8.7E: Constant Coefficient Equations with Impulses (Exercises)
( \newcommand{\kernel}{\mathrm{null}\,}\)
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=6e2t+2δ(t−1),y(0)=2,y′(0)=−6
2. y″+y′−2y=−10e−t+5δ(t−1),y(0)=7,y′(0)=−9
3. y″−4y=2e−t+5δ(t−1),y(0)=−1,y′(0)=2
4. y″+y=sin3t+2δ(t−π/2),y(0)=1,y′(0)=−1
5. y″+4y=4+δ(t−3π),y(0)=0,y′(0)=1
6. y″−y=8+2δ(t−2),y(0)=−1,y′(0)=1
7. y″+y′=et+3δ(t−6),y(0)=−1,y′(0)=4
8. y″+4y=8e2t+δ(t−π/2),y(0)=8,y′(0)=0
9. y″+3y′+2y=1+δ(t−1),y(0)=1,y′(0)=−1
10. y″+2y′+y=et+2δ(t−2),y(0)=−1,y′(0)=2
11. y″+4y=sint+δ(t−π/2),y(0)=0,y′(0)=2
12. y″+2y′+2y=δ(t−π)−3δ(t−2π),y(0)=−1,y′(0)=2
13. y″+4y′+13y=δ(t−π/6)+2δ(t−π/3),y(0)=1,y′(0)=2
14. 2y″−3y′−2y=1+δ(t−2),y(0)=−1,y′(0)=2
15. 4y″−4y′+5y=4sint−4cost+δ(t−π/2)−δ(t−π),y(0)=1,y′(0)=1
16. y″+y=cos2t+2δ(t−π/2)−3δ(t−π),y(0)=0,y′(0)=−1
17. y″−y=4e−t−5δ(t−1)+3δ(t−2),y(0)=0,y′(0)=0
18. y″+2y′+y=et−δ(t−1)+2δ(t−2),y(0)=0,y′(0)=−1
19. y″+y=f(t)+δ(t−2π),y(0)=0,y′(0)=1,
f(t)={sin2t,0≤t<π,0,t≥π.
20. y″+4y=f(t)+δ(t−π)−3δ(t−3π/2),y(0)=1,y′(0)=−1,
f(t)={1,0≤t<π/2,2,t≥π/2
Q8.7.2
21. y″+y=δ(t),y(0)=1,y′−(0)=−2
22. y″−4y=3δ(t),y(0)=−1,y′−(0)=7
23. y″+3y′+2y=−5δ(t),y(0)=0,y′−(0)=0
24. y″+4y′+4y=−δ(t),y(0)=1,y′−(0)=5
25. 4y″+4y′+y=3δ(t),y(0)=1,y′−(0)=−6
Q8.7.3
In Exercises 8.7.26-8.7.28, solve the initial value problem ay″h+by′h+cyh={0,0≤t<t01/h,t0≤t<t0+h,0,t≥t0+h,yh(0)=0,y′h(0)=0 where t0>0 and h>0. Then find w=L−1(1as2+bs+c) and verify Theorem 8.7.1 by graphing w and yh on the same axes, for small positive values of h.
26. y″+2y′+2y=fh(t),y(0)=0,y′(0)=0
27. y″+2y′+y=fh(t),y(0)=0,y′(0)=0
28. y″+3y′+2y=fh(t),y(0)=0,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,y(0)=y0,y′(0)=v0,
and that y can be written as
y=Re−ct/2mcos(ω1t−ϕ)
if the motion is underdamped. Suppose y(τ)=0. Find the impulse that would have to be applied to the object at t=τ 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,∞) on which the forcing function is zero.
- y″−y=∑∞k=1δ(t−k),y(0)=0,y′(0)=1
- y″+y=∑∞k=1δ(t−2kπ),y(0)=0,y′(0)=1
- y″−3y′+2y=∑∞k=1δ(t−k),y(0)=0,y′(0)=1
- y″+y=∑∞k=1δ(t−kπ),y(0)=0,y′(0)=0