x

Solution

To solve this differential equation, we first find the complementary function by

solving the associated homogeneous equation:

y′′ −4y′+ 4y= 0

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

So, we have a repeated root r= 2.

Therefore, the complementary function ycis given by yc=c1e2x+c2xe2x.

Next, we find the particular solution by assuming yp=Ae2xfor the nonho-

mogeneous term 6e2x.

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

yp=Ae2x

y′

p= 2Ae2x

y′′

p= 4Ae2x

Step 2: Substitute ypand its derivatives back into the original differential

equation:

y′′ −4y′+ 4y= 6e2x

4Ae2x−4(2Ae2x) + 4(Ae2x)=6e2x

4Ae2x−8Ae2x+ 4Ae2x= 6e2x

0=6e2x

Since 0 = 6e2x, we try a different form for yp. Let’s try yp=Ax2e2xthis

time.

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

yp=Ax2e2x

y′

p= (2Ax + 2x2)e2x

y′′

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

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

equation:

y′′ −4y′+ 4y= 6e2x

(4A+ 4x+ 4x)e2x−4(2Ax + 2x2)e2x+ 4(Ax2e2x)=6e2x

(4A+ 8x)e2x−8Ax −8x2e2x+ 4Ax2e2x= 6e2x

Equating the coefficients of e2xand x2e2xon both sides, we find:

4A= 6 and 4A−8=0

Solving these equations, we get A=3

Therefore, the particular solution is yp=3

2x2e2x.

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

the particular solution:

y=yc+yp=c1e2x+c2xe2x+3

2x2e2x

Question 18

Question

Find the particular solution to the differential equation y′′ −4y′+4y= 8e2x+6x

using the method of undetermined coefficients.

Solution

Step 1: Find the homogeneous solution to the differential equation y′′ −4y′+4y=

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

repeated root r= 2. So, the general solution to the homogeneous equation is

yh(x) = c1e2x+c2xe2x.

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

hand side of the differential equation contains terms of the form e2xand x, the

particular solution can be assumed in the form: yp(x) = Ae2x+Bx +C

Step 3: Find the derivatives of yp(x). y′

p(x)=2Ae2x+B y′′

p(x)=4Ae2x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) back into the differential equation

and solve for A,B, and C. Substituting into the differential equation, we get:

4Ae2x−4(2Ae2x+B) + 4(Ae2x+Bx +C)=8e2x+ 6x

Simplifying the equation, we get: 4Ae2x−8Ae2x−4B+4Ae2x+4Bx+4C=

8e2x+ 6x

−4B+ 4Bx + 4C= 8e2x+ 6x

Comparing coefficients on both sides, we get: 4B= 6 and 4C= 8

Step 5: Calculate the values of Band C. From 4B= 6, we find B=3

From 4C= 8, we find C= 2.

Therefore, the particular solution is: yp(x) = Ae2x+3

2x+ 2 where Ais an

arbitrary constant.

Thus, the general solution to the differential equation is: y(x) = yh(x) +

yp(x) = c1e2x+c2xe2x+Ae2x+3

2x+ 2 where c1,c2, and Aare arbitrary

constants.

Question 19

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −y′−2y= 3ex+ 2 cos(x)

Given that y1(x) = exand y2(x) = e−2xare solutions to the homogeneous

equation y′′ −y′−2y= 0.

Solution

Step 1: Write down the general form of the particular solution yp(x):

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

Step 2: Find the derivatives of yp(x).

y′

p(x) = Aex−Bsin(x) + Ccos(x)

y′′

p(x) = Aex−Bcos(x)−Csin(x)

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential equa-

tion.

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

Step 4: Group like terms and solve for A,B, and C.

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

To solve for A,B, and C, we set the coefficients of ex, cos(x), and sin(x) on

both sides equal:











−2A+B= 3

B−2C= 2

C−2A= 0

By solving this system of equations, we find A=−3

5,B=−4

5, and C=−6

Therefore, the particular solution is:

yp(x) = −3

5ex−4

5cos(x)−6

5sin(x)

Step 5: The general solution to the non-homogeneous differential equation

is given by:

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

y(x) = c1ex+c2e−2x−3

5ex−4

5cos(x)−6

5sin(x)

where c1and c2are constants determined by initial conditions or boundary

conditions.

Question 20

Question

Find a particular solution to the following differential equation using the method

of undetermined coefficients:

y′′ −3y′+ 2y= 12e2x.

Solution

To find a 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 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)+2Ae2x= 12e2x.

Step 3: Simplify and solve for A.

4Ae2x−6Ae2x+ 2Ae2x= 12e2x,

(4 −6 + 2)Ae2x= 12e2x,

0 = 12e2x.

Since 0 = 12e2xfor any value of x, our assumption was incorrect. To fix

this, we need to multiply our assumed solution by xand try again.

Step 4: Revise the assumed particular solution. Assume the partic-

ular solution has the form:

yp=x(Ae2x),

where Ais a constant to be determined.

Step 5: Find the first and second derivatives of the revised yp.

yp=x(Ae2x),

y′

p=Ae2x+ 2Axe2x,

y′′

p= 2Ae2x+ 2Ae2x+ 4Axe2x.

Question 21

Question

Find the particular solution to the differential equation y′′ −4y= 6xe2xusing

the method of undetermined coefficients.

Solution

Step 1: To use the method of undetermined coefficients, we first write the com-

plementary solution to the homogeneous equation y′′ −4y= 0. The character-

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

solution is yc=Ae2x+Be−2x.

Step 2: Next, we consider the form of the particular solution yp. Since 6xe2x

has the same form as xe2x, we try yp= (Ax +B)e2x.

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

yp= (Ax +B)e2x

y′

p=Ae2x+ 2(Ax +B)e2x= (2Ax + (2A+ 2B))e2x

y′′

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

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

tion:

(4Ax + 4A+ 4B)e2x−4(Ax +B)e2x= 6xe2x

Step 5: Simplify and compare coefficients:

(4A−4A)x= 6x

4B−4A= 0

Step 6: Solve the system of equations to find Aand B: From the first

equation, 4A= 6 so A=3

From the second equation, 4B= 4Aso B=3

Therefore, the particular solution is yp=3

2x+3

2e2x.

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

solutions:

y=yc+yp=Ae2x+Be−2x+3

2x+3

2e2x

So, the particular solution to the differential equation y′′ −4y= 6xe2xis

y=Ae2x+Be−2x+3

2x+3

2e2x.

Question 22

Question

Find a particular solution to the differential equation y′′ −5y′+6y=e2xcos(3x)

using the method of undetermined coefficients.

Solution

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

y′′ −5y′+6y= 0. The characteristic equation is r2−5r+ 6 = (r−2)(r−3) = 0,

which gives us roots r1= 2 and r2= 3. Therefore, the complementary solution

is yc=c1e2x+c2e3x.

Step 2: Now, we look for a particular solution to the non-homogeneous

equation y′′ −5y′+ 6y=e2xcos(3x). Assume the particular solution has the

form yp=Ae2xcos(3x) + Be2xsin(3x).

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

p= (2Ae2xcos(3x)−

3Be2xsin(3x)) y′′

p= (4Ae2xcos(3x)+9Be2xsin(3x))

Step 4: Substitute yp,y′

p, and y′′

pback into the original differential equation:

(4Ae2xcos(3x)+9Be2xsin(3x))−5(2Ae2xcos(3x)−3Be2xsin(3x))+6(Ae2xcos(3x)+

Be2xsin(3x)) = e2xcos(3x)

Step 5: Equate coefficients of e2xcos(3x) on both sides: 4A−10A+6A= 1,

which gives A=−1

Step 6: Equate coefficients of e2xsin(3x) on both sides: 9B+15B+6B= 0,

which gives B= 0.

Step 7: Hence, the particular solution is yp=−1

8e2xcos(3x).

Step 8: The general solution is y=yc+yp=c1e2x+c2e3x−1

8e2xcos(3x),

where c1and c2are arbitrary constants.

Question 23

Question

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

the differential equation:

y′′ −4y′+ 4y= 4e2x+ 3x2

Solution

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

equation r2−4r+ 4 = 0:

r2−4r+ 4 = (r−2)2= 0

r= 2 (Repeated root)

yc=C1e2x+C2xe2x

Step 2: Write the particular solution in the form yp=Ae2x+Bx2+Cx +D

where A, B, C, D are constants to be determined.

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

y′

p= 2Ae2x+ 2Bx +C

y′′

p= 4Ae2x+ 2B

Step 4: Substitute yp,y′

p, and y′′

pinto the original differential equation:

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

Step 5: Simplify and equate coefficients of like terms:

(4A−8A+ 4A)e2x+ (2B−4B+ 4B)x+ (−4C+ 4C)x+ 4D= 4e2x+ 3x2

Step 6: Solve for A,B,C, and Dby comparing coefficients:

0 = 4,2=3,0 = 0,4D= 0

Step 7: From the system of equations, we find D= 0.

Therefore, the particular solution is:

yp= 0 + x2+ 0 + 0 = x2

Step 8: Finally, the general solution is:

y=yc+yp=C1e2x+C2xe2x+x2

Question 24

Question

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

undetermined coefficients:

y′′ −2y′+y= 6xex

Solution

Step 1: First, find the complementary solution ycof the corresponding ho-

mogeneous equation y′′ −2y′+y= 0, which is the characteristic equation

r2−2r+ 1 = 0. This gives us the repeated root r= 1, so the complementary

solution is:

yc(x) = c1ex+c2xex

Step 2: Next, guess the particular solution ypin the form yp(x) = (Ax2+

Bx +C)ex. Calculate y′

pand y′′

y′

p= (2Ax +B+A)xex

y′′

p= (2A+ 2)xex+ (2Ax +B+A)xex

Step 3: Substitute yp,y′

p, and y′′

pinto the original differential equation and

solve for A,B, and C:

(2A+2)xex+(2Ax+B+A)xex−2[(2Ax+B+A)xex]+(Ax2+Bx+C)ex= 6xex

Simplify and equate coefficients to find A,B, and C.

Step 4: Substitute the found values of A,B, and Cback into ypto get the

particular solution.

Step 5: The general solution is given by the sum of the complementary

solution ycand the particular solution yp:

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

y(x) = c1ex+c2xex+ (Ax2+Bx +C)ex

Question 25

Question

Find the particular solution to the differential equation y′′ −y′−6y= 2e3x+ 4x,

using the method of undetermined coefficients.

Solution

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

we first find the complementary solution by solving the characteristic equation

m2−m−6 = 0.

Step 1: Find the complementary solution. The characteristic equation is

m2−m−6 = 0. Factoring this equation, we get (m−3)(m+ 2) = 0, so

the roots are m= 3 and m=−2. Therefore, the complementary solution is

yc=c1e3x+c2e−2x.

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

the form yp=Ae3x+Bx +C. We then find the derivatives needed to substitute

into the differential equation.

Step 3: Find the derivatives of the particular solution.

y′

p= 3Ae3x+Band y′′

p= 9Ae3x.

Step 4: Substitute the particular solution and its derivatives into the dif-

ferential equation. Substitute yp,y′

p, and y′′

pinto the differential equation

y′′ −y′−6y= 2e3x+ 4x, and simplify to get:

9Ae3x−3Ae3x−B−6(Ae3x+Bx +C)=2e3x+ 4x.

Step 5: Solve for the coefficients. Equating the coefficients of like terms,

we get the following equations:

(9A−3A−6A)e3x−6B= 2e3x,

−B−6B= 4,

which simplifies to

0 = 2e3x,

−7B= 4.

From the second equation, we find that B=−4

7. Since the first equation has

no solution, we adjust our guess for yp.

Step 6: Adjust the guess for the particular solution. With yp=Ax2e3x+

Bx +Cas our new guess, we find the derivatives and substitute them back into

the differential equation.

Step 7: Find the derivatives for the adjusted particular solution.

y′

p= (2Ax +B)e3x+Ax2e3x,

y′′

p= (6Ax + 2A)e3x+ (2Ax +B)e3x.

Step 8: Substitute the adjusted particular solution and its derivatives into

the differential equation. Substitute yp,y′

p, and y′′

pinto the differential equation

y′′ −y′−6y= 2e3x+ 4x, and simplify. This will lead to equations involving A,

B, and C.

Step 9: Solve for the coefficients. By equating the coefficients of like terms,

we can solve for A,B, and C. Once these coefficients are found, the particular

solution is complete.

By following these steps carefully, we can find the particular solution to the

given differential equation using the method of undetermined coefficients.

Question 26

Question

Find the general solution to the differential equation y′′ −4y′+ 4y= 6x2e2x+

15xe2x.

Solution

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

method of undetermined coefficients.

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

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

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

Step 2: Next, we need to find a particular solution to the nonhomogeneous

equation y′′ −4y′+ 4y= 6x2e2x+ 15xe2x. Assume a particular solution of the

form yp=Ax2e2x+Bxe2x.

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

y′

p= (2Ax +B)e2x+ 2Ax2e2x,

y′′

p= (2A+ 2Ax +B)e2x+ 4Axe2x.

Step 4: Substitute yp,y′

p, and y′′

pback into the differential equation and

simplify:

(2A+ 2Ax +B−4A−4Ax −4B+ 4Ax)e2x= 6x2e2x+ 15xe2x,

(−2A−4B)e2x= 6x2e2x+ 15xe2x.

Matching coefficients, we get the following system of equations:

−2A= 6,

−4B= 15.

Solving this system, we find A=−3 and B=−15

Step 5: The particular solution is yp=−3x2e2x−15

4xe2x.

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

of the complementary and particular solutions:

y=yc+yp=c1e2x+c2xe2x−3x2e2x−15

4xe2x.

Question 27

Question

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

undetermined coefficients: y′′ −y′−6y= 2e3x+ 5x2.

Solution

To solve the non-homogeneous differential equation y′′ −y′−6y= 2e3x+ 5x2,

we first find the complementary function ycby solving the corresponding homo-

geneous equation y′′ −y′−6y= 0.

Step 1: Find the complementary function yc.

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

is r2−r−6 = 0. Factoring gives (r−3)(r+ 2) = 0, so the roots are r= 3 and

r=−2.

Therefore, the complementary function ycis given by yc=c1e3x+c2e−2x,

where c1and c2are constants to be determined.

Step 2: Find the particular solution yp.

Since our non-homogeneous equation contains terms involving e3xand x2,

we make the educated guess that yphas the form:

yp=Ae3x+Bx2+Cx +D

Taking the derivatives:

y′

p= 3Ae3x+ 2Bx +C

y′′

p= 9Ae3x+ 2B

Substitute ypand its derivatives back into the original equation:

9Ae3x+ 2B−(3Ae3x+ 2Bx +C)−6(Ae3x+Bx2+Cx +D) = 2e3x+ 5x2

Matching coefficients of like terms gives a system of equations to solve for

A,B,C, and D.

Step 3: Solve for the coefficients A,B,C, and D.

By solving the system of equations, we find:

A=1

3, B =1

12, C =1

4, D =−1

Thus, the particular solution ypis:

yp=1

3e3x+1

12x2+1

4x−1

Step 4: Write the general solution.

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

of the complementary function ycand the particular solution yp:

y=yc+yp=c1e3x+c2e−2x+1

3e3x+1

12x2+1

4x−1

Question 28

Question

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

the method of undetermined coefficients.

Solution

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

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

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

Step 2: Next, find a particular solution ypby assuming that yp= (Ax+B)ex.

Compute the derivatives needed for substitution into the original differential

equation.

y′

p= (A+Ax +B)ex

y′′

p= (A+ 2A+Ax +B)ex

Step 3: Substitute yp, y′

p,and y′′

pinto the differential equation and then solve

for Aand B.

(A+ 2A+Ax +B)ex−3(A+Ax +B)ex+ 2(Ax +B)ex= 3xex

(3A−3B)ex= 3xex

Comparing coefficients, we get 3A−3B= 3, which simplifies to A−B= 1.

Step 4: Solve for Aand Bby setting A=B+ 1. Let B= 0 for simplicity,

then A= 1.

Step 5: Therefore, the particular solution is yp= (x+ 1)ex.

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

complementary and particular solutions, so y=yc+yp=c1ex+c2e2x+(x+1)ex.

Question 29

Question

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

differential equation

y′′ −3y′+ 2y= 12e2x.

Solution

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

the particular solution has the form yp=Ae2xwhere Ais a constant to be

determined.

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

ypare:

y′

p= 2Ae2x, y′′

p= 4Ae2x.

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation.

Substitute yp,y′

p, and y′′

pinto the differential equation to get:

4Ae2x−3(2Ae2x)+2Ae2x= 12e2x.

Step 3: Solve for A.Simplifying the equation, we get:

4Ae2x−6Ae2x+ 2Ae2x= 12e2x.

Ae2x= 12e2x.

A= 12.

Step 4: Form the particular solution. Therefore, the particular solution

is yp= 12e2x.

Thus, a particular solution to the differential equation is yp= 12e2x.

Question 30

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 8e2x.

Solution

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

efficients, we first find the complementary function and particular solution.

Step 1: Find the complementary function solution The complemen-

tary function solves the homogeneous version of the differential equation, which

y′′ −4y′+ 4y= 0.

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

Therefore, the complementary function is

yc(x) = (C1+C2x)e2x.

Step 2: Assume a particular solution For the particular solution, since

the nonhomogeneous term is in the form e2x, we assume the particular solution

has the form

yp(x) = Ae2x.

Step 3: Find the derivatives of the assumed particular solution

yp(x) = Ae2x,

y′

p(x) = 2Ae2x,

y′′

p(x) = 4Ae2x.

Step 4: Substitute the particular solution and its derivatives into

the differential equation Substitute yp(x), y′

p(x), and y′′

p(x) into the differ-

ential equation:

4Ae2x−4(2A)e2x+ 4(A)e2x= 8e2x,

4Ae2x−8Ae2x+ 4Ae2x= 8e2x.

Step 5: Solve for the unknown coefficients Solving the equation ob-

tained in the previous step:

0=8e2x⇒0 = 8.

Since 0 = 8, there is no solution for A. Therefore, to fix this issue, we

multiply the assumed particular solution by x:

yp(x) = Ax2e2x.

Step 6: Find the derivatives of the corrected assumed particular

solution

yp(x) = Ax2e2x,

y′

p(x) = (2Ax + 2x2)e2x,

y′′

p(x) = (4A+ 4x+ 4x)e2x= (4A+ 8x)e2x.

Step 7: Substitute the corrected particular solution and its deriva-

tives into the differential equation Substitute yp(x), y′

p(x), and y′′

p(x) into

the differential equation:

(4A+ 8x)e2x−4(2Ax + 2x2)e2x+ 4Ax2e2x= 8e2x,

(4A+ 8x)e2x−8Axe2x−8x2e2x+ 4Ax2e2x= 8e2x.

Step 8: Solve for the unknown coefficients Comparing coefficients of

e2xon both sides:

−8x2e2x+ 4Ax2e2x= 8e2x.

This implies 4A= 8, so A= 2. Therefore, the particular solution is

yp(x) = 2x2e2x.

Step 9: Form the general solution The general solution is the sum of

the complementary function and the particular solution:

y(x) = yc(x) + yp(x)=(C1+C2x)e2x+ 2x2e2x.

Question 31

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x−3x

Solution

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

by solving the associated homogeneous equation y′′ −3y′+ 2y= 0. Then, we

will find a particular solution for the non-homogeneous part 6e2x−3xusing the

method of undetermined coefficients.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous part is r2−3r+ 2 = 0. This can be factored 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: Find the particular solution Let’s assume the particular solution

has the form:

yp=Ae2x+Bx +C

Calculate the derivatives and substitute them into the differential equation:

y′

p= 2Ae2x+B

y′′

p= 4Ae2x

Substitute yp,y′

p, and y′′

pinto the differential equation and equate coefficients

of like terms:

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

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

Simplify and equate coefficients:

0+(−3B+ 2) + 0 = −3 =⇒B=−1

0+0+(−3(−1) + 2C) = 0 =⇒3+2C= 0 =⇒C=−3

Therefore, the particular solution is yp=Ae2x−x−3

Step 3: Construct the general solution The general solution is the sum

of the complementary and particular solutions:

y=yc+yp=c1ex+c2e2x+Ae2x−x−3

Question 32

Question

Find the particular solution to the differential equation y′′ −3y′+ 2y= 4x2ex

using the method of undetermined coefficients.

Solution

Step 1: Find the complementary solution to 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(x) = c1ex+c2e2x.

Step 2: Assume a particular solution of the form yp(x)=(Ax2+Bx +C)ex.

Then, calculate the first and second derivatives: y′

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

and y′′

p(x) = (2A+ 2Aex+ 2Aex+B+Aex)ex= (4A+ 3Aex+B)ex.

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation and

solve for A,B, and C. (4A+ 3Aex+B)ex−3[(2Ax +B+Aex)ex] + 2[(Ax2+

Bx +C)ex] = 4x2ex

This simplifies to: (4A+ 3B−3A)ex+ (−6A−3B+ 2B)ex+(2A)ex= 4x2ex

Step 4: Equate coefficients of like terms: exterms: 4A+ 3B−3A−6A−

3B+ 2B= 0 exterms simplify to: −5A−B= 0 or 5A+B= 0 (1) exterms:

2A= 4 ⇒A= 2 Substitute A= 2 into equation (1): 5(2) + B= 0 ⇒B=−10

Therefore, the particular solution is yp(x) = (2x2−10x+C)ex.

Step 5: Combine the complementary and particular solutions to get the

general solution: y(x) = yc(x) + yp(x) = c1ex+c2e2x+ (2x2−10x+C)ex.

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

c2e2x+ (2x2−10x+C)ex.

Question 33

Question

Find the general solution to the non-homogeneous differential equation:

y′′ −4y= 2xe2x

Solution

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

ential equation:

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

Therefore, the characteristic roots are r= 2 and r=−2, so the general solution

to the associated homogeneous differential equation is:

yh(x) = c1e2x+c2e−2x

where c1and c2are arbitrary constants.

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

equation using the method of undetermined coefficients.

Assume a particular solution of the form yp(x) = ax2e2x+bx +c, where a,

b, and care constants to be determined.

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

y′

p(x) = (2ax +b)e2x+ax2e2x

y′′

p(x) = (4ax + 2a+b)e2x+ (2ax +b)2e2x+ 2axe2x

Substitute these derivatives into the non-homogeneous differential equation

and simplify:

−(4ax + 2a+b)e2x−(2ax +b)2e2x−2axe2x−4(ax2e2x+bx +c) = 2xe2x

−4ax −2a−b−4bx −4c= 2x

Step 4: Equate coefficients of like terms to solve for a,b, and c:

From the constant term: −2a−b−4c= 0 From the xterm: −4a−4b= 2

From the x2e2xterm: −4a= 0

Solving these equations gives a= 0, b=−1

2, and c= 0.

Step 5: Combine the homogeneous and particular solutions to get the gen-

eral solution:

y(x) = yh(x) + yp(x) = c1e2x+c2e−2x−1

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

is:

y(x) = c1e2x+c2e−2x−1

Question 34

Question

Solve the following second-order linear differential equation using the method

of undetermined coefficients:

y′′ −3y′+ 2y=x2ex

Solution

We first find the general solution to the corresponding homogeneous 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. So, the general solution to

the homogeneous equation is

yh=c1ex+c2e2x

Next, we guess the particular solution for the given non-homogeneous equa-

tion based on the form of the right-hand side. Since x2exis a product of a

polynomial of degree 2 and an exponential function, we guess the particular

solution has the form

yp=Ax2ex+Bxex+C

We now find y′

p, y′′

pand substitute these into the original differential equation:

y′

p= 2Axex+ (A+B)ex, y′′

p= 2Aex+ 2Aex+ (A+B)ex

Substitute yp, y′

p, y′′

pinto the differential equation y′′ −3y′+ 2y=x2ex:

(2Aex+2Aex+(A+B)ex)−3(2Axex+(A+B)ex)+2(Ax2ex+Bxex+C) = x2ex

Simplify the equation and equate coefficients of like terms. After simplifica-

tion, we get the values of A,B, and C.

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

y=yh+yp

Question 35

Question

Find the particular solution to the differential equation y′′ −4y′+4y= 3e2x+5x2

using the method of undetermined coefficients.

Solution

To find the particular solution, we first need to 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, giving us a repeated

root r= 2. So, the complementary solution is yc(x)=(c1+c2x)e2x.

Next, we guess the form of the particular solution based on the form of

the nonhomogeneous terms. Since the right-hand side contains 3e2x, we guess

yp(x) = Ax2e2x. Additionally, since the right-hand side contains 5x2, we include

another term Bx2in our guess.

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

yp(x) = Ax2e2x+Bx2

Step 4: Substitute into the original equation and solve for ASub-

stitute yp(x), y′

p(x), and y′′

p(x) into the original equation:

4Ae2x−3(2Ae2x) + 2(Ae2x)=6e2x

Simplify to get:

3Ae2x= 6e2x

Therefore, A= 2.

Step 5: Write the particular solution The particular solution is:

yp(x)=2e2x

Step 6: Write the general solution The general solution is the sum of

the complementary and particular solutions:

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

Question 2

Question

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

efficients: y′′ −2y′+y=x2ex.

Solution

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

efficients, we will first find the complementary function yc(x) by solving the

characteristic equation. Then, we will find a particular solution yp(x) by as-

suming a suitable form based on the non-homogeneous term of the differential

equation.

Step 1: Find the complementary function The characteristic equation

for y′′ −2y′+y= 0 is r2−2r+ 1 = 0. Solving this, we get (r−1)2= 0 which

gives us a repeated root r= 1. Therefore, the complementary function is of the

form yc(x) = c1ex+c2xex.

Step 2: Assume the particular solution For y′′ −2y′+y=x2ex, we

will assume a particular solution of the form yp(x)=(Ax2+Bx +C)ex, where

A,B, and Care constants to be determined.

Step 3: Find the derivative of yp(x)

y′

p(x) = (2Ax +B+A)x2ex

y′′

p(x) = (2A+ 2)x2ex+ 4Ax2ex

Step 4: 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) into the differential equation y′′ −2y′+

y=x2ex:

(2A+ 2)x2ex+ 4Ax2ex−2(2Ax +B+A)x2ex+ (Ax2+Bx +C)ex=x2ex

Step 5: Simplify and solve for constants Equating coefficients of like

terms on both sides of the equation, we get the following system of equations:

2A−2A+ 2 = 0 (coefficients of x2ex)

4A−2B+A= 1 (coefficients of xex)

−2B+B+C= 0 (coefficients of ex)

Solving this system of equations, we find A= 1, B= 2, and C= 2.

Step 6: Form the general solution Therefore, the general solution to

the differential equation y′′ −2y′+y=x2exis:

y(x) = yc(x) + yp(x) = c1ex+c2xex+ (x2+ 2x+ 2)ex

Question 3

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x+ 5x2−1

Solution

Step 1: First, we find the complementary solution ycby solving the corre-

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

is r2−3r+ 2 = (r−1)(r−2) = 0, so the solutions are yc=c1ex+c2e2x.

Step 2: Next, we determine the particular solution ypfor the nonho-

mogeneous equation. Since the right-hand side is a sum of exponential and

polynomial functions, we assume the particular solution has the form yp=

Ae2x+Bx2+Cx +D. We then substitute ypback into the differential equation

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

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

y′

p= 2Ae2x+ 2Bx +C

y′′

p= 4Ae2x+ 2B

Substitute yp,y′

p, and y′′

pinto the differential equation:

4Ae2x+ 2B−3(2Ae2x+ 2Bx +C) + 2(Ae2x+Bx2+Cx +D)=6e2x+ 5x2−1

Step 4: Simplify the equation and match coefficients of like terms. This

will give us a system of equations to solve for A,B,C, and D.

Let’s simplify and equate coefficients:

A−2B+ 2C= 0

4A−3(2B)+2C= 6

2B+ 2D=−1

4A−3(2B)+2C= 0

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

substitute these values back into ypto get the particular solution.

After solving the system of equations, we find A= 2, B=−1, C= 1, and

D=−1

2. Thus, the particular solution is yp= 2e2x−x2+x−1

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

solutions: y=yc+yp. Therefore, the final solution to the differential equation

is:

y=c1ex+c2e2x+ 2e2x−x2+x−1

Question 4

Question

Find the particular solution to the differential equation y′′ −2y′−3y= 2e3x+x

using the method of undetermined coefficients.

Solution

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

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

factors as (r−3)(r+ 1) = 0. So, the roots are r= 3 and r=−1, making the

complementary solution yc=c1e3x+c2e−x.

Step 2: Next, we consider the form of the particular solution. Since the right-

hand side of the given equation includes terms 2e3xand xwhich are similar to

the terms in the complementary solution, we need to use a modification of the

form of the complementary solution for the particular solution. Therefore, we

assume the particular solution has the form:

yp=Ax2e3x+Bx +C

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

y′

p= (2Ax +B)e3x+ 3Ax2e3x

y′′

p= (6Ax2+ 6Ax + 2B)e3x+ 6Ax2e3x

Step 4: Substitute yp,y′

p, and y′′

pinto the original differential equation and

simplify:

(6Ax2+6Ax+2B−2(2Ax+B)−3(Ax2e3x+Bx+C))e3x−3(Ax2e3x+Bx+C)=2e3x+x

Step 5: Collecting like terms, we get the following equations:

(−6A+ 4B−3A)x2e3x= 0

(6A−6A−6B−3B)x= 1

(6A+ 6A+ 2B−3C)=0

Step 6: Solve the system of equations to find the values of A,B, and C.

Step 7: After finding the values of A,B, and C, the particular solution yp

can be written as:

yp= (value of A)x2e3x+ (value of B)x+ (value of C)

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

(value of A)x2e3x+ (value of B)x+ (value of C).

Question 5

Question

Find the particular solution to the following non-homogeneous second-order

differential equation using the method of undetermined coefficients:

y′′ −4y′+ 4y= 3e2x+ 10x2

Solution

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

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

which factors to (r−2)2= 0. So the complementary solution is of the form

yc(x) = c1e2x+c2xe2x.

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

equation. Since 3e2xis of the same form as the complementary solution, we

need to use xtimes a function of xfor that term. Since the repeated root has

multiplicity 2, we take yp(x) = Ax2e2x.

Step 3: Now, we find the derivative terms needed for the equation: y′

p(x) =

(2Ax + 2A)e2xand y′′

p(x) = (4A+ 4A)xe2x+ 4Ae2x.

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the original non-homogeneous

equation:

y′′

p−4y′

p+ 4yp= 3e2x+ 10x2

(4A+ 4A)xe2x+ 4Ae2x−4(2Ax + 2A)e2x+ 4Ax2e2x= 3e2x+ 10x2

Step 5: Simplify the equation and match coefficients on both sides:

(4A+ 4A)xe2x+ 4Ae2x−8Axe2x−8Ae2x+ 4Ax2e2x= 3e2x+ 10x2

Step 6: Equating the coefficients of like terms:

For e2x:

4A−8A= 3 =⇒ −4A= 3 =⇒A=−3

For x2e2x:

4A= 10 =⇒4(−3

4) = 10 =⇒ −3 = 10

There is no solution for A, so we correct the form of ypby multiplying by x:

yp(x) = −3

4x2e2x

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

ticular solutions:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x−3

4x2e2x

Question 6

Question

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

3y′+ 2y= 3e2xusing the method of undetermined coefficients.

Solution

Step 1: First, we find the complementary solution of the corresponding homoge-

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

which factors as (r−1)(r−2) = 0. So the complementary solution is yc(x) =

c1ex+c2e2x.

Step 2: Next, we assume the particular solution has the form yp(x) = Ae2x,

where Ais a constant to be determined.

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

y′′ −3y′+ 2y= (A·4e2x)−3(A·2e2x) + 2(Ae2x)

= 4Ae2x−6Ae2x+ 2Ae2x= 3e2x

Step 4: Simplify the equation 3Ae2x= 3e2xand solve for A. This gives

A= 1.

Step 5: Therefore, the particular solution is yp(x) = e2x.

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

the particular solution: y(x) = yc(x) + yp(x) = c1ex+c2e2x+e2x, where c1and

c2are arbitrary constants.

Question 7

Question

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

3y′+ 2y= 2e−t+ 3 sin(2t) using the method of undetermined coefficients.

Solution

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

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

which factors to (r+ 1)(r+ 2) = 0. Therefore, the complementary solution is

yc(t) = c1e−t+c2e−2t.

Step 2: Next, we assume a particular solution of the form yp(t) = Ae−t+

Bsin(2t) + Ccos(2t). Then calculate the first and second derivatives: y′

p(t) =

−Ae−t+ 2Bcos(2t)−2Csin(2t), y′′

p(t) = Ae−t−4Bsin(2t)−4Ccos(2t).

Step 3: Substitute yp(t), y′

p(t), and y′′

p(t) into the differential equation:

(Ae−t−4Bsin(2t)−4Ccos(2t))+3(−Ae−t+2Bcos(2t)−2Csin(2t))+2(Ae−t+

Bsin(2t) + Ccos(2t)) = 2e−t+ 3 sin(2t).

Step 4: Group terms with the same functions together and solve for A,B,

and C: (A+3A+2A)e−t+(−4B+6B+2B) sin(2t)+(−4C−6C+2C) cos(2t) =

2e−t+ 3 sin(2t). Simplify to get: 6Ae−t+ 4Bsin(2t)−8Ccos(2t) = 2e−t+

3 sin(2t).

Step 5: Equate coefficients to find A,B, and C: 6A= 2 =⇒A=1

4B= 3 =⇒B=3

4,−8C= 0 =⇒C= 0.

Step 6: Finally, the particular solution is yp(t) = 1

3e−t+3

4sin(2t). Therefore,

the general solution is y(t) = yc(t) + yp(t) = c1e−t+c2e−2t+1

3e−t+3

4sin(2t).

Question 8

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 first need to find the complementary solution.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous part of the differential equation is r2−5r+ 6 = 0. This

factors as (r−2)(r−3) = 0, so the characteristic roots are r1= 2 and r2= 3.

Therefore, the complementary solution is

yc(x) = c1e2x+c2e3x

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

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

Step 3: Calculate the derivatives We have y′

p(x) = 2Ae2xand y′′

p(x) =

4Ae2x.

Step 4: Substitute into the differential equation Substitute yp(x),

y′

p(x), and y′′

p(x) into the differential equation y′′ −5y′+ 6y= 3e2x:

4Ae2x−5(2Ae2x) + 6(Ae2x)=3e2x

Step 5: Solve for A Solving the above equation for Agives:

A=3

6−4−5=−1

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

solution is

yp(x) = −e2x

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

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

y(x) = yc(x) + yp(x) = c1e2x+c2e3x−e2x

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

c1e2x+c2e3x−e2x.

Question 9

Question

Solve the following second order linear differential equation using the method

of undetermined coefficients:

y′′ −y= 2 sin(3x)+3ex

Solution

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

efficients, we first find the complementary function yc(x) by solving the corre-

sponding homogeneous equation y′′ −y= 0.

The characteristic equation is r2−1 = 0, which has roots r=±1. Therefore,

the complementary function is given by:

yc(x) = C1ex+C2e−x

Next, we look for the particular solution yp(x) based on the form of the

forcing term. Since the right-hand side includes 2 sin(3x) and 3ex, we assume

the particular solution to have the form:

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

Now, we find the derivatives of yp(x) needed to substitute back into the

original differential equation:

y′

p(x) = 3Acos(3x) + Bex

y′′

p(x) = −9Asin(3x) + Bex

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

tion:

−9Asin(3x) + Bex−Asin(3x)−Bex= 2 sin(3x)+3ex

Matching coefficients of like terms, we get the following system of equations:

−9A−A= 2

B−B= 3

Solving this system, we find A=−2

10 =−1

5and B= 3.

Therefore, the particular solution is:

yp(x) = −1

5sin(3x)+3ex

The general solution to the original differential equation is the sum of the

complementary function and the particular solution:

y(x) = yc(x) + yp(x) = C1ex+C2e−x−1

5sin(3x)+3ex

Thus, the solution to the differential equation is:

y(x) = C1ex+C2e−x−1

5sin(3x)+3ex

Question 10

Question

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

3y′+ 2y= 4exusing the method of undetermined coefficients.

Solution

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

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

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

is yc=c1ex+c2e2x.

Step 2: Now, we guess the form of the particular solution ypto be of the

form Aex. Substitute yp=Aexback into the differential equation: (Aex)′′ −

3(Aex)′+ 2(Aex)=4ex.

Step 3: Calculate the derivatives and substitute them back into the dif-

ferential equation. We have Aex−3Aex+ 2Aex= 4ex. Simplifying, we get

Aex= 4exwhich implies A= 4.

Step 4: Therefore, the particular solution is yp= 4ex.

Step 5: The general solution is given by adding the complementary and

particular solutions: y=yc+yp. Hence, y=c1ex+c2e2x+ 4exis the general

solution to the non-homogeneous differential equation.

Question 11

Question

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

undetermined coefficients:

y′′ −4y′+ 4y= 2e2x.

Solution

Step 1: Find the complementary solution yc(x) by solving the associated homo-

geneous 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 then given by yc(x) = c1e2x+c2xe2x.

Step 2: Assume a particular solution of the form yp(x) = Ae2x. Now, we

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

y′

p(x)=2Ae2xand y′′

p(x) = 4Ae2x.

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential

equation:

4Ae2x−4(2Ae2x)+4Ae2x= 2e2x.

Simplify this equation to get 4Ae2x−8Ae2x+4Ae2x= 2e2x, which simplifies

to 0 = 2e2x, which is not true.

Step 4: Since Ae2xdoes not satisfy the original differential equation, we

need to modify our guess to include a factor of xto account for the repeated

root:

yp(x) = Ax2e2x.

Step 5: Now, find the first and second derivatives of yp(x):

y′

p(x) = (2Ax + 2A)e2xand y′′

p(x) = (4Ax2+ 8Ax + 2A)e2x.

Step 6: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential

equation:

(4Ax2+ 8Ax + 2A)e2x−4(2Ax + 2A)e2x+ 4Ax2e2x= 2e2x.

Step 7: Simplify this equation to get (4Ax2+ 8Ax + 2A−8Ax −8A+

4Ax2)e2x= 2e2x, which simplifies to (8Ax2−6A)e2x= 2e2x.

Step 8: Equate the coefficients of like terms:

8Ax2−6A= 2.

Step 9: Solve for A:

8A= 8 ⇒A= 1.

Step 10: The particular solution is yp(x) = x2e2x. Thus, the general

solution to the original differential equation is:

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

Question 12

Question

Find the particular solution to the nonhomogeneous differential equation:

y′′ −4y′+ 4y= 2e2x+ 5x

using the method of undetermined coefficients.

Solution

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

we first need to find the complementary solution. Let’s assume yp(x) is the

particular solution we are trying to find.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous part of the differential equation is:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2

Thus, the complementary solution is:

yc(x) = c1e2x+c2xe2x

Step 2: Find the particular solution Our particular solution yp(x)

should have the same form as the nonhomogeneous terms, so we can guess

that yp(x) = Ae2x+Bx +C.

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

Ae2x+Bx +Cand its derivatives into the differential equation:

y′′ −4y′+ 4y= 2e2x+ 5x

(4A−4B)e2x+ 5 = 2e2x+ 5x

Step 4: Solve for the coefficients Equating coefficients, we get:

4A−4B= 2

5=5

Solving the equations, we find A=1

2and B= 0. We can disregard the constants

here.

Step 5: Assemble the particular solution Therefore, the particular

solution is:

yp(x) = 1

2e2x

Step 6: Find the general solution The general solution is the sum of

the complementary solution and the particular solution:

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

2e2x

So, the particular solution to the nonhomogeneous differential equation is

y(x) = c1e2x+c2xe2x+1

2e2x.

Question 13

Question

Find the particular solution to the differential equation x′′ −4x′+4x=e2tcos(t)

using the method of undetermined coefficients.

Solution

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

we need to guess a form for the particular solution based on the form of the

non-homogeneous term. Since e2tcos(t) is a product of an exponential and

trigonometric function, we guess the form of the particular solution to be xp(t) =

(At2+Bt +C)e2tcos(t) + (Dt2+Et +F)e2tsin(t), where A,B,C,D,E, and

Fare constants to be determined.

Step 1: Calculate the derivatives of xp(t).

x′

p(t) = (2At+B)e2tcos(t)−(At2+Bt+C)e2tsin(t)+(2Dt+E)e2tsin(t)+(Dt2+Et+F)e2tcos(t)

x′′

p(t) = (2A)e2tcos(t)+(4At+2B)e2tsin(t)−(4At+2B+2C)e2tcos(t)+(2Dt+E)e2tcos(t)−(2Dt+E)e2tsin(t)+(2D+2E)e2tsin(t)−(Dt2+Et+F)e2tsin(t)−(2Dt+E)e2tcos(t)

Step 2: Substitute xp(t), x′

p(t), and x′′

p(t) into the differential equation

x′′ −4x′+ 4x=e2tcos(t) and simplify.

LHS = (2A)e2tcos(t)+(4At+2B)e2tsin(t)−(4At+2B+2C)e2tcos(t)+(2Dt+E)e2tcos(t)−(2Dt+E)e2tsin(t)+(2D+2E)e2tsin(t)−(Dt2+Et+F)e2tsin(t)−(2Dt+E)e2tcos(t)

RHS = e2tcos(t)

Step 3: Equate the coefficients of like terms on both sides to determine the

values of A,B,C,D,E, and F.

2A−4C= 1 (e2tcos(t) terms)

4A+ 2B−4C+ 2E= 0 (e2tsin(t) terms)

−4A+ 2B+ 2E= 0 (e2tcos(t) terms)

2D+ 2E= 0 (e2tsin(t) terms)

−2D−E= 0 (e2tcos(t) terms)

D= 0 (e2tsin(t) terms)

Step 4: Solve the system of equations to find the values of A,B,C,D,E,

and F, then substitute back into xp(t) to get the particular solution.

Step 5: Finally, the particular solution is

xp(t) = 3

10t2+1

5t+1

10e2tcos(t)

Question 14

Question

Find the particular solution of the nonhomogeneous differential equation:

y′′ −4y′+ 4y= 6e2x

Solution

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

we assume the particular solution has the form yp(x) = Ae2x.



Step 1: Find the derivatives of yp.

y′

p= 2Ae2x

y′′

p= 4Ae2x



Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation.

4Ae2x−4(2Ae2x) + 4(Ae2x)=6e2x



Step 3: Simplify the equation and solve for A.

4Ae2x−8Ae2x+ 4Ae2x= 6e2x

0=6e2x

This equation has no solutions, so we need to add a multiplying factor of

xto yp(x).



Step 4: Update the form of yp(x) to include the additional factor. Let’s

try yp(x) = Axe2x.

yp=Axe2x

y′

p=Ae2x+ 2Axe2x

y′′

p= 2Ae2x+ 2Ae2x+ 4Axe2x



Step 5: Substitute yp,y′

p, and y′′

pinto the differential equation.

2Ae2x+ 2Ae2x+ 4Axe2x−4(Ae2x+ 2Axe2x) + 4(Axe2x)=6e2x



Step 6: Simplify the equation and solve for A.

6Ae2x= 6e2x

A= 1



Step 7: The particular solution is yp(x) = xe2x.

Question 15

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6ex−5x2+ 2

Solution

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

0 to find the complementary function yc(x). The characteristic equation is

r2−3r+2 = 0, which factors as (r−1)(r−2) = 0. Thus, the roots are r1= 1 and

r2= 2. Therefore, the complementary function is given by yc(x) = C1ex+C2e2x,

where C1and C2are arbitrary constants.

Step 2: Next, we assume the particular solution has the form yp(x) =

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

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

y′

p(x) = Aex+ 2Bx +C

y′′

p(x) = Aex+ 2B

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential

equation and simplify to solve for A,B,C, and D.

Aex+ 2B−3(Aex+ 2Bx +C) + 2(Aex+Bx2+Cx +D)=6ex−5x2+ 2

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

to find the values of A,B,C, and D. Solving these simultaneous equations will

give the particular solution yp(x).

Step 6: Finally, the general solution is the sum of the complementary func-

tion and particular solution:

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

y(x) = C1ex+C2e2x+ (Aex+Bx2+Cx +D)

Question 16

Question

Solve the following nonhomogeneous second-order differential equation using the

method of undetermined coefficients:

y′′ −3y′+ 2y= 5e2x+ 4 sin x

Solution

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

y′′ −3y′+ 2y= 0

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

Thus, the characteristic roots are r1= 1 and r2= 2. The complementary

solution is:

yc(x) = c1ex+c2e2x

Step 2: Determine the form of the particular solution based on the form

of the nonhomogeneous terms. Since the right-hand side contains e2x(which

is also a solution to the homogeneous equation), we multiply by xto make it

linearly independent:

yp(x) = Axe2x+Bsin x+Ccos x

Now, find the derivatives of yp(x):

y′

p(x) = Ae2x+ 2Axe2x+Bcos x−Csin x

y′′

p(x)=3Ae2x+ 4Axe2x−Bsin x−Ccos x

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential equa-

tion to determine the coefficients A,B, and C. The equation becomes:

3Ae2x+4Axe2x−Bsin x−Ccos x−3(Ae2x+2Axe2x+Bcos x−Csin x)+2(Axe2x+Bsin x+Ccos x)=5e2x+4 sin x

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

to solve for A,B, and C. This will lead to a system of equations which can be

solved to find the values of the constants.

Step 5: Once you have the values of A,B, and C, the particular solution

yp(x) can be written. Add the complementary and particular solutions to get

the general solution:

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

Question 17

Question

Solve the following second order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 6e2x

Solution

To solve this differential equation, we first find the complementary function by

solving the associated homogeneous equation:

y′′ −4y′+ 4y= 0

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

So, we have a repeated root r= 2.

Therefore, the complementary function ycis given by yc=c1e2x+c2xe2x.

Next, we find the particular solution by assuming yp=Ae2xfor the nonho-

mogeneous term 6e2x.

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

yp=Ae2x

y′

p= 2Ae2x

y′′

p= 4Ae2x

Step 2: Substitute ypand its derivatives back into the original differential

equation:

y′′ −4y′+ 4y= 6e2x

4Ae2x−4(2Ae2x) + 4(Ae2x)=6e2x

4Ae2x−8Ae2x+ 4Ae2x= 6e2x

0=6e2x

Since 0 = 6e2x, we try a different form for yp. Let’s try yp=Ax2e2xthis

time.

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

yp=Ax2e2x

y′

p= (2Ax + 2x2)e2x

y′′

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

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

equation:

y′′ −4y′+ 4y= 6e2x

(4A+ 4x+ 4x)e2x−4(2Ax + 2x2)e2x+ 4(Ax2e2x)=6e2x

(4A+ 8x)e2x−8Ax −8x2e2x+ 4Ax2e2x= 6e2x

Equating the coefficients of e2xand x2e2xon both sides, we find:

4A= 6 and 4A−8=0

Solving these equations, we get A=3

Therefore, the particular solution is yp=3

2x2e2x.

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

the particular solution:

y=yc+yp=c1e2x+c2xe2x+3

2x2e2x

Question 18

Question

Find the particular solution to the differential equation y′′ −4y′+4y= 8e2x+6x

using the method of undetermined coefficients.

Solution

Step 1: Find the homogeneous solution to the differential equation y′′ −4y′+4y=

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

repeated root r= 2. So, the general solution to the homogeneous equation is

yh(x) = c1e2x+c2xe2x.

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

hand side of the differential equation contains terms of the form e2xand x, the

particular solution can be assumed in the form: yp(x) = Ae2x+Bx +C

Step 3: Find the derivatives of yp(x). y′

p(x)=2Ae2x+B y′′

p(x)=4Ae2x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) back into the differential equation

and solve for A,B, and C. Substituting into the differential equation, we get:

4Ae2x−4(2Ae2x+B) + 4(Ae2x+Bx +C)=8e2x+ 6x

Simplifying the equation, we get: 4Ae2x−8Ae2x−4B+4Ae2x+4Bx+4C=

8e2x+ 6x

−4B+ 4Bx + 4C= 8e2x+ 6x

Comparing coefficients on both sides, we get: 4B= 6 and 4C= 8

Step 5: Calculate the values of Band C. From 4B= 6, we find B=3

From 4C= 8, we find C= 2.

Therefore, the particular solution is: yp(x) = Ae2x+3

2x+ 2 where Ais an

arbitrary constant.

Thus, the general solution to the differential equation is: y(x) = yh(x) +

yp(x) = c1e2x+c2xe2x+Ae2x+3

2x+ 2 where c1,c2, and Aare arbitrary

constants.

Question 19

Question

Solve the following non-homogeneous second order differential equation using

the method of undetermined coefficients:

y′′ −y′−2y= 3ex+ 2 cos(x)

Given that y1(x) = exand y2(x) = e−2xare solutions to the homogeneous

equation y′′ −y′−2y= 0.

Solution

Step 1: Write down the general form of the particular solution yp(x):

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

Step 2: Find the derivatives of yp(x).

y′

p(x) = Aex−Bsin(x) + Ccos(x)

y′′

p(x) = Aex−Bcos(x)−Csin(x)

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) into the original differential equa-

tion.

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

Step 4: Group like terms and solve for A,B, and C.

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

To solve for A,B, and C, we set the coefficients of ex, cos(x), and sin(x) on

both sides equal:











−2A+B= 3

B−2C= 2

C−2A= 0

By solving this system of equations, we find A=−3

5,B=−4

5, and C=−6

Therefore, the particular solution is:

yp(x) = −3

5ex−4

5cos(x)−6

5sin(x)

Step 5: The general solution to the non-homogeneous differential equation

is given by:

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

y(x) = c1ex+c2e−2x−3

5ex−4

5cos(x)−6

5sin(x)

where c1and c2are constants determined by initial conditions or boundary

conditions.

Question 20

Question

Find a particular solution to the following differential equation using the method

of undetermined coefficients:

y′′ −3y′+ 2y= 12e2x.

Solution

To find a 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 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)+2Ae2x= 12e2x.

Step 3: Simplify and solve for A.

4Ae2x−6Ae2x+ 2Ae2x= 12e2x,

(4 −6 + 2)Ae2x= 12e2x,

0 = 12e2x.

Since 0 = 12e2xfor any value of x, our assumption was incorrect. To fix

this, we need to multiply our assumed solution by xand try again.

Step 4: Revise the assumed particular solution. Assume the partic-

ular solution has the form:

yp=x(Ae2x),

where Ais a constant to be determined.

Step 5: Find the first and second derivatives of the revised yp.

yp=x(Ae2x),

y′

p=Ae2x+ 2Axe2x,

y′′

p= 2Ae2x+ 2Ae2x+ 4Axe2x.

Question 21

Question

Find the particular solution to the differential equation y′′ −4y= 6xe2xusing

the method of undetermined coefficients.

Solution

Step 1: To use the method of undetermined coefficients, we first write the com-

plementary solution to the homogeneous equation y′′ −4y= 0. The character-

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

solution is yc=Ae2x+Be−2x.

Step 2: Next, we consider the form of the particular solution yp. Since 6xe2x

has the same form as xe2x, we try yp= (Ax +B)e2x.

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

yp= (Ax +B)e2x

y′

p=Ae2x+ 2(Ax +B)e2x= (2Ax + (2A+ 2B))e2x

y′′

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

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

tion:

(4Ax + 4A+ 4B)e2x−4(Ax +B)e2x= 6xe2x

Step 5: Simplify and compare coefficients:

(4A−4A)x= 6x

4B−4A= 0

Step 6: Solve the system of equations to find Aand B: From the first

equation, 4A= 6 so A=3

From the second equation, 4B= 4Aso B=3

Therefore, the particular solution is yp=3

2x+3

2e2x.

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

solutions:

y=yc+yp=Ae2x+Be−2x+3

2x+3

2e2x

So, the particular solution to the differential equation y′′ −4y= 6xe2xis

y=Ae2x+Be−2x+3

2x+3

2e2x.

Question 22

Question

Find a particular solution to the differential equation y′′ −5y′+6y=e2xcos(3x)

using the method of undetermined coefficients.

Solution

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

y′′ −5y′+6y= 0. The characteristic equation is r2−5r+ 6 = (r−2)(r−3) = 0,

which gives us roots r1= 2 and r2= 3. Therefore, the complementary solution

is yc=c1e2x+c2e3x.

Step 2: Now, we look for a particular solution to the non-homogeneous

equation y′′ −5y′+ 6y=e2xcos(3x). Assume the particular solution has the

form yp=Ae2xcos(3x) + Be2xsin(3x).

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

p= (2Ae2xcos(3x)−

3Be2xsin(3x)) y′′

p= (4Ae2xcos(3x)+9Be2xsin(3x))

Step 4: Substitute yp,y′

p, and y′′

pback into the original differential equation:

(4Ae2xcos(3x)+9Be2xsin(3x))−5(2Ae2xcos(3x)−3Be2xsin(3x))+6(Ae2xcos(3x)+

Be2xsin(3x)) = e2xcos(3x)

Step 5: Equate coefficients of e2xcos(3x) on both sides: 4A−10A+6A= 1,

which gives A=−1

Step 6: Equate coefficients of e2xsin(3x) on both sides: 9B+15B+6B= 0,

which gives B= 0.

Step 7: Hence, the particular solution is yp=−1

8e2xcos(3x).

Step 8: The general solution is y=yc+yp=c1e2x+c2e3x−1

8e2xcos(3x),

where c1and c2are arbitrary constants.

Question 23

Question

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

the differential equation:

y′′ −4y′+ 4y= 4e2x+ 3x2

Solution

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

equation r2−4r+ 4 = 0:

r2−4r+ 4 = (r−2)2= 0

r= 2 (Repeated root)

yc=C1e2x+C2xe2x

Step 2: Write the particular solution in the form yp=Ae2x+Bx2+Cx +D

where A, B, C, D are constants to be determined.

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

y′

p= 2Ae2x+ 2Bx +C

y′′

p= 4Ae2x+ 2B

Step 4: Substitute yp,y′

p, and y′′

pinto the original differential equation:

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

Step 5: Simplify and equate coefficients of like terms:

(4A−8A+ 4A)e2x+ (2B−4B+ 4B)x+ (−4C+ 4C)x+ 4D= 4e2x+ 3x2

Step 6: Solve for A,B,C, and Dby comparing coefficients:

0 = 4,2=3,0 = 0,4D= 0

Step 7: From the system of equations, we find D= 0.

Therefore, the particular solution is:

yp= 0 + x2+ 0 + 0 = x2

Step 8: Finally, the general solution is:

y=yc+yp=C1e2x+C2xe2x+x2

Question 24

Question

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

undetermined coefficients:

y′′ −2y′+y= 6xex

Solution

Step 1: First, find the complementary solution ycof the corresponding ho-

mogeneous equation y′′ −2y′+y= 0, which is the characteristic equation

r2−2r+ 1 = 0. This gives us the repeated root r= 1, so the complementary

solution is:

yc(x) = c1ex+c2xex

Step 2: Next, guess the particular solution ypin the form yp(x) = (Ax2+

Bx +C)ex. Calculate y′

pand y′′

y′

p= (2Ax +B+A)xex

y′′

p= (2A+ 2)xex+ (2Ax +B+A)xex

Step 3: Substitute yp,y′

p, and y′′

pinto the original differential equation and

solve for A,B, and C:

(2A+2)xex+(2Ax+B+A)xex−2[(2Ax+B+A)xex]+(Ax2+Bx+C)ex= 6xex

Simplify and equate coefficients to find A,B, and C.

Step 4: Substitute the found values of A,B, and Cback into ypto get the

particular solution.

Step 5: The general solution is given by the sum of the complementary

solution ycand the particular solution yp:

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

y(x) = c1ex+c2xex+ (Ax2+Bx +C)ex

Question 25

Question

Find the particular solution to the differential equation y′′ −y′−6y= 2e3x+ 4x,

using the method of undetermined coefficients.

Solution

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

we first find the complementary solution by solving the characteristic equation

m2−m−6 = 0.

Step 1: Find the complementary solution. The characteristic equation is

m2−m−6 = 0. Factoring this equation, we get (m−3)(m+ 2) = 0, so

the roots are m= 3 and m=−2. Therefore, the complementary solution is

yc=c1e3x+c2e−2x.

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

the form yp=Ae3x+Bx +C. We then find the derivatives needed to substitute

into the differential equation.

Step 3: Find the derivatives of the particular solution.

y′

p= 3Ae3x+Band y′′

p= 9Ae3x.

Step 4: Substitute the particular solution and its derivatives into the dif-

ferential equation. Substitute yp,y′

p, and y′′

pinto the differential equation

y′′ −y′−6y= 2e3x+ 4x, and simplify to get:

9Ae3x−3Ae3x−B−6(Ae3x+Bx +C)=2e3x+ 4x.

Step 5: Solve for the coefficients. Equating the coefficients of like terms,

we get the following equations:

(9A−3A−6A)e3x−6B= 2e3x,

−B−6B= 4,

which simplifies to

0 = 2e3x,

−7B= 4.

From the second equation, we find that B=−4

7. Since the first equation has

no solution, we adjust our guess for yp.

Step 6: Adjust the guess for the particular solution. With yp=Ax2e3x+

Bx +Cas our new guess, we find the derivatives and substitute them back into

the differential equation.

Step 7: Find the derivatives for the adjusted particular solution.

y′

p= (2Ax +B)e3x+Ax2e3x,

y′′

p= (6Ax + 2A)e3x+ (2Ax +B)e3x.

Step 8: Substitute the adjusted particular solution and its derivatives into

the differential equation. Substitute yp,y′

p, and y′′

pinto the differential equation

y′′ −y′−6y= 2e3x+ 4x, and simplify. This will lead to equations involving A,

B, and C.

Step 9: Solve for the coefficients. By equating the coefficients of like terms,

we can solve for A,B, and C. Once these coefficients are found, the particular

solution is complete.

By following these steps carefully, we can find the particular solution to the

given differential equation using the method of undetermined coefficients.

Question 26

Question

Find the general solution to the differential equation y′′ −4y′+ 4y= 6x2e2x+

15xe2x.

Solution

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

method of undetermined coefficients.

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

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

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

Step 2: Next, we need to find a particular solution to the nonhomogeneous

equation y′′ −4y′+ 4y= 6x2e2x+ 15xe2x. Assume a particular solution of the

form yp=Ax2e2x+Bxe2x.

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

y′

p= (2Ax +B)e2x+ 2Ax2e2x,

y′′

p= (2A+ 2Ax +B)e2x+ 4Axe2x.

Step 4: Substitute yp,y′

p, and y′′

pback into the differential equation and

simplify:

(2A+ 2Ax +B−4A−4Ax −4B+ 4Ax)e2x= 6x2e2x+ 15xe2x,

(−2A−4B)e2x= 6x2e2x+ 15xe2x.

Matching coefficients, we get the following system of equations:

−2A= 6,

−4B= 15.

Solving this system, we find A=−3 and B=−15

Step 5: The particular solution is yp=−3x2e2x−15

4xe2x.

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

of the complementary and particular solutions:

y=yc+yp=c1e2x+c2xe2x−3x2e2x−15

4xe2x.

Question 27

Question

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

undetermined coefficients: y′′ −y′−6y= 2e3x+ 5x2.

Solution

To solve the non-homogeneous differential equation y′′ −y′−6y= 2e3x+ 5x2,

we first find the complementary function ycby solving the corresponding homo-

geneous equation y′′ −y′−6y= 0.

Step 1: Find the complementary function yc.

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

is r2−r−6 = 0. Factoring gives (r−3)(r+ 2) = 0, so the roots are r= 3 and

r=−2.

Therefore, the complementary function ycis given by yc=c1e3x+c2e−2x,

where c1and c2are constants to be determined.

Step 2: Find the particular solution yp.

Since our non-homogeneous equation contains terms involving e3xand x2,

we make the educated guess that yphas the form:

yp=Ae3x+Bx2+Cx +D

Taking the derivatives:

y′

p= 3Ae3x+ 2Bx +C

y′′

p= 9Ae3x+ 2B

Substitute ypand its derivatives back into the original equation:

9Ae3x+ 2B−(3Ae3x+ 2Bx +C)−6(Ae3x+Bx2+Cx +D) = 2e3x+ 5x2

Matching coefficients of like terms gives a system of equations to solve for

A,B,C, and D.

Step 3: Solve for the coefficients A,B,C, and D.

By solving the system of equations, we find:

A=1

3, B =1

12, C =1

4, D =−1

Thus, the particular solution ypis:

yp=1

3e3x+1

12x2+1

4x−1

Step 4: Write the general solution.

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

of the complementary function ycand the particular solution yp:

y=yc+yp=c1e3x+c2e−2x+1

3e3x+1

12x2+1

4x−1

Question 28

Question

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

the method of undetermined coefficients.

Solution

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

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

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

Step 2: Next, find a particular solution ypby assuming that yp= (Ax+B)ex.

Compute the derivatives needed for substitution into the original differential

equation.

y′

p= (A+Ax +B)ex

y′′

p= (A+ 2A+Ax +B)ex

Step 3: Substitute yp, y′

p,and y′′

pinto the differential equation and then solve

for Aand B.

(A+ 2A+Ax +B)ex−3(A+Ax +B)ex+ 2(Ax +B)ex= 3xex

(3A−3B)ex= 3xex

Comparing coefficients, we get 3A−3B= 3, which simplifies to A−B= 1.

Step 4: Solve for Aand Bby setting A=B+ 1. Let B= 0 for simplicity,

then A= 1.

Step 5: Therefore, the particular solution is yp= (x+ 1)ex.

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

complementary and particular solutions, so y=yc+yp=c1ex+c2e2x+(x+1)ex.

Question 29

Question

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

differential equation

y′′ −3y′+ 2y= 12e2x.

Solution

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

the particular solution has the form yp=Ae2xwhere Ais a constant to be

determined.

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

ypare:

y′

p= 2Ae2x, y′′

p= 4Ae2x.

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation.

Substitute yp,y′

p, and y′′

pinto the differential equation to get:

4Ae2x−3(2Ae2x)+2Ae2x= 12e2x.

Step 3: Solve for A.Simplifying the equation, we get:

4Ae2x−6Ae2x+ 2Ae2x= 12e2x.

Ae2x= 12e2x.

A= 12.

Step 4: Form the particular solution. Therefore, the particular solution

is yp= 12e2x.

Thus, a particular solution to the differential equation is yp= 12e2x.

Question 30

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y= 8e2x.

Solution

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

efficients, we first find the complementary function and particular solution.

Step 1: Find the complementary function solution The complemen-

tary function solves the homogeneous version of the differential equation, which

y′′ −4y′+ 4y= 0.

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

Therefore, the complementary function is

yc(x) = (C1+C2x)e2x.

Step 2: Assume a particular solution For the particular solution, since

the nonhomogeneous term is in the form e2x, we assume the particular solution

has the form

yp(x) = Ae2x.

Step 3: Find the derivatives of the assumed particular solution

yp(x) = Ae2x,

y′

p(x) = 2Ae2x,

y′′

p(x) = 4Ae2x.

Step 4: Substitute the particular solution and its derivatives into

the differential equation Substitute yp(x), y′

p(x), and y′′

p(x) into the differ-

ential equation:

4Ae2x−4(2A)e2x+ 4(A)e2x= 8e2x,

4Ae2x−8Ae2x+ 4Ae2x= 8e2x.

Step 5: Solve for the unknown coefficients Solving the equation ob-

tained in the previous step:

0=8e2x⇒0 = 8.

Since 0 = 8, there is no solution for A. Therefore, to fix this issue, we

multiply the assumed particular solution by x:

yp(x) = Ax2e2x.

Step 6: Find the derivatives of the corrected assumed particular

solution

yp(x) = Ax2e2x,

y′

p(x) = (2Ax + 2x2)e2x,

y′′

p(x) = (4A+ 4x+ 4x)e2x= (4A+ 8x)e2x.

Step 7: Substitute the corrected particular solution and its deriva-

tives into the differential equation Substitute yp(x), y′

p(x), and y′′

p(x) into

the differential equation:

(4A+ 8x)e2x−4(2Ax + 2x2)e2x+ 4Ax2e2x= 8e2x,

(4A+ 8x)e2x−8Axe2x−8x2e2x+ 4Ax2e2x= 8e2x.

Step 8: Solve for the unknown coefficients Comparing coefficients of

e2xon both sides:

−8x2e2x+ 4Ax2e2x= 8e2x.

This implies 4A= 8, so A= 2. Therefore, the particular solution is

yp(x) = 2x2e2x.

Step 9: Form the general solution The general solution is the sum of

the complementary function and the particular solution:

y(x) = yc(x) + yp(x)=(C1+C2x)e2x+ 2x2e2x.

Question 31

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x−3x

Solution

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

by solving the associated homogeneous equation y′′ −3y′+ 2y= 0. Then, we

will find a particular solution for the non-homogeneous part 6e2x−3xusing the

method of undetermined coefficients.

Step 1: Find the complementary solution The characteristic equation

for the homogeneous part is r2−3r+ 2 = 0. This can be factored 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: Find the particular solution Let’s assume the particular solution

has the form:

yp=Ae2x+Bx +C

Calculate the derivatives and substitute them into the differential equation:

y′

p= 2Ae2x+B

y′′

p= 4Ae2x

Substitute yp,y′

p, and y′′

pinto the differential equation and equate coefficients

of like terms:

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

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

Simplify and equate coefficients:

0+(−3B+ 2) + 0 = −3 =⇒B=−1

0+0+(−3(−1) + 2C) = 0 =⇒3+2C= 0 =⇒C=−3

Therefore, the particular solution is yp=Ae2x−x−3

Step 3: Construct the general solution The general solution is the sum

of the complementary and particular solutions:

y=yc+yp=c1ex+c2e2x+Ae2x−x−3