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