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Mathematics LibreTexts

8.3E: Solution of Initial Value Problems (Exercises)

  • Page ID
    18507
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    In Exercises [exer:8.3.1}– [exer:8.3.31} use the Laplace transform to solve the initial value problem.

    [exer:8.3.1] \(y''+3y'+2y=e^t, \quad y(0)=1,\quad y'(0)=-6\)

    [exer:8.3.2] \(y''-y'-6y=2, \quad y(0)=1,\quad y'(0)=0\)

    [exer:8.3.3] \(y''+y'-2y=2e^{3t}, \quad y(0)=-1,\quad y'(0)=4\)

    [exer:8.3.4] \(y''-4y=2 e^{3t}, \quad y(0)=1,\quad y'(0)=-1\)

    [exer:8.3.5] \(y''+y'-2y=e^{3t}, \quad y(0)=1,\quad y'(0)=-1\)

    [exer:8.3.6] \(y''+3y'+2y=6e^t, \quad y(0)=1,\quad y'(0)=-1\)

    [exer:8.3.7] \(y''+y=\sin2t, \quad y(0)=0,\quad y'(0)=1\)

    [exer:8.3.8] \(y''-3y'+2y=2e^{3t}, \quad y(0)=1,\quad y'(0)=-1\)

    [exer:8.3.9] \(y''-3y'+2y=e^{4t}, \quad y(0)=1,\quad y'(0)=-2\)

    [exer:8.3.10] \(y''-3y'+2y=e^{3t}, \quad y(0)=-1,\quad y'(0)=-4\)

    [exer:8.3.11] \(y''+3y'+2y=2e^t, \quad y(0)=0,\quad y'(0)=-1\)

    [exer:8.3.12] \(y''+y'-2y=-4, \quad y(0)=2,\quad y'(0)=3\)

    [exer:8.3.13] \(y''+4y=4, \quad y(0)=0,\quad y'(0)=1\)

    [exer:8.3.14] \(y''-y'-6y=2, \quad y(0)=1,\quad y'(0)=0\)

    [exer:8.3.15] \(y''+3y'+2y=e^t, \quad y(0)=0,\quad y'(0)=1\)

    [exer:8.3.16] \(y''-y=1, \quad y(0)=1,\quad y'(0)=0\)

    [exer:8.3.17] \(y''+4y=3\sin t, \quad y(0)=1,\quad y'(0)=-1\)

    [exer:8.3.18] \(y''+y'=2e^{3t}, \quad y(0)=-1,\quad y'(0)=4\)

    [exer:8.3.19] \(y''+y=1, \quad y(0)=2,\quad y'(0)=0\)

    [exer:8.3.20] \(y''+y=t, \quad y(0)=0,\quad y'(0)=2\)

    [exer:8.3.21] \(y''+y=t-3\sin2t, \quad y(0)=1,\quad y'(0)=-3\)

    [exer:8.3.22] \(y''+5y'+6y=2e^{-t}, \quad y(0)=1,\quad y'(0)=3\)

    [exer:8.3.23] \(y''+2y'+y=6\sin t-4\cos t, \quad y(0)=-1,\; y'(0)=1\)

    [exer:8.3.24] \(y''-2y'-3y=10\cos t, \quad y(0)=2,\quad y'(0)=7\)

    [exer:8.3.25] \(y''+y=4\sin t+6\cos t, \quad y(0)=-6,\; y'(0)=2\)

    [exer:8.3.26] \(y''+4y=8\sin2t+9\cos t, \quad y(0)=1,\quad y'(0)=0\)

    [exer:8.3.27] \(y''-5y'+6y=10e^t\cos t, \quad y(0)=2,\quad y'(0)=1\)

    [exer:8.3.28] \(y''+2y'+2y=2t, \quad y(0)=2,\quad y'(0)=-7\)

    [exer:8.3.29] \(y''-2y'+2y=5\sin t+10\cos t, \quad y(0)=1,\; y'(0)=2\)

    [exer:8.3.30] \(y''+4y'+13y=10e^{-t}-36e^t, \quad y(0)=0,\; y'(0)=-16\)

    [exer:8.3.31] \(y''+4y'+5y=e^{-t}(\cos t+3\sin t), \quad y(0)=0,\quad y'(0)=4\)

    [exer:8.3.32] \(2y''-3y'-2y=4e^t, \quad y(0)=1,\; y'(0)=-2\)

    [exer:8.3.33] \(6y''-y'-y=3e^{2t}, \quad y(0)=0,\; y'(0)=0\)

    [exer:8.3.34] \(2y''+2y'+y=2t, \quad y(0)=1,\; y'(0)=-1\)

    [exer:8.3.35] \(4y''-4y'+5y=4\sin t-4\cos t, \quad y(0)=0,\; y'(0)=11/17\)

    [exer:8.3.36] \(4y''+4y'+y=3\sin t+\cos t, \quad y(0)=2,\; y'(0)=-1\)

    [exer:8.3.37] \(9y''+6y'+y=3e^{3t}, \quad y(0)=0,\; y'(0)=-3\)

    [exer:8.3.38] Suppose \(a,b\), and \(c\) are constants and \(a\ne0\). Let \[y_1={\cal L}^{-1}\left(as+b\over as^2+bs+c\right)\quad \text{and} \quad y_2={\cal L}^{-1}\left(a\over as^2+bs+c\right). \nonumber \]

    Show that \[y_1(0)=1,\quad y_1'(0)=0\quad \text{and} \quad y_2(0)=0,\quad y_2'(0)=1.\nonumber\]