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Diff Eq Question - Impulse Functions

  • #1

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.
 

Answers and Replies

  • #2
vela
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The total response of the system is the sum of the individual responses to each impulse. This is just the principle of superposition. Write down the response to ##k\delta(t-t_0)## and add it to the response from the first impulse and set the sum equal to 0.
 

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