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Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Geoffrey W. Cox, Ph.D., ?

Section 7.6 Summary & Exercises

Summary of the Key Ideas.

  • The method of undetermined coefficients is used to solve non-homogeneous linear differential equations.
  • The general solution to a non-homogeneous equation is the sum of the general solution to the corresponding homogeneous equation and a particular solution.
  • The method involves guessing the form of the particular solution based on the form of the non-homogeneous term and solving for the coefficients.
  • The method is applicable when the non-homogeneous term can be expressed as a linear combination of known functions.

Exercises Exercises

1.

Describe the difference between a homogeneous and a non-homogeneous differential equation.

Solve the IVP.

(43)2z″+z=9e2t,z(0)=3,z′(0)=−1

2.

Find the homogeneous part, zh, of the general solution of (43).
Answer. Answer
zh=c1cos⁡(12t)+c2sin⁡(12t)

3.

Select the initial form of the particular solution, zp, by looking at the non-homogenous term of equation (43).
Answer. Answer
zp=Ae2t

4.

Substitute zp into (43) to find the coefficient, A. Then give zp.
Answer. Answer
zp=e2t

5.

Form the general solution z=zc+zp.
Answer. Answer
z=c1cos⁡(12t)+c2sin⁡(12t)+e2t

6.

Use the initial conditions to find the particular solution to this IVP.
Answer. Answer
z(t)=−32cos⁡(12t)+2sin⁡(12t)+e2t

Find the Initial Form of yp..

yp.DO NOT determine the values of the coefficients A,B, etc.

7.

y″−3y′−17y=x2+cos⁡x
Answer. Answer
yp=[Ax2+Bx+C]+[Dcos⁡x+Esin⁡x]

8.

y″+y′−12y=4e3t+cos⁡(2t)
Answer. Answer
yp=[Ate3t]+[Bcos⁡(2t)+Csin⁡(2t)]

9.

y″−4y=2t−e2tsin⁡(3t)
Answer. Answer
yp=[At+B]+[Ce2tcos⁡(3t)+De2tsin⁡(3t)]

10.

x3sin⁡(4x)+x2sin⁡(4x)+7y″−y′+2y=0
Answer. Answer

11.

w″−4w′+13w=e3x+x2e−x
Answer. Answer
wp=[Ae3x]+[(Bx2+Cx+D)e−x]

12.

z″−4z′+5z=te2t+2e2tsin⁡t
Answer. Answer
zp=[(At+B)e2t]+[Cet2tcos⁡t+Dte2tsin⁡t]

13.

y″−2y′+y=x2+ex
Answer. Answer
yp=[Ax2+Bx+C]+[Dx2ex]

14.

w″−4w′+13=e3x+x2e−x
Answer. Answer
wp=[Ae3x]+[(Bx2+Cx+D)e−x]

15.

3z″−4z′−12z=17+2cos⁡(2θ)
Answer. Answer
zp=[A]+[Bcos⁡(2θ)+Csin⁡(2θ)]

Find the general solution for each differential equation.

16.

2x′+x=3t2
Answer. Answer
x=c1e−12t+3t2−12t+24

17.

y″−4y′−5y=t+2e−t
Answer. Answer
y=c1e5t+c2e−t−15t+425−13te−t

18.

z″−650sin⁡(6x)+34z=6z′
Answer. Answer
z=c1e3xcos⁡(5x)+c2e3xsin⁡(5x)+18cos⁡(6x)−sin⁡(6x)

19.

y″−4y′+4y=8cos⁡x+12e2x
Answer. Answer
y=c1e2x+c2xe2x+89cos⁡x−3227sin⁡x+6x2e2x

20.

y″−y′−6y=2t+3e3t−e−2t
Answer. Answer
y=c1e3t+c2e−2t−13t+118+35te3t+15te−2t

Exercise Group.

21.

Find the solution to the IVP:
2x′+x=3t2,x(0)=15.
Answer. Answer
x=−9e−12t+3t2−12t+24

Exercise Group.

22.

Notice that the DE in Question Exercise 7.6.21 is actually a first-order linear differential equation, so we can use another method of solution to solve it (and we should get the same answer!). Solve the same IVP using an integrating factor and compareyour answers.
Answer. Answer
x=3t2−12t+24−9e−12t
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