Find the inverse laplace transform of $F(s) = \frac{a(s^{2}-2a^{2})}{s^{4}+4a^{4}}$

Question

Find the inverse laplace transform of $F(s) = \frac{a(s^{2}-2a^{2})}{s^{4}+4a^{4}}$

I factored denominator by completing square but how do i proceed next .

Thanks

Answer 1

Since: $$\mathcal{L}\left(\sinh(at)\right) = \frac{a}{s^2-a^2}\tag{1}$$ by computing the partial fraction decomposition of $F(s)$ it follows that: $$\mathcal{L}^{-1}(F(s)) = \cos(at)\sinh(at).\tag{2}$$

Answer 2

By expressing the denominator of your function as a difference of squares, where the constant term is complex, you have $$F(s) = \frac{a(s^{2}-2a^{2})}{s^{4}-(-4a^{4})} = \frac{a(s^{2}-2a^{2})}{(s^{2}-2a^{2}i)(s^{2}+2a^{2}i)}$$ A partial decomposition will yield $$\begin{align}F(s)&=\frac{a(1+i)}{2(s^2-2a^2i)}+\frac{a(1-i)}{2(s^2+2a^2i)}\\&=\frac{\sqrt{2a^2i}}{2(s^2-2a^2i)}+\frac{\sqrt{2a^2i}}{2i(s^2+2a^2i)}\\&=\frac{g}{2(s^2-g^2)}+\frac{g}{2i(s^2+g^2)}\end{align}$$ where complex constant $g=\sqrt{2a^2i}=a\sqrt{2}(1+i)$. Now $F(s)$ is in a form where the inverse Laplace transform of each term is more straightforward to evaluate, as $$\mathcal{L}^{-1}\left(\frac{g}{s^2-g^2}\right)=\sinh(gt)\\\mathcal{L}^{-1}\left(\frac{g}{s^2+g^2}\right)=\sin(gt)$$