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Mathematics LibreTexts

8.7E: Constant Coefficient Equations with Impulses (Exercises)

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    In Exercises [exer:8.7.1}– [exer:8.7.20} solve the initial value problem. Where indicated by , graph the solution.

    [exer:8.7.1] \(y''+3y'+2y=6e^{2t}+2\delta(t-1), \quad y(0)=2,\quad y'(0)=-6\)

    [exer:8.7.2] \(y''+y'-2y=-10e^{-t}+5\delta(t-1), \quad y(0)=7,\quad y'(0)=-9\)

    [exer:8.7.3] \(y''-4y=2e^{-t}+5\delta(t-1), \quad y(0)=-1,\quad y'(0)=2\)

    [exer:8.7.4] \(y''+y=\sin3t+2\delta(t-\pi/2), \quad y(0)=1,\quad y'(0)=-1\)

    [exer:8.7.5] \(y''+4y=4+\delta(t-3\pi), \quad y(0)=0,\quad y'(0)=1\)

    [exer:8.7.6] \(y''-y=8+2\delta(t-2), \quad y(0)=-1,\quad y'(0)=1\)

    [exer:8.7.7] \(y''+y'=e^t+3\delta(t-6), \quad y(0)=-1,\quad y'(0)=4\)

    [exer:8.7.8] \(y''+4y=8e^{2t}+\delta(t-\pi/2), \quad y(0)=8,\quad y'(0)=0\)

    [exer:8.7.9] \(y''+3y'+2y=1+\delta(t-1), \quad y(0)=1,\quad y'(0)=-1\)

    [exer:8.7.10] \(y''+2y'+y=e^t+2\delta(t-2), \quad y(0)=-1,\quad y'(0)=2\)

    [exer:8.7.11] \(y''+4y=\sin t+\delta(t-\pi/2), \quad y(0)=0,\quad y'(0)=2\)

    [exer:8.7.12] \(y''+2y'+2y=\delta(t-\pi)-3\delta(t-2\pi), \quad y(0)=-1,\quad y'(0)=2\)

    [exer:8.7.13] \(y''+4y'+13y=\delta(t-\pi/6)+2\delta(t-\pi/3), \quad y(0)=1,\quad y'(0)=2\)

    [exer:8.7.14] \(2y''-3y'-2y=1+\delta(t-2), \quad y(0)=-1,\quad y'(0)=2\)

    [exer:8.7.15] \(4y''-4y'+5y=4\sin t-4\cos t+\delta(t-\pi/2)-\delta(t-\pi), \quad y(0)=1,\quad y'(0)=1\)

    [exer:8.7.16] \(y''+y=\cos2t+2\delta(t-\pi/2)-3\delta(t-\pi), \quad y(0)=0,\quad y'(0)=-1\)

    [exer:8.7.17] \(y''-y=4e^{-t}-5\delta(t-1)+3\delta(t-2), \quad y(0)=0,\quad y'(0)=0\)

    [exer:8.7.18] \(y''+2y'+y=e^t-\delta(t-1)+2\delta(t-2), \quad y(0)=0,\quad y'(0)=-1\)

    [exer:8.7.19] \(y''+y=f(t)+\delta(t-2\pi), \quad y(0)=0,\quad y'(0)=1\),

    \(f(t)=\left\{\begin{array}{cl} \sin2t,&0\le t<\pi,\\[5pt]0,&t\ge \pi.\end{array}\right.\)

    [exer:8.7.20] \(y''+4y=f(t)+\delta(t-\pi)-3\delta(t-3\pi/2), \quad y(0)=1,\quad y'(0)=-1\),

    \(f(t)=\left\{\begin{array}{cl}1,&0\le t<\pi/2,\\[5pt]2,&t\ge \pi/2\end{array}\right.\)

    [exer:8.7.21] \(y''+y=\delta(t), \quad y(0)=1,\quad y_-'(0)=-2\)

    [exer:8.7.22] \(y''-4y=3\delta(t), \quad y(0)=-1,\quad y_-'(0)=7\)

    [exer:8.7.23] \(y''+3y'+2y=-5\delta(t), \quad y(0)=0,\quad y_-'(0)=0\)

    [exer:8.7.24] \(y''+4y'+4y=-\delta(t), \quad y(0)=1,\quad y_-'(0)=5\)

    [exer:8.7.25] \(4y''+4y'+y=3\delta(t), \quad y(0)=1,\quad y_-'(0)=-6\)

    [exer:8.7.26] \(y''+2y'+2y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)

    [exer:8.7.27] \(y''+2y'+y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)

    [exer:8.7.28] \(y''+3y'+2y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)


    Recall from Section 6.2 that the displacement of an object of mass \(m\) in a spring–mass system in free damped oscillation is

    \[my''+cy'+ky=0, \quad y(0)=y_0,\quad y'(0)=v_0,\]

    and that \(y\) can be written as


    if the motion is underdamped. Suppose \(y(\tau)=0\). Find the impulse that would have to be applied to the object at \(t=\tau\) to put it in equilibrium.

    [exer:8.7.30] Solve the initial value problem. Find a formula that does not involve step functions and represents \(y\) on each subinterval of \([0,\infty)\) on which the forcing function is zero.

    1. \(y''-y=\sum_{k=1}^\infty\delta(t-k), \quad y(0)=0,\quad y'(0)=1\)
    2. \(y''+y=\sum_{k=1}^\infty\delta(t-2k\pi), \quad y(0)=0,\quad y'(0)=1\)
    3. \(y''-3y'+2y=\sum_{k=1}^\infty\delta(t-k), \quad y(0)=0,\quad y'(0)=1\)
    4. \(y''+y=\sum_{k=1}^\infty\delta(t-k\pi), \quad y(0)=0,\quad y'(0)=0\)
    \(f(t)\) \(F(s)\)  
    1 \( 1\over s\) \((s > 0)\)
    \(t^n\) \( n!\over s^{n+1}\) \((s > 0)\)
    (\(n = \mbox{ integer } > 0\))    
    \(t^p,\; p > -1\) \( \Gamma (p+1) \over s^{(p+1)}\) \((s>0)\)
    \(e^{at}\) \( 1 \over s-a\) \((s > a)\)
    \(t^ne^{at}\) \( n! \over (s-a)^{n+1}\) \((s > 0)\)
    (\(n= \text{ integer } > 0\))    
    \(\cos \omega t\) \( s \over s^2+\omega^2}\) \((s > 0)\)
    \(\sin \omega t\) \( \omega \over s^2+\omega^2\) \((s > 0)\)
    \(e^{\lambda t} \cos \omega t\) \( s - \lambda \over (s-\lambda)^2+\omega^2\) \((s > \lambda)\)
    \(e^{\lambda t} \sin \omega t\) \( \omega \over (s-\lambda)^2+\omega^2\) \((s > \lambda)\)
    \(\cosh bt\) \( s \over s^2-b^2\) \((s > |b|)\)
    \(\sinh bt\) \( b \over s^2-b^2\) \((s > |b|)\)
    \(t \cos \omega t\) \( s^2-\omega^2 \over (s^2+\omega^2)^2\) \((s>0)\)
    \(t \sin \omega t\) \( 2\omega s \over (s^2+\omega^2)^2\) \((s>0)\)
    \(\sin \omega t -\omega t\cos \omega t\) \( 2\omega^3\over (s^2+\omega^2)^2\) \((s>0)\)
    \(\omega t - \sin \omega t\) \( \omega^3 \over s^2(s^2+\omega^2)^2\) \((s>0)\)
    \( 1 \over t} \sin \omega t }\) \(\arctan \left({\omega \over s}\right)\) \((s>0)\)
    \(e^{at}f(t)\) \(F(s-a)\)  
    \(t^kf(t)\) \( (-1)^kF^{(k)}(s)}}\)  
    \(f(\omega t)\) \( 1\over \omega} F\left({s \over \omega}\right), \quad \omega > 0}\)  
    \(u(t-\tau)\) \( e^{-\tau s} \over s\) \((s>0)\)
    \(u(t-\tau)f(t-\tau)\, (\tau > 0)\) \(e^{-\tau s}F(s)\)  
    \(\displaystyle{\int^t_o f(\tau)g(t-\tau)\, d\tau}\) \(F(s) \cdot G(s)\)  
    \(\delta(t-a)\) \(e^{-as}\) \((s>0)\)