8.7E: Constant Coefficient Equations with Impulses (Exercises)
- Page ID
- 18508
Q8.7.1
In Exercises 8.7.1-8.7.20 solve the initial value problem. Graph the solution for Exercises 8.7.2, 8.7.4, 8.7.9, and 8.7.19.
1. \(y''+3y'+2y=6e^{2t}+2\delta(t-1), \quad y(0)=2,\quad y'(0)=-6\)
2. \(y''+y'-2y=-10e^{-t}+5\delta(t-1), \quad y(0)=7,\quad y'(0)=-9\)
3. \(y''-4y=2e^{-t}+5\delta(t-1), \quad y(0)=-1,\quad y'(0)=2\)
4. \(y''+y=\sin3t+2\delta(t-\pi/2), \quad y(0)=1,\quad y'(0)=-1\)
5. \(y''+4y=4+\delta(t-3\pi), \quad y(0)=0,\quad y'(0)=1\)
6. \(y''-y=8+2\delta(t-2), \quad y(0)=-1,\quad y'(0)=1\)
7. \(y''+y'=e^t+3\delta(t-6), \quad y(0)=-1,\quad y'(0)=4\)
8. \(y''+4y=8e^{2t}+\delta(t-\pi/2), \quad y(0)=8,\quad y'(0)=0\)
9. \(y''+3y'+2y=1+\delta(t-1), \quad y(0)=1,\quad y'(0)=-1\)
10. \(y''+2y'+y=e^t+2\delta(t-2), \quad y(0)=-1,\quad y'(0)=2\)
11. \(y''+4y=\sin t+\delta(t-\pi/2), \quad y(0)=0,\quad y'(0)=2\)
12. \(y''+2y'+2y=\delta(t-\pi)-3\delta(t-2\pi), \quad y(0)=-1,\quad y'(0)=2\)
13. \(y''+4y'+13y=\delta(t-\pi/6)+2\delta(t-\pi/3), \quad y(0)=1,\quad y'(0)=2\)
14. \(2y''-3y'-2y=1+\delta(t-2), \quad y(0)=-1,\quad y'(0)=2\)
15. \(4y''-4y'+5y=4\sin t-4\cos t+\delta(t-\pi/2)-\delta(t-\pi), \quad y(0)=1,\quad y'(0)=1\)
16. \(y''+y=\cos2t+2\delta(t-\pi/2)-3\delta(t-\pi), \quad y(0)=0,\quad y'(0)=-1\)
17. \(y''-y=4e^{-t}-5\delta(t-1)+3\delta(t-2), \quad y(0)=0,\quad y'(0)=0\)
18. \(y''+2y'+y=e^t-\delta(t-1)+2\delta(t-2), \quad y(0)=0,\quad y'(0)=-1\)
19. \(y''+y=f(t)+\delta(t-2\pi), \quad y(0)=0,\quad y'(0)=1\),
\(f(t)=\left\{\begin{array}{cl} \sin2t,&0\le t<\pi,\\[4pt]0,&t\ge \pi.\end{array}\right.\)
20. \(y''+4y=f(t)+\delta(t-\pi)-3\delta(t-3\pi/2), \quad y(0)=1,\quad y'(0)=-1\),
\(f(t)=\left\{\begin{array}{cl}1,&0\le t<\pi/2,\\[4pt]2,&t\ge \pi/2\end{array}\right.\)
Q8.7.2
21. \(y''+y=\delta(t), \quad y(0)=1,\quad y_-'(0)=-2\)
22. \(y''-4y=3\delta(t), \quad y(0)=-1,\quad y_-'(0)=7\)
23. \(y''+3y'+2y=-5\delta(t), \quad y(0)=0,\quad y_-'(0)=0\)
24. \(y''+4y'+4y=-\delta(t), \quad y(0)=1,\quad y_-'(0)=5\)
25. \(4y''+4y'+y=3\delta(t), \quad y(0)=1,\quad y_-'(0)=-6\)
Q8.7.3
In Exercises 8.7.26-8.7.28, solve the initial value problem \[ay_{h}'' + by_{h}'+cy_{h}=\left\{\begin{array}{ll} {0,}&{0\leq t<t_{0}}\\{1/h, }&{t_{0}\leq t< t_{0} +h, }\\{0,}&{t\geq t_{0}+h, } \end{array} \right. \quad y_{h}(0)=0, y_{h}'(0)=0\nonumber \] where \(t_{0}>0\) and \(h>0\). Then find \[w=\cal{L}^{-1}\left(\frac{1}{as^{2}+bs+c} \right) \nonumber \] and verify Theorem 8.7.1 by graphing \(w\) and \(y_{h}\) on the same axes, for small positive values of \(h\).
26. \(y''+2y'+2y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)
27. \(y''+2y'+y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)
28. \(y''+3y'+2y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)
Q8.7.4
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
\[my''+cy'+ky=0, \quad y(0)=y_0,\quad y'(0)=v_0,\nonumber \]
and that \(y\) can be written as
\[y=Re^{-ct/2m}\cos(\omega_1t-\phi)\nonumber\]
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.
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.
- \(y''-y=\sum_{k=1}^\infty\delta(t-k), \quad y(0)=0,\quad y'(0)=1\)
- \(y''+y=\sum_{k=1}^\infty\delta(t-2k\pi), \quad y(0)=0,\quad y'(0)=1\)
- \(y''-3y'+2y=\sum_{k=1}^\infty\delta(t-k), \quad y(0)=0,\quad y'(0)=1\)
- \(y''+y=\sum_{k=1}^\infty\delta(t-k\pi), \quad y(0)=0,\quad y'(0)=0\)