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# 5.4E: The Method of Undetermined Coefficients I (Exercises)

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University
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## Q5.4.1

In Exercises 5.4.1-5.4.14 find a particular solution.

1. $$y''-3y'+2y=e^{3x}(1+x)$$

2. $$y''-6y'+5y=e^{-3x}(35-8x)$$

3. $$y''-2y'-3y=e^x(-8+3x)$$

4. $$y''+2y'+y=e^{2x}(-7-15x+9x^2)$$

5. $$y''+4y=e^{-x}(7-4x+5x^2)$$

6. $$y''-y'-2y=e^x(9+2x-4x^2)$$

7. $$y''-4y'-5y=-6xe^{-x}$$

8. $$y''-3y'+2y=e^x(3-4x)$$

9. $$y''+y'-12y=e^{3x}(-6+7x)$$

10. $$2y''-3y'-2y=e^{2x}(-6+10x)$$

11. $$y''+2y'+y=e^{-x}(2+3x)$$

12. $$y''-2y'+y=e^x(1-6x)$$

13. $$y''-4y'+4y=e^{2x}(1-3x+6x^2)$$

14. $$9y''+6y'+y=e^{-x/3}(2-4x+4x^2)$$

## Q5.4.2

In Exercises 5.4.15-5.4.19 find the general solution.

15. $$y''-3y'+2y=e^{3x}(1+x)$$

16. $$y''-6y'+8y=e^x(11-6x)$$

17. $$y''+6y'+9y=e^{2x}(3-5x)$$

18. $$y''+2y'-3y=-16xe^x$$

19. $$y''-2y'+y=e^x(2-12x)$$

## Q5.4.3

In Exercises 5.4.20-5.4.23 solve the initial value problem and plot the solution.

20. $$y''-4y'-5y=9e^{2x}(1+x), \quad y(0)=0,\quad y'(0)=-10$$

21. $$y''+3y'-4y=e^{2x}(7+6x), \quad y(0)=2,\quad y'(0)=8$$

22. $$y''+4y'+3y=-e^{-x}(2+8x), \quad y(0)=1,\quad y'(0)=2$$

23. $$y''-3y'-10y=7e^{-2x}, \quad y(0)=1,\quad y'(0)=-17$$

## Q5.4.4

In Exercises 5.4.24-5.4.29 use the principle of superposition to find a particular solution.

24. $$y''+y'+y=xe^x+e^{-x}(1+2x)$$

25. $$y''-7y'+12y=-e^x(17-42x)-e^{3x}$$

26. $$y''-8y'+16y=6xe^{4x}+2+16x+16x^2$$

27. $$y''-3y'+2y=-e^{2x}(3+4x)-e^x$$

28. $$y''-2y'+2y=e^x(1+x)+e^{-x}(2-8x+5x^2)$$

29. $$y''+y=e^{-x}(2-4x+2x^2)+e^{3x}(8-12x-10x^2)$$

## Q5.4.5

30.

1. Prove that $$y$$ is a solution of the constant coefficient equation $ay''+by'+cy=e^{\alpha x}G(x) \tag{A}$ if and only if $$y=ue^{\alpha x}$$, where $$u$$ satisfies $au''+p'(\alpha)u'+p(\alpha)u=G(x) \tag{B}$ and $$p(r)=ar^2+br+c$$ is the characteristic polynomial of the complementary equation $ay''+by'+cy=0.\nonumber$ For the rest of this exercise, let $$G$$ be a polynomial. Give the requested proofs for the case where $G(x)=g_0+g_1x+g_2x^2+g_3x^3.\nonumber$
2. Prove that if $$e^{\alpha x}$$ isn’t a solution of the complementary equation then (B) has a particular solution of the form $$u_p=A(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, as in Example 5.4.4. Conclude that (A) has a particular solution of the form $$y_p=e^{\alpha x}A(x)$$.
3. Show that if $$e^{\alpha x}$$ is a solution of the complementary equation and $$xe^{\alpha x}$$ isn’t , then (B) has a particular solution of the form $$u_p=xA(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, as in Example 5.4.5. Conclude that (A) has a particular solution of the form $$y_p=xe^{\alpha x}A(x)$$.
4. Show that if $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are both solutions of the complementary equation then (B) has a particular solution of the form $$u_p=x^2A(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, and $$x^2A(x)$$ can be obtained by integrating $$G/a$$ twice, taking the constants of integration to be zero, as in Example 5.4.6. Conclude that (A) has a particular solution of the form $$y_p=x^2e^{\alpha x}A(x)$$.

## Q5.4.6

Exercises 5.4.31–5.4.36 treat the equations considered in Examples 5.4.1–5.4.6. Substitute the suggested form of $$y_{p}$$ into the equation and equate the resulting coefficients of like functions on the two sides of the resulting equation to derive a set of simultaneous equations for the coefficients in $$y_{p}$$. Then solve for the coefficients to obtain $$y_{p}$$. Compare the work you’ve done with the work required to obtain the same results in Examples 5.4.1–5.4.6.

31. Compare with Example 5.4.1:

$y''-7y'+12y=4e^{2x};\quad y_p=Ae^{2x}\nonumber$

32. Compare with Example 5.4.2:

$y''-7y'+12y=5e^{4x};\quad y_p=Axe^{4x}\nonumber$

33. Compare with Example 5.4.3:

$y''-8y'+16y=2e^{4x};\quad y_p=Ax^2e^{4x}\nonumber$

34. Compare with Example 5.4.4:

$y''-3y'+2y=e^{3x}(-1+2x+x^2),\quad y_p=e^{3x}(A+Bx+Cx^2)\nonumber$

35. Compare with Example 5.4.5:

$y''-4y'+3y=e^{3x}(6+8x+12x^2),\quad y_p=e^{3x}(Ax+Bx^2+Cx^3)\nonumber$

36. Compare with Example 5.4.6:

$4y''+4y'+y=e^{-x/2}(-8+48x+144x^2),\quad y_p=e^{-x/2}(Ax^2+Bx^3+Cx^4)\nonumber$

## Q5.4.7

37. Write $$y=ue^{\alpha x}$$ to find the general solution.

1. $$y''+2y'+y={e^{-x}\over\sqrt x}$$
2. $$y''+6y'+9y=e^{-3x}\ln x$$
3. $$y''-4y'+4y={e^{2x}\over1+x}$$
4. $$4y''+4y'+y={4e^{-x/2}\left({1\over x}+x\right)}$$

38. Suppose $$\alpha\ne0$$ and $$k$$ is a positive integer. In most calculus books integrals like $$\int x^k e^{\alpha x}\,dx$$ are evaluated by integrating by parts $$k$$ times. This exercise presents another method. Let

$y=\int e^{\alpha x}P(x)\,dx\nonumber$

with

$P(x)=p_0+p_1x+\cdots+p_kx^k\nonumber$

(where $$p_k \neq 0$$).

1.  Show that $$y=e^{\alpha x}u$$, where $u'+\alpha u=P(x). \tag{A}$
2. Show that (A) has a particular solution of the form $u_p=A_0+A_1x+\cdots+A_kx^k,\nonumber$ where $$A_k$$, $$A_{k-1}$$, …, $$A_0$$ can be computed successively by equating coefficients of $$x^k,x^{k-1}, \dots,1$$ on both sides of the equation $u_p'+\alpha u_p=P(x).\nonumber$
3. Conclude that $\int e^{\alpha x}P(x)\,dx=\left(A_0+A_1x+\cdots+A_kx^k\right)e^{\alpha x} +c,\nonumber$  where $$c$$ is a constant of integration.

39. Use the method of Exercise 5.4.38 to evaluate the integral.

1. $$\int e^{x}(4+x)dx$$
2. $$\int e^{-x}(-1+x^{2})dx$$
3. $$\int x^{3}e^{-2x}dx$$
4. $$\int e^{x}(1+x)^{2}dx$$
5. $$\int e^{3x}(-14+30x+27x^{2})dx$$
6. $$\int e^{-x}(1+6x^{2}-14x^{3}+3x^{4})dx$$

40. Use the method suggested in Exercise 5.4.38 to evaluate $$\int x^ke^{\alpha x}\,dx$$, where $$k$$ is an arbitrary positive integer and $$\alpha\ne0$$.