MATH 334 - DIFFERENTIAL EQUATIONS Method of undetermined coefficients Question Bank Set 5

MATH 334 - DIFFERENTIAL

EQUATIONS - Method of

undetermined coefficients

Question Bank - Set 5

Liberty University

Question 1

Question

Consider the differential equation:

y′′ −4y′+ 4y=e2x

Find the particular solution using the method of undetermined coefficients.

Solution

Step 1: Find the complementary solution.

The characteristic equation for the homogeneous part is:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2

Therefore, the complementary solution is

yc(x) = c1e2x+c2xe2x

Step 2: Assume the particular solution has the form

yp(x) = Ae2x

where Ais a constant to be determined.

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

y′

p(x)=2Ae2x

y′′

p(x)=4Ae2x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) back into the original differential

equation:

4Ae2x−4(2Ae2x)+4Ae2x=e2x

Step 5: Simplify the equation and solve for A:

4Ae2x−8Ae2x+ 4Ae2x=e2x

0 = e2x

There is no solution for Ain this case.

Step 6: Adjust the assumption for the particular solution. Since e2xis

also present in the complementary solution, we need to multiply our previous

assumption by x:

yp(x) = Axe2x

Step 7: Calculate the first and second derivatives of the new assumed par-

ticular solution:

y′

p(x) = Ae2x+ 2Axe2x

y′′

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

Step 8: Substitute the new yp(x), y′

p(x), and y′′

p(x) back into the original

differential equation:

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

Step 9: Simplify the equation and solve for A:

4Axe2x−8Axe2x+ 4Axe2x=e2x

0 = e2x

There is no solution for Ain this case either.

Step 10: Since the form of our particular solution matches terms in the

homogeneous solution, we need to multiply our assumed particular solution by

x2:

yp(x) = Ax2e2x

Step 11: Repeat steps 3 to 9 using this new assumed form of the particular

solution to find the value of A.

Question 2

Question

Solve the following second-order linear nonhomogeneous differential equation

using the method of undetermined coefficients:

y′′ + 3y′+ 2y= 2e−x+ 4 cos(x).

Solution

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

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

homogeneous equation:

y′′

h+ 3y′

h+ 2yh= 0.

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

Therefore, the complementary solution is given by:

yc(x) = c1e−x+c2e−2x,

where c1and c2are arbitrary constants.

Next, we find the particular solution of the nonhomogeneous equation by

assuming the form:

yp(x) = Ae−x+Bcos(x) + Csin(x).

Taking derivatives, we have:

y′

p(x) = −Ae−x−Bsin(x) + Ccos(x),

y′′

p(x) = Ae−x+Bcos(x)−Csin(x).

Substitute these back into the original differential equation:

Ae−x+Bcos(x)−Csin(x)+3(−Ae−x−Bsin(x)+Ccos(x))+2(Ae−x+Bcos(x)+Csin(x)) = 2e−x+4 cos(x).

Solving for A,B, and C, we find A=−1

3,B=8

5, and C= 0. Therefore,

the particular solution is:

yp(x) = −1

3e−x+8

5cos(x).

The general solution to the nonhomogeneous equation is then given by com-

bining the complementary and particular solutions:

y(x) = yc(x) + yp(x) = c1e−x+c2e−2x−1

3e−x+8

5cos(x).

Question 3

Question

Solve the following second-order differential equation using the method of un-

determined coefficients:

y′′ −5y′+ 6y= 4e2x

Solution

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

r2−5r+ 6 = 0. The characteristic equation is (r−2)(r−3) = 0, so the roots

are r= 2 and r= 3. Therefore, the complementary solution is:

yc(x) = C1e2x+C2e3x

Step 2: Since the right-hand side 4e2xcontains the term e2xand e2xis

already present in the complementary solution, we need to modify the proposed

particular solution to avoid redundancy. We can use the form yp(x) = Ax2e2x.

Step 3: Calculate the derivatives needed to substitute into the differential

equation:

yp(x) = Ax2e2x

y′

p(x) = 2Axe2x+ 2Ax2e2x

y′′

p(x)=4Ae2x+ 8Axe2x+ 4Ax2e2x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) back into the differential equation:

4Ae2x+ 8Axe2x+ 4Ax2e2x−5(2Axe2x+ 2Ax2e2x)+6Ax2e2x= 4e2x

Step 5: Simplify the equation above and solve for A.

4Ae2x+ 8Axe2x+ 4Ax2e2x−10Axe2x−10Ax2e2x+ 6Ax2e2x= 4e2x

Step 6: Combine like terms and solve for Aby setting the coefficients of e2x

equal to 4:

A(4 −10x+ 6x2)e2x= 4e2x

A(6x2−10x+ 4) = 4

Step 7: The above equation implies A=4

6x2−10x+4 . Therefore, the particu-

lar solution is yp(x) = 4x2

6x2−10x+4 e2x.

Step 8: The general solution is the sum of the complementary solution and

the particular solution:

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

y(x) = C1e2x+C2e3x+4x2

6x2−10x+ 4e2x

Question 4

Question

Solve the following non-homogeneous 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

characteristic equation for the homogeneous part of the differential equation is:

r2−2r+ 1 = 0.

Solving this quadratic equation, we find that r= 1 is a repeated root. Therefore,

the general solution to the homogeneous equation is:

yh(x) = c1ex+c2xex,

where c1and c2are arbitrary constants.

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

tion Since we have 3exon the right-hand side, we will assume the particular

solution has the form:

yp(x) = Aex,

where Ais a constant to be determined.

Substitute yp(x) into the original differential equation:

(Aex)′′ −2(Aex)′+Aex= 3ex.

Simplify and solve for A:

Aex−2Aex+Aex= 3ex.

Aex= 3ex.

A= 3.

Thus, the particular solution is yp(x) = 3ex.

Step 3: Find the general solution to the non-homogeneous equa-

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

general solution to the homogeneous equation and the particular solution:

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

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

y(x) = c1ex+c2xex+ 3ex,

where c1and c2are arbitrary constants.

Question 5

Question

Determine the particular solution to the nonhomogeneous differential equation

y′′ + 4y= 5 sin(2x) using the method of undetermined coefficients.

Solution

Step 1: Find the complementary solution by solving the homogeneous dif-

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

which has solutions r=±2i. Therefore, the complementary solution is yc=

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

Step 2: Assume the particular solution has the form yp=asin(2x) +

bcos(2x).

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

y′

p= 2acos(2x)−2bsin(2x)

y′′

p=−4asin(2x)−4bcos(2x)

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

solve for aand b.

−4asin(2x)−4bcos(2x) + 4(asin(2x) + bcos(2x)) = 5 sin(2x)

(−4a+ 4b) sin(2x) + (4a+ 4b) cos(2x) = 5 sin(2x)

Step 5: Equate coefficients of like terms to get two equations:

−4a+ 4b= 5

4a+ 4b= 0

Step 6: Solve the system of equations to find aand b. From the second

equation, b=−a. Substitute this into the first equation:

−4a+ 4(−a)=5

−8a= 5

a=−5

Step 7: Find busing b=−a.

b=−−5

8=5

Therefore, the particular solution is yp=−5

8sin(2x) + 5

8cos(2x).

The general solution to the nonhomogeneous differential equation is the sum

of the complementary and particular solutions:

y=yc+yp=c1cos(2x) + c2sin(2x)−5

8sin(2x) + 5

8cos(2x)

Question 6

Question

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

efficients:

y′′ −5y′+ 6y= 6e2x

Solution

To solve the given second order homogeneous differential equation, we first find

the complementary function, and then the particular solution using the method

of undetermined coefficients.

Step 1: Find the complementary function The characteristic equation

for the homogeneous differential equation y′′ −5y′+ 6y= 0 is:

r2−5r+ 6 = 0

(r−2)(r−3) = 0

r= 2,3

So, the complementary function is:

yc=c1e2x+c2e3x

Step 2: Find the particular solution Let’s assume the particular solution

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

Now, plug ypback into the original differential equation:

y′′ −5y′+ 6y= 6e2x

(2A)e2x−5(2A)e2x+ 6(A)e2x= 6e2x

2A−10A+ 6A= 6

−2A= 6

A=−3

Therefore, the particular solution is yp=−3e2x.

Step 3: Write the general solution The general solution to the dif-

ferential equation is given by combining the complementary function and the

particular solution:

y(x) = yc+yp

y(x) = c1e2x+c2e3x−3e2x

Thus, the solution to the given differential equation y′′ −5y′+ 6y= 6e2xis

y(x) = c1e2x+c2e3x−3e2x.

Question 7

Question

Find the particular solution to the nonhomogeneous differential equation

y′′ −3y′+ 2y= 5e2t.

Solution

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

will use the method of undetermined coefficients.

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. This gives us the characteristic roots r1= 1 and

r2= 2. Therefore, the complementary solution is

yc(t) = c1et+c2e2t,

where c1and c2are constants.

Step 2: Now, we assume the particular solution has the form yp(t) = Ae2t,

where Ais a constant to be determined.

Step 3: Calculate the derivatives of yp(t) needed for substitution into the

nonhomogeneous equation:

y′

p(t)=2Ae2tand y′′

p(t)=4Ae2t.

Step 4: Substitute yp(t), y′

p(t), and y′′

p(t) into the nonhomogeneous equa-

tion: 4Ae2t−3(2Ae2t) + 2(Ae2t)=5e2t

4Ae2t−6Ae2t+ 2Ae2t= 5e2t

0=5e2t.

Step 5: Since 0 = 5e2t, we adjust the form of yp(t) to yp(t) = Ate2t.

Step 6: Recalculate derivatives and substitute into the nonhomogeneous

equation:

y′

p(t) = Ae2t+ 2Ate2tand y′′

p(t) = 2Ae2t+ 2Ae2t+ 4Ate2t.

Substitute into the equation:

2Ae2t+ 2Ae2t+ 4Ate2t−3(Ae2t+ 2Ate2t) + 2(Ate2t)=5e2t.

Step 7: After simplification, we get 3A= 5, so A=5

3. Therefore, the

particular solution is yp(t) = 5

3te2t.

Step 8: The general solution is the sum of the complementary and particular

solutions:

y(t) = yc(t) + yp(t) = c1et+c2e2t+5

3te2t.

So the particular solution to the given nonhomogeneous differential equation is

3te2t.

Question 8

Question

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

using the method of undetermined coefficients.

Solution

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

y′′ −4y′+ 4y= 0 by solving the characteristic equation r2−4r+ 4 = 0. The

characteristic equation has a repeated root at r= 2, so the complementary

solution is yc=c1e2x+c2xe2x.

Step 2: Now, we find the particular solution by assuming that yp=Ax2e2x.

Calculate y′

pand y′′

y′

p= (2Ax +A)x2e2xand y′′

p= (2A+ 4Ax + 2A)x2e2x

Step 3: Substitute yp,y′

p, and y′′

pback into the original differential equation

to find A.

(2A+ 4Ax + 2A)x2e2x−4[(2Ax +A)x2e2x] + 4(Ax2e2x) = xe2x

Step 4: Simplify the equation and solve for A. Then, substitute Aback

into ypto find the particular solution.

Step 5: Finally, the general solution to the differential equation is the sum

of the complementary and particular solutions.

y=yc+yp=c1e2x+c2xe2x+Ax2e2x

Question 9

Question

Find the general solution to the differential equation

y′′ −3y′+ 2y= 3et+ 4 cos t.

Solution

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

method of undetermined coefficients.

Step 1: Homogeneous Solution

First, we find the 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.

Thus, r= 1,2.

So, the homogeneous solution is

yh(t) = c1et+c2e2t,

where c1and c2are arbitrary constants.

Step 2: Particular Solution

Now, we seek a particular solution to the non-homogeneous equation y′′ −

3y′+ 2y= 3et+ 4 cos tof the form

yp(t) = Atet+Bcos t+Csin t.

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

y′

p(t) = Aet+Bcos t−Csin t,

y′′

p(t) = Aet−Bcos t−Csin t.

Substitute yp(t) and its derivatives back into the differential equation, and

equate coefficients of the terms with the same function:

A−3(Aet+Bcos t−Csin t) + 2(Atet+Bcos t+Csin t)

= 3et+ 4 cos t.

Then, solve for A,B, and C.

Step 3: General Solution

The general solution to the differential equation y′′ −3y′+ 2y= 3et+ 4 cos t

is the sum of the homogeneous and particular solutions:

y(t) = yh(t) + yp(t).

Therefore, the general solution is

y(t) = c1et+c2e2t+Atet+Bcos t+Csin t.

Question 10

Question

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

the method of undetermined coefficients:

y′′ + 4y= 9 cos(2x)+3e3x

Solution

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

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

homogeneous equation, then find a particular solution to the non-homogeneous

equation.

Step 1: Find the general solution to the homogeneous equation

The homogeneous form of the differential equation is y′′ + 4y= 0. The char-

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

general solution to the homogeneous equation is:

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

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

tion

For the particular solution, we assume:

yp(x) = Acos(2x) + Be3x

Taking the derivatives and substituting them into the differential equation,

we get:

y′′

p(x)+ 4yp(x) = −4Acos(2x)+12Be3x+ 4Acos(2x)+ 4Be3x= 9 cos(2x) +3e3x

Solving the system of equations −4A= 9 and 12B+ 4B= 3 gives A=−9

and B=3

16 . Therefore, the particular solution is:

yp(x) = −9

4cos(2x) + 3

16e3x

Step 3: Form the general solution

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

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

y(x) = yh(x) + yp(x) = c1cos(2x) + c2sin(2x)−9

4cos(2x) + 3

16e3x

Question 11

Question

Find a particular solution to the differential equation: y′′ −4y′+4y= 8e2x−3x2

using the method of undetermined coefficients.

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

to the homogeneous equation.

Homogeneous Equation: The characteristic equation is r2−4r+ 4 = 0.

Factoring the characteristic equation gives (r−2)2= 0. This implies that the

characteristic root is r= 2.

Therefore, the complementary solution is yc(x) = c1e2x+c2xe2x, where c1

and c2are constants to be determined.

Particular Solution: We propose a particular solution of the form yp(x) =

Ax2e2x+Bx +Cfor the nonhomogeneous part of the equation.

Step 1: Calculate the derivatives of yp(x)

y′

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

y′′

p(x) = (2A+ 4Ax + 2B)e2x+ 4Ax2e2x

Step 2: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation

and simplify:

(2A+ 4Ax + 2B−4(2Ax +B) + 4(Ax2+Bx +C))e2x= 8e2x−3x2

Grouping like terms gives:

(4A−4B)e2x+ (8A−4B+ 4C)e2xx+ (4A)x2e2x= 8e2x−3x2

Step 3: Compare coefficients of like terms: From the term e2x: 4A−4B= 8

(1) From the term xe2x: 8A−4B+ 4C= 0 (2) From the term x2e2x: 4A=−3

Solving these equations simultaneously, we find A=−3

4,B=−11

2, and

C=89

Step 4: The particular solution is:

yp(x) = −3

4x2e2x−11

2x+89

Therefore, the general solution to the differential equation is:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x−3

4x2e2x−11

2x+89

Question 12

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x.

Solution

We first find the general solution to the associated homogeneous equation, y′′ −

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

(r−1)(r−2) = 0. Hence, the characteristic roots are r1= 1 and r2= 2. The

general solution to the homogeneous equation is then

yh=c1ex+c2e2x,

where c1and c2are arbitrary constants.

Now, we can find a particular solution to the non-homogeneous equation.

Since the right-hand side is an exponential function, we guess a particular solu-

tion of the form

yp=Ae2x,

where Ais a constant to be determined.

Step 1: Compute the derivatives of yp.

y′

p= 2Ae2x,

y′′

p= 4Ae2x.

Step 2: Substitute ypand its derivatives into the non-homogeneous equa-

tion.

y′′

p−3y′

p+ 2yp= 4Ae2x−3(2Ae2x)+2Ae2x= 6e2x.

Step 3: Simplify the equation and solve for A.

4Ae2x−6Ae2x+ 2Ae2x= 6e2x.

0=6e2x.

Since there is no solution to the equation 0 = 6e2x, we try a more general

form for yp.

yp=Axe2x.

Step 4: Compute the derivatives of yp.

y′

p= (2A+Ax)e2x,

y′′

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

Step 5: Substitute ypand its derivatives into the non-homogeneous equa-

tion.

(4A+ 2A+Ax)e2x−3(2A+Ax)e2x+ 2Axe2x= 6e2x.

Simplify the equation and solve for A.

Step 6: Solve for Aafter simplifying the equation.

6Ae2x= 6e2x.

A= 1.

Step 7: Write down the particular solution. The particular solution to the

non-homogeneous equation is

yp=xe2x.

Step 8: Write down the general solution. Therefore, the general solution to

the non-homogeneous equation is

y=yh+yp=c1ex+c2e2x+xe2x,

where c1and c2are arbitrary constants.

Question 13

Question

Find the particular solution to the differential equation y′′ −4y= 6e2xusing

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 of the homogeneous equation

y′′ −4y= 0.

Step 1: Find the Complementary Solution

The characteristic equation is r2−4 = 0, which gives roots r=±2. There-

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

Step 2: Guess the Form of the Particular Solution

For the particular solution, we guess a form based on the form of the non-

homogeneous term. Since 6e2xhas the same form as Ae2x, we guess the partic-

ular solution to be of the form yp=Ae2x.

Step 3: Find the Particular Solution

Substitute yp=Ae2xinto the differential equation:

y′′ −4y= 6e2x

(4A−4A)e2x= 6e2x

Solving for A, we get A=3

Therefore, the particular solution is yp=3

2e2x.

Step 4: Write the General Solution

The general solution is given by y=yc+yp, thus the general solution to the

differential equation y′′ −4y= 6e2xis:

y=c1e2x+c2e−2x+3

2e2x

Question 14

Question

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

4y= 8x2−3 sin(x) using the method of undetermined coefficients.

Solution

To find the general solution to the non-homogeneous differential equation, we

will first find the complementary function by solving the corresponding ho-

mogeneous equation y′′ −4y= 0, and then find a particular solution for the

non-homogeneous part of the equation.

Step 1: Find the complementary function The complementary func-

tion satisfies the homogeneous equation y′′ −4y= 0. The characteristic equation

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

of the form:

yc=c1e2x+c2e−2x

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

assume the form yp=Ax2+Bx+C+Dsin(x). Now, we calculate the derivatives

needed for substitution:

y′

p= 2Ax +B+Dcos(x)

y′′

p= 2A−Dsin(x)

Substitute these derivatives back into the non-homogeneous equation to get:

2A−Dsin(x)−4(Ax2+Bx +C+Dsin(x)) = 8x2−3 sin(x)

This simplifies to:

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

Now, we match coefficients on both sides to get the system of equations:

2A−4D= 8

2B−4D= 0

2A−4C= 0

Solving this system of equations gives A=−1, B= 0, C=−1, and D=−2.

Therefore, the particular solution is:

yp=−x2−2 sin(x)

Step 3: General Solution Finally, the general solution to the non-homogeneous

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

ular solution:

y=yc+yp=c1e2x+c2e−2x−x2−2 sin(x)

where c1and c2are constants determined by initial conditions.

Question 15

Question

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

the method of undetermined coefficients: y′′ −4y′+ 4y= 2e2x+x2.

Solution

To solve the non-homogeneous 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 com-

plementary solution is yc(x)=(c1+c2x)e2x,where c1and c2are arbitrary

constants.

Next, we find the particular solution yp(x) for the non-homogeneous equation

y′′ −4y′+ 4y= 2e2x+x2.

1. For the term 2e2x: Assume yp(x) = Ae2x,where Ais a constant to be

determined. Substituting yp(x) into the differential equation, we get:

(2A−4A+ 4A)e2x= 2e2x

2A= 2

A= 1

Therefore, the particular solution for 2e2xis yp1(x) = e2x.

2. For the term x2: Assume yp(x) = Bx2+Cx +D, where B,C, and D

are constants to be determined. Substituting yp(x) into the differential

equation, we get:

2B−4(2Bx +C) + 4(Bx2+Cx +D) = x2

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

Equating coefficients, we get:

4B−4B= 1 ⇒0 = 1

(no solution) Since there is no solution for the term x2in yp(x), we need

to modify our assumption.

3. Revised assumption for x2: Assume yp(x) = Bx2+Cx +Dx +E, where

B,C,D, and Eare constants to be determined. Substituting yp(x) into

the differential equation, we get:

2B−4(2Bx +C) + 4(Bx2+Cx +Dx +E) = x2

Equating coefficients, we get:

4B= 0 ⇒B= 0

−8B+ 4C+ 4B= 1 ⇒4C= 1 ⇒C=1

2B−4C+ 4D= 0 ⇒4D= 1 ⇒D=1

−4C+ 4E= 0 ⇒4E= 1 ⇒E=1

Therefore, the particular solution for x2is yp2(x) = 1

4x2+1

4x+1

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

is:

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

4x2+1

4x+1

Question 16

Question

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

undetermined coefficients:

y′′ −y′= 6x2+ 1

Solution

We first find the solution to the associated homogeneous equation:

y′′ −y′= 0

Characteristics equation: r2−r=r(r−1) = 0 Therefore, the characteristic

roots are r1= 0 and r2= 1.

The homogeneous solution is:

yh=c1+c2ex

Next, we need to find a particular solution of the non-homogeneous equation:

yp=Ax2+Bx +C

y′

p= 2Ax +B

y′′

p= 2A

Substitute these expressions back into the original differential equation:

2A−(2Ax +B)=6x2+ 1

2A−2Ax −B= 6x2+ 1

By comparing coefficients, we find:

A=−3, B = 0, C =−1

Therefore, the particular solution is:

yp=−3x2−1

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

homogeneous and particular solutions:

y=yh+yp=c1+c2ex−3x2−1

Question 17

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y=e2x−x

Solution

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

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

homogeneous equation:

y′′ −4y′+ 4y= 0

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

Therefore, the complementary solution is:

yc(x)=(c1+c2x)e2x

Now, let’s find the particular solution yp(x) for the nonhomogeneous equa-

tion y′′ −4y′+ 4y=e2x−x.

Step 1: Guess the form of yp(x) based on the nonhomogeneous terms:

yp(x) = Ae2x+Bx +C

Step 2: Calculate the first and second derivatives of yp(x):

y′

p(x)=2Ae2x+Band y′′

p(x) = 4Ae2x

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) back into the nonhomogeneous

equation:

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

4Ae2x−8Ae2x−4B+ 4Ae2x+ 4Bx + 4C=e2x−x

Step 4: Equate coefficients of like terms to solve for A,B, and C:

4A−8A+ 4A= 1 =⇒A=−1

−4B+ 4B=−1 =⇒0 = −1 (No solution)

4C=−1 =⇒C=−1

Therefore, the particular solution is:

yp(x) = −1

4e2x−1

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

solutions:

y(x) = yc(x) + yp(x)=(c1+c2x)e2x−1

4e2x−1

Hence, the solution to the differential equation is:

y(x)=(c1−1

4)e2x+c2xe2x−1

Question 18

Question

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

the method of undetermined coefficients: y′′ −4y′+ 4y= 2xe2x+ 3.

Solution

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

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

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

Step 2: Identify the form of the particular solution based on the non-

homogeneous terms. Since the right-hand side contains xe2xand a constant

term, we assume the particular solution has the form yp= (Ax +B)xe2x+C,

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

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

p= (2A(x+ 1) +

B)xe2x+Ae2xy′′

p= (2A(x+ 2) + 2A+B)xe2x+ 4Ae2x

Step 4: Substitute yp,y′

p, and y′′

pback into the original differential equation

and simplify to form an equation in terms of A,B, and C:

(2A(x+ 2) + 2A+B)xe2x+ 4Ae2x−4[(2A(x+ 1) + B)xe2x+Ae2x] + 4[(Ax +

B)xe2x+C] = 2xe2x+ 3

Step 5: Equate coefficients of terms with the same factors on both sides of

the equation. This will give you a system of equations to solve for A,B, and C.

Step 6: Once A,B, and Care found, the particular solution ypcan be

determined. Finally, the general solution is given by y=yc+yp.

Question 19

Question

Use the method of undetermined coefficients to solve the following non-homogeneous

differential equation:

y′′ −4y′+ 4y= 4x2ex

Solution

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

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

the particular solution based on the form of the non-homogeneous term, and

finally combine both solutions.

Step 1: Find the complementary solution The complementary solution

for the given differential equation y′′ −4y′+ 4y= 0 can be found by solving the

characteristic equation:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2 (repeated root)

So, the complementary solution is:

yc(x) = c1e2x+c2xe2x

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

for the non-homogeneous part 4x2exto be of the form:

yp(x) = (ax2+bx +c)ex

We will now differentiate yp(x) twice and substitute it back into the differ-

ential equation to determine the coefficients a,b, and c.

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

yp(x) into the differential equation y′′ −4y′+ 4y= 4x2ex:

y′′

p= 2a+ 2b+ 2(ax +b+c)ex

y′

p= 2ax +a+b+ (2ax + 2b+c)ex

y′′ −4y′+4y= 2a+2b+2(ax+b+c)ex−4(2ax+a+b+(2ax+2b+c)ex)+4(ax2+bx+c)ex

Equating the coefficients of similar terms, we get:

2a−4(2a)+4a= 4x2=⇒ −6a= 4 =⇒a=−2

2b−4(2b)+4b= 4x=⇒ −6b= 4 =⇒b=−2

2+2b−4(2a+a+b)+4c= 0 =⇒2−12a−6b+ 4c= 0 =⇒c=−10

Therefore, the particular solution is:

yp(x) = −2

3x2−2

3x−10

9ex

Step 4: Write the general solution The general solution of the differ-

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

solution:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x−2

3x2ex−2

3xex−10

9ex

So, the solution to the non-homogeneous differential equation y′′ −4y′+4y=

4x2exis:

y(x) = c1e2x+c2xe2x−2

3x2ex−2

3xex−10

9ex

Question 20

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x.

Solution

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

we will first find the complementary solution to the associated homogeneous

equation, then determine the particular solution for the non-homogeneous term.

Step 1: Find the complementary solution

The associated homogeneous equation is:

y′′ −3y′+ 2y= 0.

Characteristics equation: 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=c1et+c2e2t.

Step 2: Find the particular solution

We assume the particular solution has the form:

yp=Ae2x.

Taking derivatives:

y′

p= 2Ae2xand y′′

p= 4Ae2x.

Substitute these derivatives and ypback into the non-homogeneous equation:

4Ae2x−3(2Ae2x)+2Ae2x= 6e2x.

Simplify the left-hand side:

4Ae2x−6Ae2x+ 2Ae2x= 6e2x.

Solving for Agives A= 3.

Therefore, the particular solution is:

yp= 3e2x.

Step 3: Determine the general solution

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

y=yc+yp=c1ex+c2e2x+ 3e2x.

So the general solution to the differential equation is:

y=c1ex+c2e2x+ 3e2x.

Question 21

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −2y′−3y=ex+ sin(2x)

Solution

To solve the given nonhomogeneous differential equation using the method of

undetermined coefficients, we first need to find the complementary function and

particular solution.

Step 1: Find the complementary function

The complementary function yc(x) is found by solving the corresponding

homogeneous equation:

y′′ −2y′−3y= 0

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

0. So, the characteristic roots are r1= 3 and r2=−1.

Therefore, the complementary function is:

yc(x) = C1e3x+C2e−x

where C1and C2are constants to be determined.

Step 2: Find the particular solution

Since the right-hand side of the given differential equation includes terms

like exand sin(2x), we assume the particular solution has the form:

yp(x) = Aex+Bsin(2x) + Ccos(2x)

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

Step 3: Substitute the particular solution and its derivatives into

the differential equation

Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation and simplify

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

Step 4: Write the general solution

The general solution to the nonhomogeneous differential equation is the sum

of the complementary function and the particular solution:

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

Substitute the found constants C1,C2,A,B, and Cinto this expression to

obtain the final solution.

Question 22

Question

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

the method of undetermined coefficients.

Solution

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

we assume that the particular solution has the form yp=Asin(3x) + Bcos(3x),

where Aand Bare constants to be determined.

Step 1: Find the derivatives of yp.

y′

p= 3Acos(3x)−3Bsin(3x)

y′′

p=−9Asin(3x)−9Bcos(3x)

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation.

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

Step 3: Equate coefficients of terms with the same trigonometric

function. For the sin(3x) terms:

−9A−A= 1 =⇒ −10A= 1 =⇒A=−1

For the cos(3x) terms:

−9B−B= 0 =⇒ −10B= 0 =⇒B= 0

Step 4: Write down the particular solution. The particular solution

is:

yp=−1

10 sin(3x)

Question 23

Question

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

following differential equation:

y′′ +y= 3 cos(2x)+4x2.

Solution

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

a form for yp(x) that includes all terms from the right-hand side of the equation.

In this case, we will guess that yp(x) = Acos(2x) + Bx2+Cx +D, where A,B,

C, and Dare constants to be determined.

Step 1: Find y′

p(x)and y′′

p(x).

y′

p(x) = −2Asin(2x)+2Bx +C,

y′′

p(x) = −4Acos(2x)+2B.

Step 2: 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 given differential equation:

−4Acos(2x)+2B+Acos(2x) + Bx2+Cx +D= 3 cos(2x)+4x2.

Step 3: Equate coefficients of like terms. Grouping like terms, we

have:

(B−4A)x2+ (C−2A)x+ (−A+ 2B+D)=4x2+ 3 cos(2x).

Since both sides must be equal for all values of x, we equate the coefficients

of each term on both sides to get the system of equations:











B−4A= 4,

C−2A= 0,

−A+ 2B+D= 0.

Step 4: Solve the system of equations. Solving the system of equations,

we find:

A=−2, B = 2, C =−4, D = 8.

Step 5: Write the particular solution. Therefore, the particular solu-

tion yp(x) is:

yp(x) = −2 cos(2x)+2x2−4x+ 8.

Question 24

Question

Find the particular solution to the differential equation y′′ −4y′+ 4y= 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(x) = Ae2x, where Ais a

constant to be determined.

Step 1: Find y′

p(x)and y′′

p(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 differential equa-

tion.

4Ae2x−4(2Ae2x) + 4(Ae2x)=3e2x

Step 3: Solve for A.

4Ae2x−8Ae2x+ 4Ae2x= 3e2x

−4Ae2x= 3e2x

A=−3

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

yp(x) = −3

4e2x.

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

3e2xis yp(x) = −3

4e2x.

Question 25

Question

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

using the method of undetermined coefficients:

y′′ −4y′+ 4y= 3e2x+ 2x2−1

Solution

The general form of the non-homogeneous differential equation is given by:

y′′ −4y′+ 4y= 3e2x+ 2x2−1

Step 1: Find the complementary solution First, we solve the corre-

sponding homogeneous equation:

y′′ −4y′+ 4y= 0

The characteristic equation is:

r2−4r+ 4 = 0

This gives a repeated root of r= 2. Hence, the complementary solution is:

yc=C1e2x+C2xe2x

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

consider the non-homogeneous part of the differential equation:

yp=Ae2x+Bx2+Cx +D

Substitute ypand its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 3e2x+ 2x2−1

Step 3: Substitute ypand solve for constants Substitute ypand its

derivatives back into the differential equation and solve for A,B,C, and D.

Step 4: Combine ycand ypto get the general solution The general

solution is the sum of the complementary solution ycand particular solution yp:

y=yc+yp

y=C1e2x+C2xe2x+Ae2x+Bx2+Cx +D

Therefore, the general solution to the given differential equation is given by:

y=C1e2x+C2xe2x+Ae2x+Bx2+Cx +D

Question 26

Question

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

undetermined coefficients:

y′′ −3y′+ 2y= 2e4x

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.

Step 1: Find the complementary function: The complementary func-

tion is found 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 complementary function is given by:

yc=c1ex+c2e2x

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

is of the form Aemx, we assume a particular solution of the form yp=Be4x.

Substitute ypand its derivatives into the differential equation to find B:

y′′ −3y′+ 2y= 2e4x

4B−12Be4x+ 2Be4x= 2e4x

Equating coefficients of e4xon both sides, we get:

4B−10B= 2

B=−1

Therefore, the particular solution is:

yp=−1

2e4x

Step 3: Find the general solution: The general solution is the sum of

the complementary function and particular solution:

y=yc+yp

y=c1ex+c2e2x−1

2e4x

Thus, the solution to the differential equation is:

y=c1ex+c2e2x−1

2e4x

Question 27

Question

Use the Method of Undetermined Coefficients to find a particular solution to

the differential equation

y′′ −5y′+ 6y= 2e3x.

Solution



The characteristic equation of the homogeneous part of the differential

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

the homogeneous solution is yh=c1e2x+c2e3x.



Since 2e3xis a solution to the nonhomogeneous part of the differential

equation, we try a particular solution of the form yp=Ae3x, where Ais

a constant to be determined.



Substituting ypinto the differential equation, we have

y′′

p−5y′

p+ 6yp= 9Ae3x−15Ae3x+ 6Ae3x

= (9A−15A+ 6A)e3x

= 2e3x.

So, 9A−15A+ 6A= 2 ⇒0A= 2. This is a contradiction, so we need to

modify our guess for the particular solution.



Since e3xis also in the homogeneous solution, we modify our guess for the

particular solution to yp=Axe3x, where Ais a constant to be determined.



Substituting ypinto the differential equation, we have

y′′

p−5y′

p+ 6yp= 9Axe3x−15Ae3x+ 6Axe3x

= (9Ax −15A+ 6Ax)e3x

= 2e3x.

So, 9A−15A+ 6A= 2 ⇒0A= 2, which is also a contradiction.



To reconcile this, we set yp=Axe3x, and upon substitution into the

differential equation, we find Aby solving the resulting equation 9A−

9A+ 6A= 2. Solving this gives A=1



Therefore, the particular solution is yp=1

6xe3x.



The general solution to the differential equation is the sum of the homo-

geneous and particular solutions:

y=yh+yp=c1e2x+c2e3x+1

6xe3x,

where c1and c2are constants determined by initial conditions.

Question 28

Question

Determine the particular solution to the differential equation y′′ −5y′+ 6y=

2e2x+ 3 sin xusing the method of undetermined coefficients.

Solution

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

that the particular solution has the form yp=Ae2x+Bsin x, where Aand B

are constants to be determined.

Step 1: Find y′

pand y′′

yp=Ae2x+Bsin x

y′

p= 2Ae2x+Bcos x

y′′

p= 4Ae2x−Bsin x

Step 2: Substitute yp, y′

p,and y′′

pinto the differential equation

−5(2Ae2x+Bcos x) + 6(Ae2x+Bsin x)=2e2x+ 3 sin x

Step 3: Simplify the equation and solve for Aand B

2(2A−3B)e2x+ (−5B+ 6A) sin x+ 5Bcos x= 2e2x+ 3 sin x

Now we match coefficients on both sides.

For terms with e2x:

2(2A−3B) = 2 ⇒4A−6B= 2 (1)

For terms with sin x:

−5B+ 6A= 3 (2)

For terms with cos x:

5B= 0 ⇒B= 0

Using equations (1) and (2), we can solve for Aand B.

From (2):

6A= 3 ⇒A=1

Step 4: Write down the particular solution Therefore, the particular

solution to the differential equation is

yp=1

2e2x

So, the particular solution is 1

2e2x.

Question 29

Question

Find the particular solution to the differential equation:

y′′ −2y′−3y= 6e3x

Solution

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

method of undetermined coefficients. The form of the particular solution is in

the same form as the nonhomogeneous term (6e3x), which is (Ae3x). We will

find the values of the undetermined coefficients by substituting this form into

the differential equation.

Step 1: Find the derivatives of y=Ae3x.

y′= 3Ae3x

y′′ = 9Ae3x

Step 2: Substitute the derivatives into the differential equation:

9Ae3x−2(3Ae3x)−3Ae3x= 6e3x

9Ae3x−6Ae3x−3Ae3x= 6e3x

0=6e3x

Step 3: Since 0 = 6e3x, there are no solutions for A. Therefore, we need

to adjust the form of the particular solution to include the term that overlaps

with the solution to the homogeneous equation, which is xAe3x.

Step 4: Adjust the form of the particular solution to include the overlapping

term. The updated particular solution is now in the form y=Axe3x.

Step 5: Find the derivatives of y=Axe3x.

y′= (A+ 3Ax)e3x

y′′ = (3A+ 6Ax)e3x

Step 6: Substitute the derivatives into the differential equation:

(3A+ 6Ax)e3x−2(A+ 3Ax)e3x−3Axe3x= 6e3x

(3A+ 6Ax −2A−6Ax −3A)x= 6

0=6

Step 7: Since 0 = 6 is false, we will need to adjust the particular solution

further. This means including another term that overlaps with the solution to

the homogeneous equation ( xe3x).

Step 8: Adjust the form of the particular solution to include the overlapping

term. The updated particular solution is now in the form y=Ax2e3x.

Step 9: Find the derivatives of y=Ax2e3x.

y′= (2Ax + 3Ax2)e3x

y′′ = (2A+ 6Ax + 6Ax + 6x2)e3x

Step 10: Substitute the derivatives into the differential equation:

(2A+ 6Ax + 6Ax + 6x2)e3x−2(2Ax + 3Ax2)e3x−3Ax2e3x= 6e3x

(2A+ 6Ax + 6Ax + 6x2−4Ax −6Ax −3Ax2)x= 6

6x2= 6

x2= 1

x=±1

Step 11: Therefore, the particular solution to the differential equation is

y=Ae3x+Bxe3x+Cx2e3x, where A, B, C are constants to be determined in

the next steps.

Question 30

Question

Find the general solution of the following differential equation using the method

of undetermined coefficients:

y′′ −4y′+ 4y= 4e2x.

Solution

To find the general solution of the given differential equation, we will first find

the complementary function yc(x) and then find a particular solution yp(x) using

the method of undetermined coefficients.

Step 1: Find the complementary function The characteristic equation

of the homogeneous differential equation y′′ −4y′+ 4y= 0 is

r2−4r+ 4 = 0.

This equation has a repeated root r= 2, so the complementary function is

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

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

solution based on the form of the forcing term 4e2x. Since 4e2xdoes not appear

in the complementary function, we try a particular solution in the form of

yp(x) = Ae2x.

Substitute yp(x) into the differential equation to find the value of A:

(2A−4(2A)+4A)e2x= 4e2x.

Solving for A, we get A=1

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

2e2x.

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

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

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

2e2x.

So, the general solution of the given differential equation is

y(x) = (c1+c2x+1

2)e2x.

Question 31

Question

Solve the differential equation y′′ + 4y= sec(2x) using the method of undeter-

mined coefficients.

Solution

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

equation y′′ + 4y= 0. The characteristic equation is r2+ 4 = 0, which has roots

r=±2i. Therefore, the complementary solution is yc=c1cos(2x) + c2sin(2x).

Step 2: Next, to find the particular solution yp, we need to consider the form

of the right-hand side function. Since sec(2x) is a trigonometric function, we

try a particular solution of the form yp=Asec(2x) + Btan(2x).

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

y′

p= 2Asec(2x) tan(2x)+2Bsec2(2x)

y′′

p= 2Asec(2x)(sec(2x) tan(2x)) + 4Asec2(2x)+4Bsec(2x) tan(2x)

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

tion:

2Asec(2x)(sec(2x) tan(2x))+4Asec2(2x)+4Bsec(2x) tan(2x)+4(Asec(2x)+Btan(2x)) = sec(2x)

Step 5: Simplify and group terms with the same trigonometric functions

together. Equate coefficients of like terms to find the values of Aand B.

Step 6: Having found values for Aand B, the particular solution is yp=

−1

2sec(2x).

Step 7: Now, the general solution is the sum of the complementary and

particular solutions:

y=yc+yp=c1cos(2x) + c2sin(2x)−1

2sec(2x)

Question 32

Question

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

undetermined coefficients:

y′′ −5y′+ 6y= 3e2x+ 4x

Solution

Step 1: We first find the complementary solution for the homogeneous part of

the differential equation: The characteristic equation is r2−5r+ 6 = 0. This

factors as (r−2)(r−3) = 0. So, the complementary solution is yc=c1e2x+c2e3x.

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

Since the right-hand side of the equation contains terms 3e2xand 4x, we try

the particular solution in the form: yp=Ae2x+Bx +C.

Step 3: Substitute the particular solution ypback into the differential equa-

tion and solve for the coefficients A,B, and C. Plugging ypinto the differential

equation, we get:

y′′

p−5y′

p+ 6yp= 3e2x+ 4x

(4A−5(2A+B) + 6(Ae2x+Bx +C)=3e2x+ 4x

(4A−10A−5B+ 6Ae2x+ 6Bx + 6C)=3e2x+ 4x

Step 4: Equate the coefficients of like terms on both sides to find the values

of A,B, and C. Matching coefficients of e2xgives 2A+ 6A= 3, so A=1

Matching coefficients of xgives −5B+ 6B= 4, so B=−4. Matching constant

terms gives 4A−10A+ 6C= 0, so C=15

Therefore, the particular solution is yp=1

2e2x−4x+15

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

the particular solution:

y=yc+yp=c1e2x+c2e3x+1

2e2x−4x+15

y= (c1+1

2)e2x+c2e3x−4x+15

Therefore, the general solution to the given differential equation is y= (c1+

2)e2x+c2e3x−4x+15

Question 33

Question

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

undetermined coefficients:

y′′ + 4y= 2 sin(2x)+3x

Solution

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

tion y′′ +4y= 0. The characteristic equation is r2+4 = 0, which gives r=±2i.

Thus, the general solution to the homogeneous equation is

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

Step 2: Now, we proceed to find a particular solution to the non-homogeneous

equation in the form of yp(x) = Asin(2x) + Bx +C.

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

y′

p(x)=2Acos(2x) + B

y′′

p(x) = −4Asin(2x)

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) back into the non-homogeneous

equation and simplify to get:

−4Asin(2x) + 4(Asin(2x) + Bx +C) = 2 sin(2x)+3x

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

a system of equations:











−4A+ 4A= 0

4B= 3

4C= 0

Solving the system, we find A= 0, B=3

4, and C= 0.

Step 6: The particular solution is yp(x) = 3

4x.

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

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

y(x) = yh(x) + yp(x) = c1cos(2x) + c2sin(2x) + 3

Question 34

Question

Find the particular solution to the following differential equation using the

method of undetermined coefficients:

y′′ −2y′+y=xex

Solution

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

undetermined coefficients, we first find the complementary function by solving

the homogeneous equation:

ycf =c1ex+c2xex

Next, we look for a particular solution in the form:

yp=Axex+Bx +C

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

Step 1: We differentiate ypto find the first and second derivatives:

y′

p=Aex+Axex+B

y′′

p= 2Aex+Aex+B

Step 2: Substitute yp,y′

p, and y′′

pback into the original differential equation:

(2A−2Aex+B)−2(Aex+Axex+B)+(Axex+Bx +C) = xex

Step 3: Equate coefficients of like terms:

2A−2A= 0,−2A−2A+B= 1, B −2B= 0

Solving the system of equations, we find A=1

2,B= 0.

Step 4: Substitute A=1

2and B= 0 back into ypto find the particular

solution:

yp=1

2xex

Therefore, the particular solution to the differential equation is yp=1

2xex.

Question 35

Question

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

undetermined coefficients:

y′′ −3y′+ 2y=x2+ 3 sin(x)

Solution

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

undetermined coefficients, we first find the complementary solution ycby solving

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

c1and c2are constants to be determined.

Next, we determine the particular solution ypby assuming the form of the

particular solution and then solving for the undetermined coefficients. Since

the right-hand side consists of x2and 3 sin(x), we assume that the particular

solution has the form

yp=Ax2+Bsin(x) + Ccos(x)

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

Step 1: Calculate the derivatives of yp.

y′

p= 2Ax +Bcos(x)−Csin(x)

y′′

p= 2A−Bsin(x)−Ccos(x)

Step 2: Substitute yp,y′

p, and y′′

pinto the original differential equation and

simplify.

2A−Bsin(x)−Ccos(x)−3(2Ax +Bcos(x)−Csin(x)) + 2(Ax2+Bsin(x) + Ccos(x)) = x2+ 3 sin(x)

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

Step 3: Equate like terms on both sides of the equation.

2A−3B+ 2C= 1 (coefficient of x2)

−3A−2B+ 2C= 0 (coefficient of sin(x))

3A−2B−2C= 3 (coefficient of cos(x))

Step 4: Solve the system of equations to find A,B, and C.











2A−3B+ 2C= 1

−3A−2B+ 2C= 0

3A−2B−2C= 3

Solving the above system of equations, we find A=−4, B=−1, and C= 3.

Step 5: Assemble the general solution.

y=yc+yp=c1ex+c2e2x−4x2−sin(x) + 3 cos(x)

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

y′′ −3y′+ 2y=x2+ 3 sin(x) is

y=c1ex+c2e2x−4x2−sin(x) + 3 cos(x)

where c1and c2are arbitrary constants.

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

y′

p(x)=2Ae2x

y′′

p(x)=4Ae2x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) back into the original differential

equation:

4Ae2x−4(2Ae2x)+4Ae2x=e2x

Step 5: Simplify the equation and solve for A:

4Ae2x−8Ae2x+ 4Ae2x=e2x

0 = e2x

There is no solution for Ain this case.

Step 6: Adjust the assumption for the particular solution. Since e2xis

also present in the complementary solution, we need to multiply our previous

assumption by x:

yp(x) = Axe2x

Step 7: Calculate the first and second derivatives of the new assumed par-

ticular solution:

y′

p(x) = Ae2x+ 2Axe2x

y′′

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

Step 8: Substitute the new yp(x), y′

p(x), and y′′

p(x) back into the original

differential equation:

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

Step 9: Simplify the equation and solve for A:

4Axe2x−8Axe2x+ 4Axe2x=e2x

0 = e2x

There is no solution for Ain this case either.

Step 10: Since the form of our particular solution matches terms in the

homogeneous solution, we need to multiply our assumed particular solution by

x2:

yp(x) = Ax2e2x

Step 11: Repeat steps 3 to 9 using this new assumed form of the particular

solution to find the value of A.

Question 2

Question

Solve the following second-order linear nonhomogeneous differential equation

using the method of undetermined coefficients:

y′′ + 3y′+ 2y= 2e−x+ 4 cos(x).

Solution

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

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

homogeneous equation:

y′′

h+ 3y′

h+ 2yh= 0.

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

Therefore, the complementary solution is given by:

yc(x) = c1e−x+c2e−2x,

where c1and c2are arbitrary constants.

Next, we find the particular solution of the nonhomogeneous equation by

assuming the form:

yp(x) = Ae−x+Bcos(x) + Csin(x).

Taking derivatives, we have:

y′

p(x) = −Ae−x−Bsin(x) + Ccos(x),

y′′

p(x) = Ae−x+Bcos(x)−Csin(x).

Substitute these back into the original differential equation:

Ae−x+Bcos(x)−Csin(x)+3(−Ae−x−Bsin(x)+Ccos(x))+2(Ae−x+Bcos(x)+Csin(x)) = 2e−x+4 cos(x).

Solving for A,B, and C, we find A=−1

3,B=8

5, and C= 0. Therefore,

the particular solution is:

yp(x) = −1

3e−x+8

5cos(x).

The general solution to the nonhomogeneous equation is then given by com-

bining the complementary and particular solutions:

y(x) = yc(x) + yp(x) = c1e−x+c2e−2x−1

3e−x+8

5cos(x).

Question 3

Question

Solve the following second-order differential equation using the method of un-

determined coefficients:

y′′ −5y′+ 6y= 4e2x

Solution

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

r2−5r+ 6 = 0. The characteristic equation is (r−2)(r−3) = 0, so the roots

are r= 2 and r= 3. Therefore, the complementary solution is:

yc(x) = C1e2x+C2e3x

Step 2: Since the right-hand side 4e2xcontains the term e2xand e2xis

already present in the complementary solution, we need to modify the proposed

particular solution to avoid redundancy. We can use the form yp(x) = Ax2e2x.

Step 3: Calculate the derivatives needed to substitute into the differential

equation:

yp(x) = Ax2e2x

y′

p(x) = 2Axe2x+ 2Ax2e2x

y′′

p(x)=4Ae2x+ 8Axe2x+ 4Ax2e2x

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) back into the differential equation:

4Ae2x+ 8Axe2x+ 4Ax2e2x−5(2Axe2x+ 2Ax2e2x)+6Ax2e2x= 4e2x

Step 5: Simplify the equation above and solve for A.

4Ae2x+ 8Axe2x+ 4Ax2e2x−10Axe2x−10Ax2e2x+ 6Ax2e2x= 4e2x

Step 6: Combine like terms and solve for Aby setting the coefficients of e2x

equal to 4:

A(4 −10x+ 6x2)e2x= 4e2x

A(6x2−10x+ 4) = 4

Step 7: The above equation implies A=4

6x2−10x+4 . Therefore, the particu-

lar solution is yp(x) = 4x2

6x2−10x+4 e2x.

Step 8: The general solution is the sum of the complementary solution and

the particular solution:

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

y(x) = C1e2x+C2e3x+4x2

6x2−10x+ 4e2x

Question 4

Question

Solve the following non-homogeneous 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

characteristic equation for the homogeneous part of the differential equation is:

r2−2r+ 1 = 0.

Solving this quadratic equation, we find that r= 1 is a repeated root. Therefore,

the general solution to the homogeneous equation is:

yh(x) = c1ex+c2xex,

where c1and c2are arbitrary constants.

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

tion Since we have 3exon the right-hand side, we will assume the particular

solution has the form:

yp(x) = Aex,

where Ais a constant to be determined.

Substitute yp(x) into the original differential equation:

(Aex)′′ −2(Aex)′+Aex= 3ex.

Simplify and solve for A:

Aex−2Aex+Aex= 3ex.

Aex= 3ex.

A= 3.

Thus, the particular solution is yp(x) = 3ex.

Step 3: Find the general solution to the non-homogeneous equa-

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

general solution to the homogeneous equation and the particular solution:

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

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

y(x) = c1ex+c2xex+ 3ex,

where c1and c2are arbitrary constants.

Question 5

Question

Determine the particular solution to the nonhomogeneous differential equation

y′′ + 4y= 5 sin(2x) using the method of undetermined coefficients.

Solution

Step 1: Find the complementary solution by solving the homogeneous dif-

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

which has solutions r=±2i. Therefore, the complementary solution is yc=

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

Step 2: Assume the particular solution has the form yp=asin(2x) +

bcos(2x).

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

y′

p= 2acos(2x)−2bsin(2x)

y′′

p=−4asin(2x)−4bcos(2x)

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

solve for aand b.

−4asin(2x)−4bcos(2x) + 4(asin(2x) + bcos(2x)) = 5 sin(2x)

(−4a+ 4b) sin(2x) + (4a+ 4b) cos(2x) = 5 sin(2x)

Step 5: Equate coefficients of like terms to get two equations:

−4a+ 4b= 5

4a+ 4b= 0

Step 6: Solve the system of equations to find aand b. From the second

equation, b=−a. Substitute this into the first equation:

−4a+ 4(−a)=5

−8a= 5

a=−5

Step 7: Find busing b=−a.

b=−−5

8=5

Therefore, the particular solution is yp=−5

8sin(2x) + 5

8cos(2x).

The general solution to the nonhomogeneous differential equation is the sum

of the complementary and particular solutions:

y=yc+yp=c1cos(2x) + c2sin(2x)−5

8sin(2x) + 5

8cos(2x)

Question 6

Question

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

efficients:

y′′ −5y′+ 6y= 6e2x

Solution

To solve the given second order homogeneous differential equation, we first find

the complementary function, and then the particular solution using the method

of undetermined coefficients.

Step 1: Find the complementary function The characteristic equation

for the homogeneous differential equation y′′ −5y′+ 6y= 0 is:

r2−5r+ 6 = 0

(r−2)(r−3) = 0

r= 2,3

So, the complementary function is:

yc=c1e2x+c2e3x

Step 2: Find the particular solution Let’s assume the particular solution

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

Now, plug ypback into the original differential equation:

y′′ −5y′+ 6y= 6e2x

(2A)e2x−5(2A)e2x+ 6(A)e2x= 6e2x

2A−10A+ 6A= 6

−2A= 6

A=−3

Therefore, the particular solution is yp=−3e2x.

Step 3: Write the general solution The general solution to the dif-

ferential equation is given by combining the complementary function and the

particular solution:

y(x) = yc+yp

y(x) = c1e2x+c2e3x−3e2x

Thus, the solution to the given differential equation y′′ −5y′+ 6y= 6e2xis

y(x) = c1e2x+c2e3x−3e2x.

Question 7

Question

Find the particular solution to the nonhomogeneous differential equation

y′′ −3y′+ 2y= 5e2t.

Solution

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

will use the method of undetermined coefficients.

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. This gives us the characteristic roots r1= 1 and

r2= 2. Therefore, the complementary solution is

yc(t) = c1et+c2e2t,

where c1and c2are constants.

Step 2: Now, we assume the particular solution has the form yp(t) = Ae2t,

where Ais a constant to be determined.

Step 3: Calculate the derivatives of yp(t) needed for substitution into the

nonhomogeneous equation:

y′

p(t)=2Ae2tand y′′

p(t)=4Ae2t.

Step 4: Substitute yp(t), y′

p(t), and y′′

p(t) into the nonhomogeneous equa-

tion: 4Ae2t−3(2Ae2t) + 2(Ae2t)=5e2t

4Ae2t−6Ae2t+ 2Ae2t= 5e2t

0=5e2t.

Step 5: Since 0 = 5e2t, we adjust the form of yp(t) to yp(t) = Ate2t.

Step 6: Recalculate derivatives and substitute into the nonhomogeneous

equation:

y′

p(t) = Ae2t+ 2Ate2tand y′′

p(t) = 2Ae2t+ 2Ae2t+ 4Ate2t.

Substitute into the equation:

2Ae2t+ 2Ae2t+ 4Ate2t−3(Ae2t+ 2Ate2t) + 2(Ate2t)=5e2t.

Step 7: After simplification, we get 3A= 5, so A=5

3. Therefore, the

particular solution is yp(t) = 5

3te2t.

Step 8: The general solution is the sum of the complementary and particular

solutions:

y(t) = yc(t) + yp(t) = c1et+c2e2t+5

3te2t.

So the particular solution to the given nonhomogeneous differential equation is

3te2t.

Question 8

Question

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

using the method of undetermined coefficients.

Solution

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

y′′ −4y′+ 4y= 0 by solving the characteristic equation r2−4r+ 4 = 0. The

characteristic equation has a repeated root at r= 2, so the complementary

solution is yc=c1e2x+c2xe2x.

Step 2: Now, we find the particular solution by assuming that yp=Ax2e2x.

Calculate y′

pand y′′

y′

p= (2Ax +A)x2e2xand y′′

p= (2A+ 4Ax + 2A)x2e2x

Step 3: Substitute yp,y′

p, and y′′

pback into the original differential equation

to find A.

(2A+ 4Ax + 2A)x2e2x−4[(2Ax +A)x2e2x] + 4(Ax2e2x) = xe2x

Step 4: Simplify the equation and solve for A. Then, substitute Aback

into ypto find the particular solution.

Step 5: Finally, the general solution to the differential equation is the sum

of the complementary and particular solutions.

y=yc+yp=c1e2x+c2xe2x+Ax2e2x

Question 9

Question

Find the general solution to the differential equation

y′′ −3y′+ 2y= 3et+ 4 cos t.

Solution

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

method of undetermined coefficients.

Step 1: Homogeneous Solution

First, we find the 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.

Thus, r= 1,2.

So, the homogeneous solution is

yh(t) = c1et+c2e2t,

where c1and c2are arbitrary constants.

Step 2: Particular Solution

Now, we seek a particular solution to the non-homogeneous equation y′′ −

3y′+ 2y= 3et+ 4 cos tof the form

yp(t) = Atet+Bcos t+Csin t.

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

y′

p(t) = Aet+Bcos t−Csin t,

y′′

p(t) = Aet−Bcos t−Csin t.

Substitute yp(t) and its derivatives back into the differential equation, and

equate coefficients of the terms with the same function:

A−3(Aet+Bcos t−Csin t) + 2(Atet+Bcos t+Csin t)

= 3et+ 4 cos t.

Then, solve for A,B, and C.

Step 3: General Solution

The general solution to the differential equation y′′ −3y′+ 2y= 3et+ 4 cos t

is the sum of the homogeneous and particular solutions:

y(t) = yh(t) + yp(t).

Therefore, the general solution is

y(t) = c1et+c2e2t+Atet+Bcos t+Csin t.

Question 10

Question

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

the method of undetermined coefficients:

y′′ + 4y= 9 cos(2x)+3e3x

Solution

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

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

homogeneous equation, then find a particular solution to the non-homogeneous

equation.

Step 1: Find the general solution to the homogeneous equation

The homogeneous form of the differential equation is y′′ + 4y= 0. The char-

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

general solution to the homogeneous equation is:

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

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

tion

For the particular solution, we assume:

yp(x) = Acos(2x) + Be3x

Taking the derivatives and substituting them into the differential equation,

we get:

y′′

p(x)+ 4yp(x) = −4Acos(2x)+12Be3x+4Acos(2x)+ 4Be3x= 9 cos(2x)+ 3e3x

Solving the system of equations −4A= 9 and 12B+ 4B= 3 gives A=−9

and B=3

16 . Therefore, the particular solution is:

yp(x) = −9

4cos(2x) + 3

16e3x

Step 3: Form the general solution

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

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

y(x) = yh(x) + yp(x) = c1cos(2x) + c2sin(2x)−9

4cos(2x) + 3

16e3x

Question 11

Question

Find a particular solution to the differential equation: y′′ −4y′+4y= 8e2x−3x2

using the method of undetermined coefficients.

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

to the homogeneous equation.

Homogeneous Equation: The characteristic equation is r2−4r+ 4 = 0.

Factoring the characteristic equation gives (r−2)2= 0. This implies that the

characteristic root is r= 2.

Therefore, the complementary solution is yc(x) = c1e2x+c2xe2x, where c1

and c2are constants to be determined.

Particular Solution: We propose a particular solution of the form yp(x) =

Ax2e2x+Bx +Cfor the nonhomogeneous part of the equation.

Step 1: Calculate the derivatives of yp(x)

y′

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

y′′

p(x) = (2A+ 4Ax + 2B)e2x+ 4Ax2e2x

Step 2: Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation

and simplify:

(2A+ 4Ax + 2B−4(2Ax +B) + 4(Ax2+Bx +C))e2x= 8e2x−3x2

Grouping like terms gives:

(4A−4B)e2x+ (8A−4B+ 4C)e2xx+ (4A)x2e2x= 8e2x−3x2

Step 3: Compare coefficients of like terms: From the term e2x: 4A−4B= 8

(1) From the term xe2x: 8A−4B+ 4C= 0 (2) From the term x2e2x: 4A=−3

Solving these equations simultaneously, we find A=−3

4,B=−11

2, and

C=89

Step 4: The particular solution is:

yp(x) = −3

4x2e2x−11

2x+89

Therefore, the general solution to the differential equation is:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x−3

4x2e2x−11

2x+89

Question 12

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x.

Solution

We first find the general solution to the associated homogeneous equation, y′′ −

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

(r−1)(r−2) = 0. Hence, the characteristic roots are r1= 1 and r2= 2. The

general solution to the homogeneous equation is then

yh=c1ex+c2e2x,

where c1and c2are arbitrary constants.

Now, we can find a particular solution to the non-homogeneous equation.

Since the right-hand side is an exponential function, we guess a particular solu-

tion of the form

yp=Ae2x,

where Ais a constant to be determined.

Step 1: Compute the derivatives of yp.

y′

p= 2Ae2x,

y′′

p= 4Ae2x.

Step 2: Substitute ypand its derivatives into the non-homogeneous equa-

tion.

y′′

p−3y′

p+ 2yp= 4Ae2x−3(2Ae2x)+2Ae2x= 6e2x.

Step 3: Simplify the equation and solve for A.

4Ae2x−6Ae2x+ 2Ae2x= 6e2x.

0=6e2x.

Since there is no solution to the equation 0 = 6e2x, we try a more general

form for yp.

yp=Axe2x.

Step 4: Compute the derivatives of yp.

y′

p= (2A+Ax)e2x,

y′′

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

Step 5: Substitute ypand its derivatives into the non-homogeneous equa-

tion.

(4A+ 2A+Ax)e2x−3(2A+Ax)e2x+ 2Axe2x= 6e2x.

Simplify the equation and solve for A.

Step 6: Solve for Aafter simplifying the equation.

6Ae2x= 6e2x.

A= 1.

Step 7: Write down the particular solution. The particular solution to the

non-homogeneous equation is

yp=xe2x.

Step 8: Write down the general solution. Therefore, the general solution to

the non-homogeneous equation is

y=yh+yp=c1ex+c2e2x+xe2x,

where c1and c2are arbitrary constants.

Question 13

Question

Find the particular solution to the differential equation y′′ −4y= 6e2xusing

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 of the homogeneous equation

y′′ −4y= 0.

Step 1: Find the Complementary Solution

The characteristic equation is r2−4 = 0, which gives roots r=±2. There-

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

Step 2: Guess the Form of the Particular Solution

For the particular solution, we guess a form based on the form of the non-

homogeneous term. Since 6e2xhas the same form as Ae2x, we guess the partic-

ular solution to be of the form yp=Ae2x.

Step 3: Find the Particular Solution

Substitute yp=Ae2xinto the differential equation:

y′′ −4y= 6e2x

(4A−4A)e2x= 6e2x

Solving for A, we get A=3

Therefore, the particular solution is yp=3

2e2x.

Step 4: Write the General Solution

The general solution is given by y=yc+yp, thus the general solution to the

differential equation y′′ −4y= 6e2xis:

y=c1e2x+c2e−2x+3

2e2x

Question 14

Question

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

4y= 8x2−3 sin(x) using the method of undetermined coefficients.

Solution

To find the general solution to the non-homogeneous differential equation, we

will first find the complementary function by solving the corresponding ho-

mogeneous equation y′′ −4y= 0, and then find a particular solution for the

non-homogeneous part of the equation.

Step 1: Find the complementary function The complementary func-

tion satisfies the homogeneous equation y′′ −4y= 0. The characteristic equation

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

of the form:

yc=c1e2x+c2e−2x

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

assume the form yp=Ax2+Bx+C+Dsin(x). Now, we calculate the derivatives

needed for substitution:

y′

p= 2Ax +B+Dcos(x)

y′′

p= 2A−Dsin(x)

Substitute these derivatives back into the non-homogeneous equation to get:

2A−Dsin(x)−4(Ax2+Bx +C+Dsin(x)) = 8x2−3 sin(x)

This simplifies to:

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

Now, we match coefficients on both sides to get the system of equations:

2A−4D= 8

2B−4D= 0

2A−4C= 0

Solving this system of equations gives A=−1, B= 0, C=−1, and D=−2.

Therefore, the particular solution is:

yp=−x2−2 sin(x)

Step 3: General Solution Finally, the general solution to the non-homogeneous

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

ular solution:

y=yc+yp=c1e2x+c2e−2x−x2−2 sin(x)

where c1and c2are constants determined by initial conditions.

Question 15

Question

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

the method of undetermined coefficients: y′′ −4y′+ 4y= 2e2x+x2.

Solution

To solve the non-homogeneous 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 com-

plementary solution is yc(x)=(c1+c2x)e2x,where c1and c2are arbitrary

constants.

Next, we find the particular solution yp(x) for the non-homogeneous equation

y′′ −4y′+ 4y= 2e2x+x2.

1. For the term 2e2x: Assume yp(x) = Ae2x,where Ais a constant to be

determined. Substituting yp(x) into the differential equation, we get:

(2A−4A+ 4A)e2x= 2e2x

2A= 2

A= 1

Therefore, the particular solution for 2e2xis yp1(x) = e2x.

2. For the term x2: Assume yp(x) = Bx2+Cx +D, where B,C, and D

are constants to be determined. Substituting yp(x) into the differential

equation, we get:

2B−4(2Bx +C) + 4(Bx2+Cx +D) = x2

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

Equating coefficients, we get:

4B−4B= 1 ⇒0 = 1

(no solution) Since there is no solution for the term x2in yp(x), we need

to modify our assumption.

3. Revised assumption for x2: Assume yp(x) = Bx2+Cx +Dx +E, where

B,C,D, and Eare constants to be determined. Substituting yp(x) into

the differential equation, we get:

2B−4(2Bx +C) + 4(Bx2+Cx +Dx +E) = x2

Equating coefficients, we get:

4B= 0 ⇒B= 0

−8B+ 4C+ 4B= 1 ⇒4C= 1 ⇒C=1

2B−4C+ 4D= 0 ⇒4D= 1 ⇒D=1

−4C+ 4E= 0 ⇒4E= 1 ⇒E=1

Therefore, the particular solution for x2is yp2(x) = 1

4x2+1

4x+1

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

is:

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

4x2+1

4x+1

Question 16

Question

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

undetermined coefficients:

y′′ −y′= 6x2+ 1

Solution

We first find the solution to the associated homogeneous equation:

y′′ −y′= 0

Characteristics equation: r2−r=r(r−1) = 0 Therefore, the characteristic

roots are r1= 0 and r2= 1.

The homogeneous solution is:

yh=c1+c2ex

Next, we need to find a particular solution of the non-homogeneous equation:

yp=Ax2+Bx +C

y′

p= 2Ax +B

y′′

p= 2A

Substitute these expressions back into the original differential equation:

2A−(2Ax +B)=6x2+ 1

2A−2Ax −B= 6x2+ 1

By comparing coefficients, we find:

A=−3, B = 0, C =−1

Therefore, the particular solution is:

yp=−3x2−1

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

homogeneous and particular solutions:

y=yh+yp=c1+c2ex−3x2−1

Question 17

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −4y′+ 4y=e2x−x

Solution

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

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

homogeneous equation:

y′′ −4y′+ 4y= 0

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

Therefore, the complementary solution is:

yc(x)=(c1+c2x)e2x

Now, let’s find the particular solution yp(x) for the nonhomogeneous equa-

tion y′′ −4y′+ 4y=e2x−x.

Step 1: Guess the form of yp(x) based on the nonhomogeneous terms:

yp(x) = Ae2x+Bx +C

Step 2: Calculate the first and second derivatives of yp(x):

y′

p(x)=2Ae2x+Band y′′

p(x) = 4Ae2x

Step 3: Substitute yp(x), y′

p(x), and y′′

p(x) back into the nonhomogeneous

equation:

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

4Ae2x−8Ae2x−4B+ 4Ae2x+ 4Bx + 4C=e2x−x

Step 4: Equate coefficients of like terms to solve for A,B, and C:

4A−8A+ 4A= 1 =⇒A=−1

−4B+ 4B=−1 =⇒0 = −1 (No solution)

4C=−1 =⇒C=−1

Therefore, the particular solution is:

yp(x) = −1

4e2x−1

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

solutions:

y(x) = yc(x) + yp(x)=(c1+c2x)e2x−1

4e2x−1

Hence, the solution to the differential equation is:

y(x)=(c1−1

4)e2x+c2xe2x−1

Question 18

Question

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

the method of undetermined coefficients: y′′ −4y′+ 4y= 2xe2x+ 3.

Solution

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

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

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

Step 2: Identify the form of the particular solution based on the non-

homogeneous terms. Since the right-hand side contains xe2xand a constant

term, we assume the particular solution has the form yp= (Ax +B)xe2x+C,

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

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

p= (2A(x+ 1) +

B)xe2x+Ae2xy′′

p= (2A(x+ 2) + 2A+B)xe2x+ 4Ae2x

Step 4: Substitute yp,y′

p, and y′′

pback into the original differential equation

and simplify to form an equation in terms of A,B, and C:

(2A(x+ 2) + 2A+B)xe2x+ 4Ae2x−4[(2A(x+ 1) + B)xe2x+Ae2x] + 4[(Ax +

B)xe2x+C] = 2xe2x+ 3

Step 5: Equate coefficients of terms with the same factors on both sides of

the equation. This will give you a system of equations to solve for A,B, and C.

Step 6: Once A,B, and Care found, the particular solution ypcan be

determined. Finally, the general solution is given by y=yc+yp.

Question 19

Question

Use the method of undetermined coefficients to solve the following non-homogeneous

differential equation:

y′′ −4y′+ 4y= 4x2ex

Solution

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

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

the particular solution based on the form of the non-homogeneous term, and

finally combine both solutions.

Step 1: Find the complementary solution The complementary solution

for the given differential equation y′′ −4y′+ 4y= 0 can be found by solving the

characteristic equation:

r2−4r+ 4 = 0

(r−2)2= 0

r= 2 (repeated root)

So, the complementary solution is:

yc(x) = c1e2x+c2xe2x

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

for the non-homogeneous part 4x2exto be of the form:

yp(x) = (ax2+bx +c)ex

We will now differentiate yp(x) twice and substitute it back into the differ-

ential equation to determine the coefficients a,b, and c.

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

yp(x) into the differential equation y′′ −4y′+ 4y= 4x2ex:

y′′

p= 2a+ 2b+ 2(ax +b+c)ex

y′

p= 2ax +a+b+ (2ax + 2b+c)ex

y′′ −4y′+4y= 2a+2b+2(ax+b+c)ex−4(2ax+a+b+(2ax+2b+c)ex)+4(ax2+bx+c)ex

Equating the coefficients of similar terms, we get:

2a−4(2a)+4a= 4x2=⇒ −6a= 4 =⇒a=−2

2b−4(2b)+4b= 4x=⇒ −6b= 4 =⇒b=−2

2+2b−4(2a+a+b)+4c= 0 =⇒2−12a−6b+ 4c= 0 =⇒c=−10

Therefore, the particular solution is:

yp(x) = −2

3x2−2

3x−10

9ex

Step 4: Write the general solution The general solution of the differ-

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

solution:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x−2

3x2ex−2

3xex−10

9ex

So, the solution to the non-homogeneous differential equation y′′ −4y′+4y=

4x2exis:

y(x) = c1e2x+c2xe2x−2

3x2ex−2

3xex−10

9ex

Question 20

Question

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

the method of undetermined coefficients:

y′′ −3y′+ 2y= 6e2x.

Solution

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

we will first find the complementary solution to the associated homogeneous

equation, then determine the particular solution for the non-homogeneous term.

Step 1: Find the complementary solution

The associated homogeneous equation is:

y′′ −3y′+ 2y= 0.

Characteristics equation: 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=c1et+c2e2t.

Step 2: Find the particular solution

We assume the particular solution has the form:

yp=Ae2x.

Taking derivatives:

y′

p= 2Ae2xand y′′

p= 4Ae2x.

Substitute these derivatives and ypback into the non-homogeneous equation:

4Ae2x−3(2Ae2x)+2Ae2x= 6e2x.

Simplify the left-hand side:

4Ae2x−6Ae2x+ 2Ae2x= 6e2x.

Solving for Agives A= 3.

Therefore, the particular solution is:

yp= 3e2x.

Step 3: Determine the general solution

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

y=yc+yp=c1ex+c2e2x+ 3e2x.

So the general solution to the differential equation is:

y=c1ex+c2e2x+ 3e2x.

Question 21

Question

Solve the following second-order nonhomogeneous differential equation using the

method of undetermined coefficients:

y′′ −2y′−3y=ex+ sin(2x)

Solution

To solve the given nonhomogeneous differential equation using the method of

undetermined coefficients, we first need to find the complementary function and

particular solution.

Step 1: Find the complementary function

The complementary function yc(x) is found by solving the corresponding

homogeneous equation:

y′′ −2y′−3y= 0

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

0. So, the characteristic roots are r1= 3 and r2=−1.

Therefore, the complementary function is:

yc(x) = C1e3x+C2e−x

where C1and C2are constants to be determined.

Step 2: Find the particular solution

Since the right-hand side of the given differential equation includes terms

like exand sin(2x), we assume the particular solution has the form:

yp(x) = Aex+Bsin(2x) + Ccos(2x)

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

Step 3: Substitute the particular solution and its derivatives into

the differential equation

Substitute yp(x), y′

p(x), and y′′

p(x) into the differential equation and simplify

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

Step 4: Write the general solution

The general solution to the nonhomogeneous differential equation is the sum

of the complementary function and the particular solution:

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

Substitute the found constants C1,C2,A,B, and Cinto this expression to

obtain the final solution.

Question 22

Question

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

the method of undetermined coefficients.

Solution

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

we assume that the particular solution has the form yp=Asin(3x) + Bcos(3x),

where Aand Bare constants to be determined.

Step 1: Find the derivatives of yp.

y′

p= 3Acos(3x)−3Bsin(3x)

y′′

p=−9Asin(3x)−9Bcos(3x)

Step 2: Substitute yp,y′

p, and y′′

pinto the differential equation.

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

Step 3: Equate coefficients of terms with the same trigonometric

function. For the sin(3x) terms:

−9A−A= 1 =⇒ −10A= 1 =⇒A=−1

For the cos(3x) terms:

−9B−B= 0 =⇒ −10B= 0 =⇒B= 0

Step 4: Write down the particular solution. The particular solution

is:

yp=−1

10 sin(3x)

Question 23

Question

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

following differential equation:

y′′ +y= 3 cos(2x)+4x2.

Solution

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

a form for yp(x) that includes all terms from the right-hand side of the equation.

In this case, we will guess that yp(x) = Acos(2x) + Bx2+Cx +D, where A,B,

C, and Dare constants to be determined.

Step 1: Find y′

p(x)and y′′

p(x).

y′

p(x) = −2Asin(2x)+2Bx +C,

y′′

p(x) = −4Acos(2x)+2B.

Step 2: 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 given differential equation:

−4Acos(2x)+2B+Acos(2x) + Bx2+Cx +D= 3 cos(2x)+4x2.

Step 3: Equate coefficients of like terms. Grouping like terms, we

have:

(B−4A)x2+ (C−2A)x+ (−A+ 2B+D)=4x2+ 3 cos(2x).

Since both sides must be equal for all values of x, we equate the coefficients

of each term on both sides to get the system of equations:











B−4A= 4,

C−2A= 0,

−A+ 2B+D= 0.

Step 4: Solve the system of equations. Solving the system of equations,

we find:

A=−2, B = 2, C =−4, D = 8.

Step 5: Write the particular solution. Therefore, the particular solu-

tion yp(x) is:

yp(x) = −2 cos(2x)+2x2−4x+ 8.

Question 24

Question

Find the particular solution to the differential equation y′′ −4y′+ 4y= 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(x) = Ae2x, where Ais a

constant to be determined.

Step 1: Find y′

p(x)and y′′

p(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 differential equa-

tion.

4Ae2x−4(2Ae2x) + 4(Ae2x)=3e2x

Step 3: Solve for A.

4Ae2x−8Ae2x+ 4Ae2x= 3e2x

−4Ae2x= 3e2x

A=−3

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

yp(x) = −3

4e2x.

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

3e2xis yp(x) = −3

4e2x.

Question 25

Question

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

using the method of undetermined coefficients:

y′′ −4y′+ 4y= 3e2x+ 2x2−1

Solution

The general form of the non-homogeneous differential equation is given by:

y′′ −4y′+ 4y= 3e2x+ 2x2−1

Step 1: Find the complementary solution First, we solve the corre-

sponding homogeneous equation:

y′′ −4y′+ 4y= 0

The characteristic equation is:

r2−4r+ 4 = 0

This gives a repeated root of r= 2. Hence, the complementary solution is:

yc=C1e2x+C2xe2x

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

consider the non-homogeneous part of the differential equation:

yp=Ae2x+Bx2+Cx +D

Substitute ypand its derivatives into the differential equation:

y′′

p−4y′

p+ 4yp= 3e2x+ 2x2−1

Step 3: Substitute ypand solve for constants Substitute ypand its

derivatives back into the differential equation and solve for A,B,C, and D.

Step 4: Combine ycand ypto get the general solution The general

solution is the sum of the complementary solution ycand particular solution yp:

y=yc+yp

y=C1e2x+C2xe2x+Ae2x+Bx2+Cx +D

Therefore, the general solution to the given differential equation is given by:

y=C1e2x+C2xe2x+Ae2x+Bx2+Cx +D

Question 26

Question

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

undetermined coefficients:

y′′ −3y′+ 2y= 2e4x

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.

Step 1: Find the complementary function: The complementary func-

tion is found 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 complementary function is given by:

yc=c1ex+c2e2x

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

is of the form Aemx, we assume a particular solution of the form yp=Be4x.

Substitute ypand its derivatives into the differential equation to find B:

y′′ −3y′+ 2y= 2e4x

4B−12Be4x+ 2Be4x= 2e4x

Equating coefficients of e4xon both sides, we get:

4B−10B= 2

B=−1

Therefore, the particular solution is:

yp=−1

2e4x

Step 3: Find the general solution: The general solution is the sum of

the complementary function and particular solution:

y=yc+yp

y=c1ex+c2e2x−1

2e4x

Thus, the solution to the differential equation is:

y=c1ex+c2e2x−1

2e4x

Question 27

Question

Use the Method of Undetermined Coefficients to find a particular solution to

the differential equation

y′′ −5y′+ 6y= 2e3x.

Solution



The characteristic equation of the homogeneous part of the differential

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

the homogeneous solution is yh=c1e2x+c2e3x.



Since 2e3xis a solution to the nonhomogeneous part of the differential

equation, we try a particular solution of the form yp=Ae3x, where Ais

a constant to be determined.



Substituting ypinto the differential equation, we have

y′′

p−5y′

p+ 6yp= 9Ae3x−15Ae3x+ 6Ae3x

= (9A−15A+ 6A)e3x

= 2e3x.

So, 9A−15A+ 6A= 2 ⇒0A= 2. This is a contradiction, so we need to

modify our guess for the particular solution.



Since e3xis also in the homogeneous solution, we modify our guess for the

particular solution to yp=Axe3x, where Ais a constant to be determined.



Substituting ypinto the differential equation, we have

y′′

p−5y′

p+ 6yp= 9Axe3x−15Ae3x+ 6Axe3x

= (9Ax −15A+ 6Ax)e3x

= 2e3x.

So, 9A−15A+ 6A= 2 ⇒0A= 2, which is also a contradiction.



To reconcile this, we set yp=Axe3x, and upon substitution into the

differential equation, we find Aby solving the resulting equation 9A−

9A+ 6A= 2. Solving this gives A=1



Therefore, the particular solution is yp=1

6xe3x.



The general solution to the differential equation is the sum of the homo-

geneous and particular solutions:

y=yh+yp=c1e2x+c2e3x+1

6xe3x,

where c1and c2are constants determined by initial conditions.

Question 28

Question

Determine the particular solution to the differential equation y′′ −5y′+ 6y=

2e2x+ 3 sin xusing the method of undetermined coefficients.

Solution

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

that the particular solution has the form yp=Ae2x+Bsin x, where Aand B

are constants to be determined.

Step 1: Find y′

pand y′′

yp=Ae2x+Bsin x

y′

p= 2Ae2x+Bcos x

y′′

p= 4Ae2x−Bsin x

Step 2: Substitute yp, y′

p,and y′′

pinto the differential equation

−5(2Ae2x+Bcos x) + 6(Ae2x+Bsin x)=2e2x+ 3 sin x

Step 3: Simplify the equation and solve for Aand B

2(2A−3B)e2x+ (−5B+ 6A) sin x+ 5Bcos x= 2e2x+ 3 sin x

Now we match coefficients on both sides.

For terms with e2x:

2(2A−3B) = 2 ⇒4A−6B= 2 (1)

For terms with sin x:

−5B+ 6A= 3 (2)

For terms with cos x:

5B= 0 ⇒B= 0

Using equations (1) and (2), we can solve for Aand B.

From (2):

6A= 3 ⇒A=1

Step 4: Write down the particular solution Therefore, the particular

solution to the differential equation is

yp=1

2e2x

So, the particular solution is 1

2e2x.

Question 29

Question

Find the particular solution to the differential equation:

y′′ −2y′−3y= 6e3x

Solution

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

method of undetermined coefficients. The form of the particular solution is in

the same form as the nonhomogeneous term (6e3x), which is (Ae3x). We will

find the values of the undetermined coefficients by substituting this form into

the differential equation.

Step 1: Find the derivatives of y=Ae3x.

y′= 3Ae3x

y′′ = 9Ae3x

Step 2: Substitute the derivatives into the differential equation:

9Ae3x−2(3Ae3x)−3Ae3x= 6e3x

9Ae3x−6Ae3x−3Ae3x= 6e3x

0=6e3x

Step 3: Since 0 = 6e3x, there are no solutions for A. Therefore, we need

to adjust the form of the particular solution to include the term that overlaps

with the solution to the homogeneous equation, which is xAe3x.

Step 4: Adjust the form of the particular solution to include the overlapping

term. The updated particular solution is now in the form y=Axe3x.

Step 5: Find the derivatives of y=Axe3x.

y′= (A+ 3Ax)e3x

y′′ = (3A+ 6Ax)e3x

Step 6: Substitute the derivatives into the differential equation:

(3A+ 6Ax)e3x−2(A+ 3Ax)e3x−3Axe3x= 6e3x

(3A+ 6Ax −2A−6Ax −3A)x= 6

0=6

Step 7: Since 0 = 6 is false, we will need to adjust the particular solution

further. This means including another term that overlaps with the solution to

the homogeneous equation ( xe3x).

Step 8: Adjust the form of the particular solution to include the overlapping

term. The updated particular solution is now in the form y=Ax2e3x.

Step 9: Find the derivatives of y=Ax2e3x.

y′= (2Ax + 3Ax2)e3x

y′′ = (2A+ 6Ax + 6Ax + 6x2)e3x

Step 10: Substitute the derivatives into the differential equation:

(2A+ 6Ax + 6Ax + 6x2)e3x−2(2Ax + 3Ax2)e3x−3Ax2e3x= 6e3x

(2A+ 6Ax + 6Ax + 6x2−4Ax −6Ax −3Ax2)x= 6

6x2= 6

x2= 1

x=±1

Step 11: Therefore, the particular solution to the differential equation is

y=Ae3x+Bxe3x+Cx2e3x, where A, B, C are constants to be determined in

the next steps.

Question 30

Question

Find the general solution of the following differential equation using the method

of undetermined coefficients:

y′′ −4y′+ 4y= 4e2x.

Solution

To find the general solution of the given differential equation, we will first find

the complementary function yc(x) and then find a particular solution yp(x) using

the method of undetermined coefficients.

Step 1: Find the complementary function The characteristic equation

of the homogeneous differential equation y′′ −4y′+ 4y= 0 is

r2−4r+ 4 = 0.

This equation has a repeated root r= 2, so the complementary function is

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

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

solution based on the form of the forcing term 4e2x. Since 4e2xdoes not appear

in the complementary function, we try a particular solution in the form of

yp(x) = Ae2x.

Substitute yp(x) into the differential equation to find the value of A:

(2A−4(2A)+4A)e2x= 4e2x.

Solving for A, we get A=1

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

2e2x.

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

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

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

2e2x.

So, the general solution of the given differential equation is

y(x) = (c1+c2x+1

2)e2x.

Question 31

Question

Solve the differential equation y′′ + 4y= sec(2x) using the method of undeter-

mined coefficients.

Solution

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

equation y′′ + 4y= 0. The characteristic equation is r2+ 4 = 0, which has roots

r=±2i. Therefore, the complementary solution is yc=c1cos(2x) + c2sin(2x).

Step 2: Next, to find the particular solution yp, we need to consider the form

of the right-hand side function. Since sec(2x) is a trigonometric function, we

try a particular solution of the form yp=Asec(2x) + Btan(2x).

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

y′

p= 2Asec(2x) tan(2x)+2Bsec2(2x)

y′′

p= 2Asec(2x)(sec(2x) tan(2x)) + 4Asec2(2x)+4Bsec(2x) tan(2x)

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

tion:

2Asec(2x)(sec(2x) tan(2x))+4Asec2(2x)+4Bsec(2x) tan(2x)+4(Asec(2x)+Btan(2x)) = sec(2x)

Step 5: Simplify and group terms with the same trigonometric functions

together. Equate coefficients of like terms to find the values of Aand B.

Step 6: Having found values for Aand B, the particular solution is yp=

−1

2sec(2x).

Step 7: Now, the general solution is the sum of the complementary and

particular solutions:

y=yc+yp=c1cos(2x) + c2sin(2x)−1

2sec(2x)

Question 32

Question

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

undetermined coefficients:

y′′ −5y′+ 6y= 3e2x+ 4x

Solution

Step 1: We first find the complementary solution for the homogeneous part of

the differential equation: The characteristic equation is r2−5r+ 6 = 0. This

factors as (r−2)(r−3) = 0. So, the complementary solution is yc=c1e2x+c2e3x.

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

Since the right-hand side of the equation contains terms 3e2xand 4x, we try

the particular solution in the form: yp=Ae2x+Bx +C.

Step 3: Substitute the particular solution ypback into the differential equa-

tion and solve for the coefficients A,B, and C. Plugging ypinto the differential

equation, we get:

y′′

p−5y′

p+ 6yp= 3e2x+ 4x

(4A−5(2A+B) + 6(Ae2x+Bx +C)=3e2x+ 4x

(4A−10A−5B+ 6Ae2x+ 6Bx + 6C)=3e2x+ 4x

Step 4: Equate the coefficients of like terms on both sides to find the values

of A,B, and C. Matching coefficients of e2xgives 2A+ 6A= 3, so A=1

Matching coefficients of xgives −5B+ 6B= 4, so B=−4. Matching constant

terms gives 4A−10A+ 6C= 0, so C=15

Therefore, the particular solution is yp=1

2e2x−4x+15

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

the particular solution:

y=yc+yp=c1e2x+c2e3x+1

2e2x−4x+15

y= (c1+1

2)e2x+c2e3x−4x+15

Therefore, the general solution to the given differential equation is y= (c1+

2)e2x+c2e3x−4x+15

Question 33

Question

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

undetermined coefficients:

y′′ + 4y= 2 sin(2x)+3x

Solution

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

tion y′′ +4y= 0. The characteristic equation is r2+4 = 0, which gives r=±2i.

Thus, the general solution to the homogeneous equation is

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

Step 2: Now, we proceed to find a particular solution to the non-homogeneous

equation in the form of yp(x) = Asin(2x) + Bx +C.

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

y′

p(x)=2Acos(2x) + B

y′′

p(x) = −4Asin(2x)

Step 4: Substitute yp(x), y′

p(x), and y′′

p(x) back into the non-homogeneous

equation and simplify to get:

−4Asin(2x) + 4(Asin(2x) + Bx +C) = 2 sin(2x)+3x

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

a system of equations:











−4A+ 4A= 0

4B= 3

4C= 0

Solving the system, we find A= 0, B=3

4, and C= 0.

Step 6: The particular solution is yp(x) = 3

4x.

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

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

y(x) = yh(x) + yp(x) = c1cos(2x) + c2sin(2x) + 3

Question 34

Question

Find the particular solution to the following differential equation using the

method of undetermined coefficients:

y′′ −2y′+y=xex

Solution

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

undetermined coefficients, we first find the complementary function by solving

the homogeneous equation:

ycf =c1ex+c2xex

Next, we look for a particular solution in the form:

yp=Axex+Bx +C

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

Step 1: We differentiate ypto find the first and second derivatives:

y′

p=Aex+Axex+B

y′′

p= 2Aex+Aex+B

Step 2: Substitute yp,y′

p, and y′′

pback into the original differential equation:

(2A−2Aex+B)−2(Aex+Axex+B)+(Axex+Bx +C) = xex

Step 3: Equate coefficients of like terms:

2A−2A= 0,−2A−2A+B= 1, B −2B= 0

Solving the system of equations, we find A=1

2,B= 0.

Step 4: Substitute A=1

2and B= 0 back into ypto find the particular

solution:

yp=1

2xex

Therefore, the particular solution to the differential equation is yp=1

2xex.

Question 35

Question

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

undetermined coefficients:

y′′ −3y′+ 2y=x2+ 3 sin(x)

Solution

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

undetermined coefficients, we first find the complementary solution ycby solving

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

c1and c2are constants to be determined.

Next, we determine the particular solution ypby assuming the form of the

particular solution and then solving for the undetermined coefficients. Since

the right-hand side consists of x2and 3 sin(x), we assume that the particular

solution has the form

yp=Ax2+Bsin(x) + Ccos(x)

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

Step 1: Calculate the derivatives of yp.

y′

p= 2Ax +Bcos(x)−Csin(x)

y′′

p= 2A−Bsin(x)−Ccos(x)

Step 2: Substitute yp,y′

p, and y′′

pinto the original differential equation and

simplify.

2A−Bsin(x)−Ccos(x)−3(2Ax +Bcos(x)−Csin(x)) + 2(Ax2+Bsin(x) + Ccos(x)) = x2+ 3 sin(x)

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

Step 3: Equate like terms on both sides of the equation.

2A−3B+ 2C= 1 (coefficient of x2)

−3A−2B+ 2C= 0 (coefficient of sin(x))

3A−2B−2C= 3 (coefficient of cos(x))

Step 4: Solve the system of equations to find A,B, and C.











2A−3B+ 2C= 1

−3A−2B+ 2C= 0

3A−2B−2C= 3

Solving the above system of equations, we find A=−4, B=−1, and C= 3.

Step 5: Assemble the general solution.

y=yc+yp=c1ex+c2e2x−4x2−sin(x) + 3 cos(x)

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

y′′ −3y′+ 2y=x2+ 3 sin(x) is

y=c1ex+c2e2x−4x2−sin(x) + 3 cos(x)

where c1and c2are arbitrary constants.

MATH 334 - DIFFERENTIAL EQUATIONS Method of undetermined coefficients Question Bank Set 5

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