Laplace Transform Damp Harmonic Motion

Question

http://gyazo.com/19d18f085731c6dbc304fefdaece4f3c.png

I'm currently on (a) where so far I have gotten;

$ y'' + 2y' + 5y = f(t) $

Using Laplace transforms, I get;

$ Y(s)$ = $ F(s) + s+2\over(s^2 + 2s + 5) $

Completing square gives me;

$ F(s) + s + 2\over (s+1)^2 + 4) $ and subbing $ S = s + 1 $ and then splitting them so I can inverse Laplace them gets me;

$ F(s)\over(S^2 + 2^2) $+ $ S\over(S^2 + 2^2) $ + $ 1\over(S^2 + 2^2) $

now I don't know how to apply an inverse laplace transform to the last term.

Answer 1

We are given the differential equation:

$$\tag 1 y'' + 2y' + 5y = f(t), ~ y(0) = 1, y'(0) = 0 $$

Taking the Laplace Transform of $(1)$ and simplifying, we end up with the system:

$$Y(s) = \dfrac{1}{(s+1)^2 + 2^2} \left(F(s) + s + 2 \right)$$

We need to find the Inverse Laplace Transform for those three terms on the RHS.

For the general $F(s)$ term, we need to make use of the Convolution Theorem.

The solution will be:

$$y(t) = (e^{-t} \sin t \cos t)*f(t)) + \left( \dfrac{1}{2}e^{-t}(2 \cos 2t - \sin 2t) \right) + (e^{-t} \sin 2t)$$

Thus:

$$y(t) = \int_0^t (e^{-(t-v)} \sin(t-v) \cos(t-v))*f(v)~dv + \dfrac{1}{2}e^{-t}(2 \cos 2t - \sin 2t) + e^{-t} \sin 2t$$