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

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

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

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 \[ay''+by'+cy=0,\] 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 \[\begin{array}{lcl} \quad (c-a\omega^2)A+b\omega B+2a\omega B'+bA'+aA''&=P\phantom{.}\\ -b\omega A+(c-a\omega^2)B-2a\omega A'+bB'+aB''&=Q, \end{array}\] where \((A_k,B_k)\), \((A_{k-1},B_{k-1})\), …,\((A_0,B_0)\) can be computed successively by solving the systems \[\begin{array}{lcl} \phantom{-}(c-a\omega^2)A_k+b\omega B_k&=p_k\phantom{.}\\ -b\omega A_k+(c-a\omega^2)B_k&=q_k, \end{array}\] and, if \(1\le r\le k\), \[\begin{array}{lcl} \phantom{-}(c-a\omega^2)A_{k-r}+b\omega B_{k-r}&=p_{k-r}+\cdots\phantom{.}\\ -b\omega A_{k-r}+(c-a\omega^2)B_{k-r}&=q_{k-r}+\cdots, \end{array}\] 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 \[ay''+by'+cy=P(x)\cos\omega x+Q(x)\sin\omega x.\]
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 \[A(x)=A_0x+A_1x^2+\cdots+A_kx^{k+1}\quad \text{and} \quad B(x)=B_0x+B_1x^2+\cdots+B_kx^{k+1}\] such that \[\begin{array}{rcl} a(A''+2\omega B')&=P\\ a(B''-2\omega A')&=Q, \end{array}\] where the pairs \((A_k,B_k)\), \((A_{k-1},B_{k-1})\), …, \((A_0,B_0)\) can be computed successively as follows: \[\begin{aligned} A_k&=-{q_k\over2a\omega(k+1)}\\ B_k&=\phantom{-}{p_k\over2a\omega(k+1)},\end{aligned}\] and, if \(k\ge 1\), \[\begin{aligned} A_{k-j}&=-{1\over2\omega} \left[{q_{k-j}\over a(k-j+1)}-(k-j+2)B_{k-j+1}\right]\\ B_{k-j}&=\phantom{-}{1\over2\omega} \left[{p_{k-j}\over a(k-j+1)}-(k-j+2)A_{k-j+1}\right]\end{aligned}\] 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 \(\PageIndex{1}\)

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

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

has a particular solution

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

where

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

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

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

\((\)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

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

39. This exercise presents a method for evaluating the integral

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

where \(\omega\ne0\) and

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

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 \[u_p=A(x)\cos \omega x+B(x)\sin \omega x,\] where \[A(x)=A_0+A_1x+\cdots+A_kx^k,\quad B(x)=B_0+B_1x+\cdots+B_kx^k,\] 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 \[\int e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)\,dx = e^{\lambda x}\left(A(x)\cos \omega x+B(x)\sin \omega x\right) +c,\] 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. \(\int e^{-x}[(1+x^{2})\cos x +(1_x^{2})\sin x]dx\)