5.5E: The Method of Undetermined Coefficients II (Exercises)

Q5.5.1

In Exercises 5.5.1-5.5.17 find a particular solution.

1. $y''+3y'+2y=7\cos x-\sin x$

2. $y''+3y'+y=(2-6x)\cos x-9\sin x$

3. $y''+2y'+y=e^x(6\cos x+17\sin x)$

4. $y''+3y'-2y=-e^{2x}(5\cos2x+9\sin2x)$

5. $y''-y'+y=e^x(2+x)\sin x$

6. $y''+3y'-2y=e^{-2x}\left[(4+20x)\cos 3x+(26-32x)\sin 3x\right]$

7. $y''+4y=-12\cos2x-4\sin2x$

8. $y''+y=(-4+8x)\cos x+(8-4x)\sin x$

9. $4y''+y=-4\cos x/2-8x\sin x/2$

10. $y''+2y'+2y=e^{-x}(8\cos x-6\sin x)$

11. $y''-2y'+5y=e^x\left[(6+8x)\cos 2x+(6-8x)\sin2x\right]$

12. $y''+2y'+y=8x^2\cos x-4x\sin x$

13. $y''+3y'+2y=(12+20x+10x^2)\cos x+8x\sin x$

14. $y''+3y'+2y=(1-x-4x^2)\cos2x-(1+7x+2x^2)\sin2x$

15. $y''-5y'+6y=-e^x\left[(4+6x-x^2)\cos x-(2-4x+3x^2)\sin x\right]$

16. $y''-2y'+y=-e^x\left[(3+4x-x^2)\cos x+(3-4x-x^2)\sin x\right]$

17. $y''-2y'+2y=e^x\left[(2-2x-6x^2)\cos x+(2-10x+6x^2)\sin x\right]$

Q5.5.2

In Exercises 5.5.18-5.5.21 find a particular solution and graph it.

18. $y''+2y'+y=e^{-x}\left[(5-2x)\cos x-(3+3x)\sin x\right]$

19. $y''+9y=-6\cos 3x-12\sin 3x$

20. $y''+3y'+2y=(1-x-4x^2)\cos2x-(1+7x+2x^2)\sin2x$

21. $y''+4y'+3y=e^{-x}\left[(2+x+x^2)\cos x+(5+4x+2x^2)\sin x\right]$

Q5.5.3

In Exercises 5.5.22-5.5.26 solve the initial value problem.

22. $y''-7y'+6y=-e^x(17\cos x-7\sin x), \quad y(0)=4,\; y'(0)=2$

23. $y''-2y'+2y=-e^x(6\cos x+4\sin x), \quad y(0)=1,\; y'(0)=4$

24. $y''+6y'+10y=-40e^x\sin x, \quad y(0)=2,\quad y'(0)=-3$

25. $y''-6y'+10y=-e^{3x}(6\cos x+4\sin x), \quad y(0)=2,\quad y'(0)=7$

26. $y''-3y'+2y=e^{3x}\left[21\cos x-(11+10x)\sin x\right], \; y(0)=0, \quad y'(0)=6$

Q5.5.4

In Exercises 5.5.27-5.5.32 use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem.

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

28. $y''+y=4\cos x-2\sin x+xe^x+e^{-x}$

29. $y''-3y'+2y=xe^x+2e^{2x}+\sin x$

30. $y''-2y'+2y=4xe^x\cos x+xe^{-x}+1+x^2$

31. $y''-4y'+4y=e^{2x}(1+x)+e^{2x}(\cos x-\sin x)+3e^{3x}+1+x$

32. $y''-4y'+4y=6e^{2x}+25\sin x, \quad y(0)=5,\; y'(0)=3$

Q5.5.5

In Exercises 5.5.33-5.5.35 solve the initial value problem and graph the solution.

33. $y''+4y=-e^{-2x}\left[(4-7x)\cos x+(2-4x)\sin x\right], \; y(0)=3, \quad y'(0)=1$

34. $y''+4y'+4y=2\cos2x+3\sin2x+e^{-x}, \quad y(0)=-1,\; y'(0)=2$

35. $y''+4y=e^x(11+15x)+8\cos2x-12\sin2x, \quad y(0)=3,\; y'(0)=5$

Q5.5.6

36.

1. Verify that if $$y_p=A(x)\cos\omega x+B(x)\sin\omega x \nonumber$$ where $A$ and $B$ are twice differentiable, then $$\begin{aligned} y_p'&=(A'+\omega B)\cos\omega x+(B'-\omega A) \sin\omega x\quad \mbox{ and}\\[4pt] y_p''&=(A''+2\omega B'-\omega^2A)\cos\omega x +(B''-2\omega A'-\omega^2B)\sin\omega x.\end{aligned} \nonumber$$
2. Use the results of (a) to verify that $$\begin{aligned} ay_p''+by_p'+cy_p=&\left[(c-a\omega^2)A+b\omega B+2a\omega B'+bA'+aA''\right] \cos\omega x+\\[4pt] & \left[-b\omega A+(c-a\omega^2)B-2a\omega A'+bB'+aB''\right]\sin\omega x.\end{aligned} \nonumber$$
3. Use the results of (a) to verify that $$y_p''+\omega^2 y_p=(A''+2\omega B')\cos\omega x+ (B''-2\omega A')\sin\omega x. \nonumber$$
4. Prove Theorem 5.5.2.

37. Let $a$, $b$, $c$, and $\omega$ be constants, with $a\ne0$ and $\omega>0$, and let

$$P(x)=p_0+p_1x+\cdots+p_kx^k \quad \text{and} \quad Q(x)=q_0+q_1x+\cdots+q_kx^k, \nonumber$$

where at least one of the coefficients $p_k$, $q_k$ is nonzero, so $k$ is the larger of the degrees of $P$ and $Q$.

1. Show that if $\cos\omega x$ and $\sin\omega x$ are not solutions of the complementary equation $$ay''+by'+cy=0, \nonumber$$ then there are polynomials $$A(x)=A_0+A_1x+\cdots+A_kx^k \quad \text{and} \quad B(x)=B_0+B_1x+\cdots+B_kx^k \tag{A}$$ such that $$\begin{array}{lcl} \quad (c-a\omega^2)A+b\omega B+2a\omega B'+bA'+aA''&=P\phantom{.}\\[4pt] -b\omega A+(c-a\omega^2)B-2a\omega A'+bB'+aB''&=Q, \end{array} \nonumber$$ where $(A_k,B_k)$, $(A_{k-1},B_{k-1})$, …,$(A_0,B_0)$ can be computed successively by solving the systems $$\begin{array}{lcl} \phantom{-}(c-a\omega^2)A_k+b\omega B_k&=p_k\phantom{.}\\[4pt] -b\omega A_k+(c-a\omega^2)B_k&=q_k, \end{array} \nonumber$$ and, if $1\le r\le k$, $$\begin{array}{lcl} \phantom{-}(c-a\omega^2)A_{k-r}+b\omega B_{k-r}&=p_{k-r}+\cdots\phantom{.}\\[4pt] -b\omega A_{k-r}+(c-a\omega^2)B_{k-r}&=q_{k-r}+\cdots, \end{array} \nonumber$$ where the terms indicated by “$\cdots$” depend upon the previously computed coefficients with subscripts greater than $k-r$. Conclude from this and Exercise 5.5.36b that $$y_p=A(x)\cos\omega x+B(x)\sin\omega x \tag{B}$$ is a particular solution of $$ay''+by'+cy=P(x)\cos\omega x+Q(x)\sin\omega x. \nonumber$$
2. Conclude from Exercise 5.5.36c that the equation $$a(y''+\omega^2 y)=P(x)\cos\omega x+Q(x)\sin\omega x \tag{C}$$ does not have a solution of the form (B) with $A$ and $B$ as in (A). Then show that there are polynomials $$A(x)=A_0x+A_1x^2+\cdots+A_kx^{k+1}\quad \text{and} \quad B(x)=B_0x+B_1x^2+\cdots+B_kx^{k+1} \nonumber$$ such that $$\begin{array}{rcl} a(A''+2\omega B')&=P\\[4pt] a(B''-2\omega A')&=Q, \end{array} \nonumber$$ where the pairs $(A_k,B_k)$, $(A_{k-1},B_{k-1})$, …, $(A_0,B_0)$ can be computed successively as follows: $$\begin{aligned} A_k&=-{q_k\over2a\omega(k+1)}\\[4pt] B_k&=\phantom{-}{p_k\over2a\omega(k+1)},\end{aligned} \nonumber$$ and, if $k\ge 1$, $$\begin{aligned} A_{k-j}&=-{1\over2\omega} \left[{q_{k-j}\over a(k-j+1)}-(k-j+2)B_{k-j+1}\right]\\[4pt] B_{k-j}&=\phantom{-}{1\over2\omega} \left[{p_{k-j}\over a(k-j+1)}-(k-j+2)A_{k-j+1}\right]\end{aligned} \nonumber$$ for $1\le j\le k$. Conclude that (B) with this choice of the polynomials $A$ and $B$ is a particular solution of (C).

38. Show that Theorem 5.5.1 implies the next theorem:

Theorem 5.5E.1

Suppose $\omega$ is a positive number and $P$ and $Q$ are polynomials. Let $k$ be the larger of the degrees of $P$ and $Q$. Then the equation

$$ay''+by'+cy=e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right) \nonumber$$

has a particular solution

$$y_p=e^{\lambda x}\left(A(x)\cos\omega x+B(x)\sin\omega x\right), \tag{A}$$

where

$$A(x)=A_0+A_1x+\cdots+A_kx^k \quad \text{and} \quad B(x)=B_0+B_1x+\cdots+B_kx^k, \nonumber$$

provided that $e^{\lambda x}\cos\omega x$ and $e^{\lambda x}\sin\omega x$ are not solutions of the complementary equation. The equation

$$a\left[y''-2\lambda y'+(\lambda^2+\omega^2)y\right]= e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right) \nonumber$$

$($for which $e^{\lambda x}\cos\omega x$ and $e^{\lambda x}\sin\omega x$ are solutions of the complementary equation$)$ has a particular solution of the form (A), where

$$A(x)=A_0x+A_1x^2+\cdots+A_kx^{k+1} \quad \text{and} \quad B(x)=B_0x+B_1x^2+\cdots+B_kx^{k+1}. \nonumber$$

39. This exercise presents a method for evaluating the integral

$$y=\int e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)\,dx \nonumber$$

where $\omega\ne0$ and

$$P(x)=p_0+p_1x+\cdots+p_kx^k,\quad Q(x)=q_0+q_1x+\cdots+q_kx^k. \nonumber$$

1. Show that $y=e^{\lambda x}u$, where $$u'+\lambda u=P(x)\cos \omega x+Q(x)\sin \omega x. \tag{A}$$
2. Show that (A) has a particular solution of the form $$u_p=A(x)\cos \omega x+B(x)\sin \omega x, \nonumber$$ where $$A(x)=A_0+A_1x+\cdots+A_kx^k,\quad B(x)=B_0+B_1x+\cdots+B_kx^k, \nonumber$$ and the pairs of coefficients $(A_k,B_k)$, $(A_{k-1},B_{k-1})$, …,$(A_0,B_0)$ can be computed successively as the solutions of pairs of equations obtained by equating the coefficients of $x^r\cos\omega x$ and $x^r\sin\omega x$ for $r=k$, $k-1$, …, $0$.
3. Conclude that $$\int e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)\,dx = e^{\lambda x}\left(A(x)\cos \omega x+B(x)\sin \omega x\right) +c, \nonumber$$ where $c$ is a constant of integration.

40. Use the method of Exercise 5.5.39 to evaluate the integral.

1. $\int x^{2}\cos x dx$
2. $\int x^{2} e^{x}\cos x dx$
3. $\int xe^{-x}\sin 2x dx$
4. $\int x^{2}e^{-x}\sin x dx$
5. $\int x^{3}e^{x}\sin x dx$
6. $\int e^{x}[x\cos x - (1+3x)\sin x ]dx$
7. $\int e^{-x}[(1+x^{2})\cos x +(1_x^{2})\sin x]dx$