Unsure of Inverse Laplace Transform for B/(A-s^2) I'm currently solving a differential equation via Laplace transforms and am unsure on how to apply the reverse transform to one of the terms. I have  $L[F(x)]=f(s)$ and currently trying to perform:
$$L^{-1}\left[\frac{B}{A-s^2}\right]$$
I've tried to take the negative of this and try and match it with one in a table, but have been unsuccessful. I know the answer is 
$$\frac{B e^{-\sqrt{A}}(e^{2\sqrt{A}}-A)}{2\sqrt{A}}$$
but cannot get the workings correctly (determined the answer from mathematica, but cant get the steps right).
Can anyone help with which transform to use in the table?
 A: We have three possibilities, depending on the sign of $A$. First, let's assume $A>0.$ Then we have
\begin{align*}
\mathcal{L}^{-1}\left(\frac{B}{A-s^2}\right)&=B\,\mathcal{L}^{-1}\left(\frac{1}{\left(\sqrt{A}-s\right)\left(\sqrt{A}+s\right)}\right) \\
&=B\,\mathcal{L}^{-1}\left(\frac{1}{2\sqrt{A}}\left(\frac{1}{\sqrt{A}+s}+\frac{1}{\sqrt{A}-s}\right)\right) \\
&=\frac{B}{2\sqrt{A}}\,\mathcal{L}^{-1}\left(\frac{1}{\sqrt{A}+s}+\frac{1}{\sqrt{A}-s}\right)\\
&=\frac{B}{2\sqrt{A}}\left(e^{-\sqrt{A}\,t}-e^{\sqrt{A}\,t}\right)\\
&=-\frac{B}{\sqrt{A}}\,\sinh\left(\sqrt{A}\,t\right).
\end{align*}
You probably have unit step functions in there, technically, though you might not need them. 
Now let's suppose $A<0.$ Let $C^2=|A|.$ So $C=\sqrt{|A|}.$ Then we have
\begin{align*}
\mathcal{L}^{-1}\left(\frac{B}{A-s^2}\right)&=\mathcal{L}^{-1}\left(\frac{B}{-|A|-s^2}\right) \\
&=-\frac{B}{C}\,\mathcal{L}^{-1}\left(\frac{C}{C^2+s^2}\right) \\
&=-\frac{B}{C}\,\sin(C\,t) \\
&=-\frac{B}{\sqrt{|A|}}\,\sin\left(\sqrt{|A|}\,t\right).
\end{align*}
Finally, suppose $A=0.$ Then you have merely
\begin{align*}
\mathcal{L}^{-1}\left(\frac{B}{A-s^2}\right)&=-B\,\mathcal{L}^{-1}\left(\frac{1}{s^2}\right) \\
&=-B\,t.
\end{align*}
So, your final answer can be written this way:
$$\mathcal{L}^{-1}\left(\frac{B}{A-s^2}\right)=\begin{cases}-\dfrac{B}{\sqrt{A}}\,\sinh\left(\sqrt{A}\,t\right),\quad &A>0\\ -B\, t,\quad &A=0 \\-\dfrac{B}{\sqrt{-A}}\,\sin\left(\sqrt{-A}\,t\right),\quad &A<0
\end{cases}. $$
A: $$f(t)=\mathcal{L^{-1}}(B/(A-s^2))$$
$$f(t)=\frac {B}{2\sqrt A}\mathcal{L^{-1}}(\frac 1{(\sqrt A-s)}+\frac 1{(\sqrt A+s)})$$
$$f(t)=\frac {B}{2\sqrt A}\mathcal{L^{-1}}(-\frac 1{(s-\sqrt A)}+\frac 1{(s+\sqrt A)})$$
Look at the table of lapalce transform
$$f(t)=\frac {B}{2\sqrt A}(-e^{\sqrt At}+e^{-\sqrt At})$$
$$f(t)=-\frac {B}{\sqrt A}\sinh(\sqrt At)$$
