# Laplace transform with initial value problem $y''+4y=12\sin(2t)$.

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.

• You might want to see my another answer below using convolution theorem. Apr 26, 2014 at 18:14

HINT : $$\mathcal{L}^{-1}\left[\frac{2a^3}{(s^2+a^2)^2}\right]=\sin at-at\cos at.$$
• Take $a=2$, then \begin{align} \mathcal{L}^{-1}\left[\frac{24}{(s^2+4)^2}\right]&=\frac32\mathcal{L}^{-1}\left[\frac{2\cdot 2^3}{(s^2+2^2)^2}\right] =\frac32(\sin 2t-2t\cos 2t). \end{align} Apr 26, 2014 at 17:09
• Do I need to memorize this for my exam? Because it was a previous exam question and its solution begins with $u = t - \pi$ and goes from there but I could not understand the solution. Is my solution way correct?
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 $$\mathcal{L}^{-1}\left[\frac{24}{(s^2+4)^2}\right]=24\cdot\frac14\int_{0}^{t}\sin(2t-2\tau)\,\sin2\tau\ d\tau=\frac32(\sin2t-2t\cos2t).$$