- #1

lospollos90

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## Homework Statement

From Boyce and DiPrima's Elementary Differential Equations (9th Ed.), Section 6.5, Problem 13:

Consider again the system in Example 1:

[itex]2y'' + y' + 2y = \delta(t-5); y(0)=y'(0)=0[/itex]

(Solution in text: [itex]y=\frac{2}{\sqrt{15}}u_{5}(t)e^{\frac{-(t-5)}{4}}sin\frac{\sqrt{15}}{4}(t-5)[/itex])

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 [itex]k \delta(t-t_{0})[/itex] that should be applied to the system in order to accomplish this objective. Note that k is the magnitude of the impulse and [itex]t_{0}[/itex] is the time of its application.

(b) Solve the resulting initial value problem and plot its solution to confirm that it behaves in the specified manner.

## Homework Equations

The solution to part (a) given at the back of the text is

[itex]-e^{\frac{-T}{4}}\delta(t-5-T), T=\frac{8\pi}{\sqrt{15}}[/itex]

## The Attempt at a Solution

Honestly, I'm completely lost here.

I get why T must equal 8pi/sqrt(15) - we need the stuff inside sin(blah) to equal 2pi since that will tell us when one cycle is completed; this happens when t-5=8pi/sqrt(15). But I'm confused as to how that gets us to the answer provided above and, moreover, how to proceed with part (b).

Any help would be appreciated.