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Mathematics LibreTexts

9.3.1: Undetermined Coefficients for Higher Order Equations (Exercises)

  • Page ID
    18270
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    Q9.3.1

    In Exercises 9.3.1-9.3.59 find a particular solution.

    1. \(y'''-6y''+11y'-6y=-e^{-x}(4+76x-24x^2)\)

    2. \(y'''-2y''-5y'+6y=e^{-3x}(32-23x+6x^2)\)

    3. \(4y'''+8y''-y'-2y=-e^x(4+45x+9x^2)\)

    4. \(y'''+3y''-y'-3y=e^{-2x}(2-17x+3x^2)\)

    5. \(y'''+3y''-y'-3y=e^x(-1+2x+24x^2+16x^3)\)

    6. \(y'''+y''-2y=e^x(14+34x+15x^2)\)

    7. \(4y'''+8y''-y'-2y=-e^{-2x}(1-15x)\)

    8. \(y'''-y''-y'+y=e^x(7+6x)\)

    9. \(2y'''-7y''+4y'+4y=e^{2x}(17+30x)\)

    10. \(y'''-5y''+3y'+9y=2e^{3x}(11-24x^2)\)

    11. \(y'''-7y''+8y'+16y=2e^{4x}(13+15x)\)

    12. \(8y'''-12y''+6y'-y=e^{x/2}(1+4x)\)

    13. \(y^{(4)}+3y'''-3y''-7y'+6y=-e^{-x}(12+8x-8x^2)\)

    14. \(y^{(4)}+3y'''+y''-3y'-2y=-3e^{2x}(11+12x)\)

    15. \(y^{(4)}+8y'''+24y''+32y'=-16e^{-2x}(1+x+x^2-x^3)\)

    16. \(4y^{(4)}-11y''-9y'-2y=-e^x(1-6x)\)

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

    18. \(y^{(4)}-4y'''+6y''-4y'+2y=e^{2x}(24+x+x^4)\)

    19. \(2y^{(4)}+5y'''-5y'-2y=18e^x(5+2x)\)

    20. \(y^{(4)}+y'''-2y''-6y'-4y=-e^{2x}(4+28x+15x^2)\)

    21. \(2y^{(4)}+y'''-2y'-y=3e^{-x/2}(1-6x)\)

    22. \(y^{(4)}-5y''+4y=e^x(3+x-3x^2)\)

    23. \(y^{(4)}-2y'''-3y''+4y'+4y=e^{2x}(13+33x+18x^2)\)

    24. \(y^{(4)}-3y'''+4y'=e^{2x}(15+26x+12x^2)\)

    25. \(y^{(4)}-2y'''+2y'-y=e^x(1+x)\)

    26. \(2y^{(4)}-5y'''+3y''+y'-y=e^x(11+12x)\)

    27. \(y^{(4)}+3y'''+3y''+y'=e^{-x}(5-24x+10x^2)\)

    28. \(y^{(4)}-7y'''+18y''-20y'+8y=e^{2x}(3-8x-5x^2)\)

    29. \(y'''-y''-4y'+4y=e^{-x}\left[(16+10x)\cos x+(30-10x)\sin x\right]\)

    30. \(y'''+y''-4y'-4y=e^{-x}\left[(1-22x)\cos 2x-(1+6x)\sin2x\right]\)

    31. \(y'''-y''+2y'-2y=e^{2x}[(27+5x-x^2)\cos x+(2+13x+9x^2)\sin x]\)

    32. \(y'''-2y''+y'-2y=-e^x[(9-5x+4x^2)\cos 2x-(6-5x-3x^2)\sin2x]\)

    33. \(y'''+3y''+4y'+12y=8\cos2x-16\sin2x\)

    34. \(y'''-y''+2y=e^x[(20+4x)\cos x-(12+12x)\sin x]\)

    35. \(y'''-7y''+20y'-24y=-e^{2x}[(13-8x)\cos 2x-(8-4x)\sin2x]\)

    36. \(y'''-6y''+18y'=-e^{3x}[(2-3x)\cos 3x-(3+3x)\sin3x]\)

    37. \(y^{(4)}+2y'''-2y''-8y'-8y=e^x(8\cos x+16\sin x)\)

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

    39. \(y^{(4)}-8y'''+24y''-32y'+15y=e^{2x}(15x\cos2x+32\sin2x)\)

    40. \(y^{(4)}+6y'''+13y''+12y'+4y=e^{-x}[(4-x)\cos x-(5+x)\sin x]\)

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

    42. \(y^{(4)}-5y'''+13y''-19y'+10y=e^x (\cos2x+\sin2x)\)

    43. \(y^{(4)}+8y'''+32y''+64y'+39y=e^{-2x}[(4-15x)\cos3x-(4+15x)\sin 3x]\)

    44. \(y^{(4)}-5y'''+13y''-19y'+10y=e^x[(7+8x)\cos 2x+(8-4x)\sin2x]\)

    45. \(y^{(4)}+4y'''+8y''+8y'+4y=-2e^{-x} (\cos x-2\sin x)\)

    46. \(y^{(4)}-8y'''+32y''-64y'+64y=e^{2x} (\cos2x-\sin2x)\)

    47. \(y^{(4)}-8y'''+26y''-40y'+25y=e^{2x}[3\cos x-(1+3x)\sin x]\)

    48. \(y'''-4y''+5y'-2y=e^{2x}-4e^x-2\cos x+4\sin x\)

    49. \(y'''-y''+y'-y=5e^{2x}+2e^x-4\cos x+4\sin x\)

    50. \(y'''-y'=-2(1+x)+4e^x-6e^{-x}+96e^{3x}\)

    51. \(y'''-4y''+9y'-10y=10e^{2x}+20e^x\sin2x-10\)

    52. \(y'''+3y''+3y'+y=12e^{-x}+9\cos2x-13\sin2x\)

    53. \(y'''+y''-y'-y=4e^{-x}(1-6x)-2x\cos x+2(1+x)\sin x\)

    54. \(y^{(4)}-5y''+4y=-12e^x+6e^{-x}+10\cos x\)

    55. \(y^{(4)}-4y'''+11y''-14y'+10y=-e^x(\sin x+2\cos2x)\)

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

    57. \(y^{(4)}+4y=\sinh x\cos x-\cosh x\sin x\)

    58. \(y^{(4)}+5y'''+9y''+7y'+2y=e^{-x}(30+24x)-e^{-2x}\)

    59. \(y^{(4)}-4y'''+7y''-6y'+2y=e^x(12x-2\cos x+2\sin x)\)

    Q9.3.2

    In Exercises 9.3.60-9.3.68 find the general solution.

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

    61. \(y'''+y''-2y=-e^{3x}(9+67x+17x^2)\)

    62. \(y'''-6y''+11y'-6y=e^{2x}(5-4x-3x^2)\)

    63. \(y'''+2y''+y'=-2e^{-x}(7-18x+6x^2)\)

    64. \(y'''-3y''+3y'-y=e^x(1+x)\)

    65. \(y^{(4)}-2y''+y=-e^{-x}(4-9x+3x^2)\)

    66. \(y'''+2y''-y'-2y=e^{-2x}\left[(23-2x)\cos x+(8-9x)\sin x\right]\)

    67. \(y^{(4)}-3y'''+4y''-2y'=e^x\left[(28+6x)\cos 2x+(11-12x)\sin2x\right]\)

    68. \(y^{(4)}-4y'''+14y''-20y'+25y=e^x\left[(2+6x)\cos 2x+3\sin2x\right]\)

    Q9.3.3

    In Exercises 9.3.69-9.3.74 solve the initial value problem and graph the solution.

    69. \(y'''-2y''-5y'+6y=2e^x(1-6x),\quad y(0)=2, \quad y'(0)=7,\quad y''(0)=9\)

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

    71. \(4y'''-3y'-y=e^{-x/2}(2-3x),\quad y(0)=-1, \quad y'(0)=15,\quad y''(0)=-17\)

    72. \(y^{(4)}+2y'''+2y''+2y'+y=e^{-x}(20-12x),\, y(0)=3,\; y'(0)=-4,\; y''(0)=7,\; y'''(0)=-22\)

    73. \(y'''+2y''+y'+2y=30\cos x-10\sin x, \quad y(0)=3,\quad y'(0)=-4,\quad y''(0)=16\)

    74. \(y^{(4)}-3y'''+5y''-2y'=-2e^x(\cos x-\sin x),\; y(0)=2,\; y'(0)=0,\; y''(0)~=~-1, \; y'''(0)=-5\)

    Q9.3.4

    75. Prove: A function \(y\) is a solution of the constant coefficient nonhomogeneous equation

    \[a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=e^{\alpha x}G(x) \tag{A}\]

    if and only if \(y=ue^{\alpha x}\), where \(u\) satisfies the differential equation

    \[a_0u^{(n)}+{p^{(n-1)}(\alpha)\over(n-1)!}u^{(n-1)}+ {p^{(n-2)}(\alpha)\over(n-2)!}u^{(n-2)}+\cdots+p(\alpha)u=G(x) \tag{B}\]

    and

    \[p(r)=a_0r^n+a_1r^{n-1} + \cdots + a_n\nonumber\]

    is the characteristic polynomial of the complementary equation

    \[a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=0.\nonumber\]

    76. Prove:

    1. The equation \[\begin{array}{lll}{a_{0}u^{(n)}}&{+}&{\frac{p^{(n-1)}(\alpha )}{(n-1)!}u^{(n-1)}+\frac{p^{(n-2)}(\alpha )}{(n-2)!}u^{(n-2)}+\cdots + P(\alpha )u}\\{}&{=}&{(p_{0}+p_{1}x+\cdots +p_{k}x^{k})\cos\omega x} \\ {}&{+}&{(q_{0}+q_{1}x+\cdots +q_{k}x^{k})\sin\omega x} \end{array}\tag{A}\] has a particular solution of the form \[u_p=x^m\left(u_0+u_1x+\cdots+u_kx^k\right)\cos\omega x+ \left(v_0+v_1x+\cdots+v_kx^k\right)\sin\omega x.\nonumber\]
    2. If \(\lambda+i\omega\) is a zero of \(p\) with multiplicity \(m\ge1\), then (A) can be written as \[a(u''+\omega^2 u)= \left(p_0+p_1x+\cdots+p_kx^k\right)\cos\omega x+ \left(q_0+q_1x+\cdots+q_kx^k\right)\sin\omega x,\nonumber\] which has a particular solution of the form \[u_p=U(x)\cos\omega x+V(x)\sin\omega x,\nonumber\] where \[U(x)=u_0x+u_1x^2+\cdots+u_kx^{k+1},\,V(x)=v_0x+v_1x^2+\cdots+v_kx^{k+1}\nonumber\] and \[\begin{array}{rcl} a(U''(x)+2\omega V'(x))&=&p_0+p_1x+\cdots+p_kx^k\\[10pt] a(V''(x)-2\omega U'(x))&=&q_0+q_1x+\cdots+q_kx^k. \end{array}\nonumber\]