I'm trying to get the inverse Laplace transform of the following expression:
\begin{equation} \begin{split} 2 F(s)=\frac{-\alpha ^2 A}{\left(-B -\alpha ^2 A -s^{2}\right)}F(s+2 \alpha )+\frac{-\alpha ^2 A}{\left(-B -\alpha ^2 A -s^{2}\right)} F(s-2 \alpha)\\ +\frac{-\left((s+4\alpha)^2+B\right)}{\left(-B -\alpha ^2 A -s^{2}\right)}F(s+4 \alpha )+\frac{-\left((s-4\alpha)^2+B\right)}{\left(-B -\alpha ^2 A -s^{2}\right)}F(s-4 \alpha ) \end{split} \end{equation}
being $A,B,\alpha \in \mathbb{R}$
It seems to me that the Inverse laplace transform of this expression does not exist, the terms $F(s),F(s-4 \alpha ),F(s+4 \alpha )$, etc, seems to cancel each other by the Shifting Theorem of the Laplace transform:
$\displaystyle{ \mathcal{L}\{ e^{-at} f(t) \} = F(s+a) }$
And at the end we get some kind of expression like
$0=0$
It seems to me that I am doing something wrong, since the Inverse Laplace transform seems to cancel itself. Do you have any ideas on how could I try to get the inverse transform to get $f(t)$?
A hint I suspect, it could be somehow related to the Hypergeometric differential equation.
Thanks in advance.