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in the interval 0 < t < 5. The impulse at t = 5 produces a decaying oscillation that
persists indefinitely. The response is continuous at t = 5 despite the singularity in the forcing function at that point. However, the first derivative of the solution has a jump discontinuity at t = 5 and the second derivative has an infinite discontinuity there. This is required by the differential equation (17), since a singularity on one side of the equation must be balanced by a corresponding singularity on the other side.
FIGURE 6.5.3 Solution of the initial value problem (17), (18).
PROBLEMS In each of Problems 1 through 12 find the solution of the given initial value problem and draw
ã its graph.
Ó + 2 y = S(t — ï); y(0) = 1, y (0) = 0
y = S(t — ï) — S(t — 2n); y(0) = 0, /(0) = 0
;/ + 2 y = S(t — 5) + u10(t); y(0) = 0, y (0) = 1/2
r = —20S(t — 3); y(0) = 1, y(0) = 0
y + 3 y = sin t + S(t — 3n); y(0) = 0, y(0) = 0
y = S(t — 4n); y(0) = 1/2, /(0) = 0
r = S(t — 2n) cos t; y(0) = 0, y (0) = 1
> 8. Ó + 4y = 2S(t — ï/4); y(0) = 0, /(0) = 0
> 1. ó
> 2. ó
> 3. Ó
> 4. Ó
> 5. Ó
> 6. Ó
> 7. Ó
> 8. Ó
6.5 Impulse Functions
> 9. Ó' + y = ux/2(t) + 3S(t - 3n/2) - u7n(t); y(0) = 0, 7(0) = 0
> 10. 27' + 7 + 4y = S(t - n/6) sin t; y(0) = 0, 7(0) = 0
> 11. 7' + 27 + 2y = cos t + S(t - æ/2); y(0) = 0, /(0) = 0
> 12. /v - y = S(t - 1); y(0) = 0, /(0) = 0, y"(0) = 0, /"(0) = 0
> 13. Consider again the system in Example 1 in the text in which an oscillation is excited by
a unit impulse at t = 5. Suppose that it is desired to bring the system to rest again after exactly one cycle, that is, when the response first returns to equilibrium moving in the positive direction.
(a) Determine the impulse kS(t - t0) that should be applied to the system in order to
accomplish this objective. Note that k is the magnitude of the impulse and t0 is the time of
(b) Solve the resulting initial value problem and plot its solution to confirm that it behaves in the specified manner.
> 14. Consider the initial value problem
y" + Yy + y = S(t - 1), y(0) = 0, 7(0) = 0,
where Y is the damping coefficient (or resistance).
(a) Let y = j. Find the solution of the initial value problem and plot its graph.
(b) Find the time t1 at which the solution attains its maximum value. Also find the maximum value y1 of the solution.
(c) Let y = 4 and repeat parts (a) and (b).
(d) Determine how t1 and y1 vary as Y decreases. What are the values of t1 and y1 when Y = 0?
> 15. Consider the initial value problem
7' + Y7 + y = kS(t - 1), y(0) = 0, y(0) = 0,
where k is the magnitude of an impulse at t = 1 and y is the damping coefficient (or
(a) Let y = 2. Find the value of k for which the response has a peak value of 2; call this value k1.
(b) Repeat part (a) for y = 4.
(c) Determine how k1 varies as y decreases. What is the value of k1 when y = 0?
> 16. Consider the initial value problem
Ó + y = fk(t), y(0) = 0, y(0) = 0,
where fk(t) = [u4-k(t) - u4+k(t)]/2k with 0 < k < 1.
(a) Find the solution y = ô (t, k) of the initial value problem.
(b) Calculate jim ô (t, k) from the solution found in part (a).
(c) Observe that lim fk(t) = S(t - 4). Find the solution ô0(¥) of the given initial value problem with fk(t) replaced by S(t - 4). Is it true that ô0(´) = lim ô (t, k)?
(d) Plot ô (t, 1 /2), ô (t, 1 /4), and ô0(¥) on the same axes. Describe the relation between ô (t, k) and ô().
Problems 17 through 22 deal with the effect of a sequence of impulses on an undamped oscillator. Suppose that
7' + y = f(t), y(0) = 0, 7(0) = 0.
For each of the following choices for f (t):
(a) Try to predict the nature of the solution without solving the problem.
(b) Test your prediction by finding the solution and drawing its graph.
(c) Determine what happens after the sequence of impulses ends.
Chapter 6. The Laplace Transform
> 17. f(t) =J2 S(t - kn) > 18. f(t) =J2 (-1)k+1S(t - kn)