Having problem in Laplace I have a question:
Find the Laplace transform of
$$ \frac{1}{s^2(s^2+4)}. $$
Now I also do have its solution. I wanted to know why do we Laplace inverse integral method (Please correct me if the name of the method is wrong) and not just simply take its Laplace inverse (i.e use linearity principle and find the Laplace of $\frac{1}{s^2}$ and $\frac{1}{s^2+4}$ separately and multiply them)?
 A: Use partial fractions:
$$
\frac{1}{s^2(s^2+4)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{s^2 + 4}
$$
Find $A$, $B$, $C$, $D$ and then apply linearity.
Linearity says that you can find inverse Laplace transforms of the terms in a sum and then add them, but it doesn't say you can find inverse Laplace transforms of the factors in a product and then multiply them.
A: Because that's not how linearity works. The definition of a map $T: V \to W$ being linear is that $T(a_1v_1 + a_2v_2) = a_1T(v_1) + a_2T(v_2)$. Nowhere does it speak anything about $T(v_1v_2)$ - in fact "$v_1v_2$" doesn't even have a meaningful definition for most vector spaces.
A: Partial Fractions
$$ F(s)=\frac{1}{s^2(s^2+4)} =  \frac{K_{11}}{s^2} + \frac{K_{12}}{s}  + \frac{k_2}{s^2 + 4} $$
$$ K_{11}=  F(s)\cdot s^2 = \frac{1}{s^2+4}\text{ (when $s=0$) }\rightarrow k_{11}=\frac{1}{4}  $$
$$ k_{12}=\frac{d}{ds}[ F(s)\cdot s^2]\text{ (when $s=0$) }\rightarrow k_{12}= 0 $$
$$ k_2=F(s)\cdot(s^2+4)\text{ (when $s=2i$) }\rightarrow k_2=\frac{-1}{4} $$
So :
$$ f(t)=\mathcal{L^{-1}}\left[\frac{\frac{1}{4}}{s^2}  + \frac{\frac{-1}{4}}{s^2 + 4}\right]  $$
$$ f(t)=\frac{t}{4}-\frac{\sin(2t)}{8} $$
