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.
This is another answer. Your approach using convolution theorem is correct. $$ \begin{align} \int_{0}^{t}\sin(2t-2\tau)\,\sin2\tau\ d\tau&=\int_{0}^{t}(\sin2t\cos2\tau-\cos2t\sin2\tau)\sin2\tau\ d\tau\\ &=\sin2t\int_{0}^{t}\cos2\tau\sin2\tau\ d\tau-\cos2t\int_{0}^{t}\sin^22\tau\ d\tau\\ &=\sin2t\int_{0}^{t}\frac12\sin4\tau\ d\tau-\cos2t\int_{0}^{t}\frac12(1-\cos4\tau)\ d\tau\\ &=-\frac18\sin2t\left.\cos4\tau\right|_0^t-\frac12\cos2t\left[\tau-\frac14\sin4\tau\right]_0^t\\ &=-\frac18\sin2t\cos4t+\frac18\sin2t-\frac t2\cos2t+\frac18\cos2t\sin4t\\ &=\frac18\sin(4t-2t)+\frac18\sin2t-\frac t2\cos2t\\ &=\frac14\sin2t-\frac t2\cos2t\\ &=\frac14(\sin2t-2t\cos2t). \end{align} $$ Therefore $$ \mathcal{L}^{-1}\left[\frac{24}{(s^2+4)^2}\right]=24\cdot\frac14\int_{0}^{t}\sin(2t-2\tau)\,\sin2\tau\ d\tau=\frac32(\sin2t-2t\cos2t). $$