How to find the Laplace inversion of $\frac{p}{p^4+4}$?

Question

How to calculate $$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}$$ where $\mathscr{L}$ is the Laplace transform operator? I thing need to apply some partial fraction first but am unable to work it out. Any idea or further help will be very good to me.

Answer 1

$p^4+4$ has 4 roots: $p=\pm 1 \pm i$, so it can be written as: $(p-1-i)(p-1+i)(p+1-i)(p+1+i)=(p^2-2p+2)(p+2p+2)$, so: $$\frac{p}{p^4+4}=\frac{p}{(p^2-2p+2)(p+2p+2)}=\frac{Ap+B}{p^2-2p+2}+\frac{Cp+D}{p^2+2p+2}$$ The solution is $A=0, B=\frac{1}{4}, C=0, D=-\frac{1}{4}$ (I can show more detail of the computation if you need it), so: $$\frac{p}{p^4+4}=\frac{\frac{1}{4}}{p^2-2p+2}-\frac{\frac{1}{4}}{p^2+2p+2}=\frac{\frac{1}{4}}{(p-1)^2+1}-\frac{\frac{1}{4}}{(p+1)^2+1}$$ We know that $\mathscr{L}\{e^{ax}f(x)\}=F(p-a), \mathscr{L}\{\sin{(ax)}\}=\frac{a}{p^2+a^2}, \mathscr{L}\{\cos{(ax)}\}=\frac{p}{p^2+a^2}$, so: $$\mathscr{L}^{-1}\left\{\frac{\frac{1}{4}}{(p-1)^2+1} \right\}=\frac{1}{4}e^{x}\sin{x}$$ $$\mathscr{L}^{-1}\left\{\frac{\frac{1}{4}}{(p+1)^2+1} \right\}=\frac{1}{4}e^{-x}\sin{x}$$ Finaly: $$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}=\frac{1}{4}e^{x}\sin{x}-\frac{1}{4}e^{-x}\sin{x}$$ $$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}=\frac{1}{4}\sin{x}(e^{x}-e^{-x})$$ $$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}=\frac{1}{4}\sin{x}(2\sinh{x})$$ $$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}=\frac{1}{2}\sin{x}\sinh{x}$$