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4.7E: The Method of Undetermined Coefficients II (Exercises)

  • Page ID
    43301
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    Q4.7.1

    In Exercises 4.7.1-4.7.17 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. \(y''-2y'+2y=e^x\left[(2-2x-6x^2)\cos x+(2-10x+6x^2)\sin x\right]\)

    Q4.7.2

    In Exercises 4.7.18-4.7.22 solve the initial value problem.

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

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

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

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

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

    Q4.7.3

    In Exercises 4.7.23-4.7.28 use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem.

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

    24. \(y''+y=4\cos x-2\sin x+xe^x+e^{-x}\)

    25. \(y''-3y'+2y=xe^x+2e^{2x}+\sin x\)

    26. \(y''-2y'+2y=4xe^x\cos x+xe^{-x}+1+x^2\)

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

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

    Q4.7.4

    In Exercises 4.7.29-4.7.31 solve the initial value problem and graph the solution.

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

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

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

    Q4.7.5

    32. 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.

    33. Use the method of Exercise 4.7.32 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. \(\int e^{-x}[(1+x^{2})\cos x +(1_x^{2})\sin x]dx\)

    This page titled 4.7E: The Method of Undetermined Coefficients II (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.