Using Laplace transforms solve the initial value problem.
$$y''+4y
=
12\text{sin}(2t);
\qquad\qquad
y(\pi)=-3, \quad y'(\pi)=-3$$
I have begun with writing:
- $\mathcal{L} (y'') = s^2y(s) -s y(\pi) -y'(\pi)$ = $s^2y(s)+3s+3$
- $\mathcal{L} (4y) = 4y(s)$
- $\mathcal{L} (12sin(2t)) = \frac{24}{s^2+4}$
So I got:
$$y(s) = \frac{24}{(s^2+4)^2}-\frac{3s}{s^2+4}-\frac{3}{s^2+4}$$
Using inverse laplace transforms:
- $\mathcal{L}^{-1} (\frac{3s}{s^2+4}) = 3\,\text{cos}\,(2t)$
- $\mathcal{L}^{-1} (\frac{3}{s^2+4}) = \frac{3}{2}\,\text{sin}\,(2t)$
I think what I got until here is correct (at least I hope so). The problem is:
- $\mathcal{L}^{-1} (\frac{24}{(s^2+4)^2}) = \quad?$
I think I can do this by using convolution theorem but I am not sure. Taking $24$ aside, I have
$\frac{1}{s^2+4}$ and $\frac{1}{s^2+4}$. They are both $\frac{1}{2}\text{sin}(2t)$. I think I can write them by convolution problem as:
$$\frac{1}{4}\int_{0}^{t}\text{sin}(2t-2\tau)\,\text{sin}(2\tau)\,\text{d}\tau$$
But I don't know where to go from here. Any help would be appreciated.
Best Answer
HINT : $$ \mathcal{L}^{-1}\left[\frac{2a^3}{(s^2+a^2)^2}\right]=\sin at-at\cos at. $$