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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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How did you get this result? –  Mhenni Benghorbal Dec 5 '12 at 6:57
    
@MhenniBenghorbal which result? My result or the correct answer? –  Q.matin Dec 5 '12 at 7:20
    
The correct answer. –  Mhenni Benghorbal Dec 5 '12 at 7:26
    
@MhenniBenghorbal the back of my book. I am using "differential equations by James brannan". Why, is it wrong? –  Q.matin Dec 5 '12 at 7:27
    
It is a correct answer. –  Mhenni Benghorbal Dec 5 '12 at 7:49
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1 Answer 1

up vote 2 down vote accepted

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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Thanks a lot copper, but I am confused on how you got $x_3(t) = x_2(t-\pi)u(t-\pi)$. How did you get that? –  Q.matin Dec 5 '12 at 7:19
    
Notice that $\hat{x}_3(s) = e^{-\pi s} \hat{x}_2(s)$. Then the time-shift property gives the result you asked about. Does this clarify? –  copper.hat Dec 5 '12 at 7:53
    
Yes it does, thanks a lot!! –  Q.matin Dec 6 '12 at 4:22
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