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Find the solution of the initial value problem.

$y'' +2y' +2y = \delta(t - \pi)$ with initial conditions $y(0) =1, y'(0) =0$.

What I did was take the Laplace and got:

$(s^2Y(s) - s) + 2(sY(s) -2) +2Y(s) = e^{-\pi s}$, then simplying and solving for $Y(s)$

$Y(s) = \frac{e^{-\pi s} +s + 2}{s^2 + 2s +2}$

Now taking the inverse I got:

$(e^{-\pi} + s +2) \dot\ e^{-t}sint$

But the answer is:

$y = e^{-t}cost + e^{-t}sint + u_\pi(t)e^{-(t-\pi )}sin(t-\pi )$

How did they get that?

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    It is a correct answer.2012-12-05

1 Answers 1

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Rewrite $\hat{y}$ as $\hat{y}(s) = \frac{e^{-\pi s}+(s+1)+1}{(s+1)^2+1} = \frac{e^{-\pi s}}{(s+1)^2+1} + \frac{(s+1)}{(s+1)^2+1} + \frac{1}{(s+1)^2+1}$.

I am assuming that $t \geq 0$ in the following:

If $x_1(t) = e^{-t} \cos t$, then $\hat{x}_1(s) = \frac{(s+1)}{(s+1)^2+1}$.

If $x_2(t) = e^{-t} \sin t$, then $\hat{x}_2(s) = \frac{1}{(s+1)^2+1}$.

Using time-shifting, if $x_3(t) = x_2(t-\pi)u(t-\pi)$, then $\hat{x}_3(s) = \frac{e^{-\pi s}}{(s+1)^2+1}$.

Consequently, you have $y(t) = x_1(t)+x_2(t)+x_3(t)$, or explicitly,

$y(t) = e^{-t} \cos t + e^{-t} \sin t + e^{-(t-\pi)} \sin (t-\pi) u(t-\pi)$

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    Yes it does, thanks a lot!!2012-12-06