8.7E: Constant Coefficient Equations with Impulses (Exercises)

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] \(y''+3y'+2y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)

[exer:8.7.29]

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

\[y=Re^{-ct/2m}\cos(\omega_1t-\phi)\]

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	\( 1\over s\)	\((s > 0)\)
\(t^n\)	\( n!\over s^{n+1}\)	\((s > 0)\)
(\(n = \mbox{ integer } > 0\))
\(t^p,\; p > -1\)	\( \Gamma (p+1) \over s^{(p+1)}\)	\((s>0)\)
\(e^{at}\)	\( 1 \over s-a\)	\((s > a)\)
\(t^ne^{at}\)	\( n! \over (s-a)^{n+1}\)	\((s > 0)\)
(\(n= \text{ integer } > 0\))
\(\cos \omega t\)	\( s \over s^2+\omega^2}\)	\((s > 0)\)
\(\sin \omega t\)	\( \omega \over s^2+\omega^2\)	\((s > 0)\)
\(e^{\lambda t} \cos \omega t\)	\( s - \lambda \over (s-\lambda)^2+\omega^2\)	\((s > \lambda)\)
\(e^{\lambda t} \sin \omega t\)	\( \omega \over (s-\lambda)^2+\omega^2\)	\((s > \lambda)\)
\(\cosh bt\)	\( s \over s^2-b^2\)	\((s > |b|)\)
\(\sinh bt\)	\( b \over s^2-b^2\)	\((s > |b|)\)
\(t \cos \omega t\)	\( s^2-\omega^2 \over (s^2+\omega^2)^2\)	\((s>0)\)

\(t \sin \omega t\)	\( 2\omega s \over (s^2+\omega^2)^2\)	\((s>0)\)
\(\sin \omega t -\omega t\cos \omega t\)	\( 2\omega^3\over (s^2+\omega^2)^2\)	\((s>0)\)
\(\omega t - \sin \omega t\)	\( \omega^3 \over s^2(s^2+\omega^2)^2\)	\((s>0)\)
\( 1 \over t} \sin \omega t }\)	\(\arctan \left({\omega \over s}\right)\)	\((s>0)\)
\(e^{at}f(t)\)	\(F(s-a)\)
\(t^kf(t)\)	\( (-1)^kF^{(k)}(s)}}\)
\(f(\omega t)\)	\( 1\over \omega} F\left({s \over \omega}\right), \quad \omega > 0}\)
\(u(t-\tau)\)	\( e^{-\tau s} \over s\)	\((s>0)\)
\(u(t-\tau)f(t-\tau)\, (\tau > 0)\)	\(e^{-\tau s}F(s)\)
\(\displaystyle{\int^t_o f(\tau)g(t-\tau)\, d\tau}\)	\(F(s) \cdot G(s)\)
\(\delta(t-a)\)	\(e^{-as}\)	\((s>0)\)