Finding the Inverse Laplace function.

Question

I have been given $F (s)=\frac {2(3s^2+1 )}{(s^2-1)^3} $ and I want to find $ f (t) $ so that $L(f (t))=F (s) $. Can handle powers in denominator till $2$ but here they will go till $6$ . The only hint given is use exponentials for $\sinh (at),\cosh (at) $ but that doesn't seem very helpful to begin with. Also partial fractions seem very tedious. Is there any simple way to do it or even a simple partial fractions will help. Thanks!

Answer 1

$$ \frac {2(3s^2+1 )}{(s^2-1)^3} = \frac{K_1}{(s^2-1)^3} + \frac{K_2}{(s^2-1)^2} + \frac{K_3}{(s^2-1)} \tag{1} $$

To find $K_1$, we multiply Eq(1) by $(s^2-1)^3$, we get

$$ 2(3s^2+1 ) = K_1 + K_2(s^2-1) + K_3 (s^2-1)^2 \tag{2} $$

In Eq(2), let $s=-1$, we obtain $K_1=8$.

To isolate $K_2$, we differentiate Eq(2) with respect to $s$, we get

$$ 2(6s) = K_2(2s) + K_3 4s(s^2-1) \tag{3} $$

In Eq(3), let $s=-1$, we obtain $K_2=6$.

Again, we differentiate Eq(3) with respect to $s$ and $K_2=6$, we get

$$ \begin{align} 12 &= (6)(2) + K_3 4(3s^2-1) \\ 12 &= 12K_3s^2 + 12 -4K_3 \end{align} $$

As a result, $12K_3=0 \rightarrow K_3 =0$ or $(12-4K_3)=12 \rightarrow K_3=0$.