I have this problem: $F(s)=\frac{a_{1}s+a_{2}}{s^{2}(s^{2}+12s+37)}$
I thought in Taylor's expansion of a function f(s) in s=0:
$f(s)= f(0)+f'(0)\cdot s+f''(0)\cdot s^{2}+\cdots$ and then I defined:
$f(s)=\frac{a_{1}s+a_{2}}{(s^{2}+12s+37)}$
Then: $F(s)=\frac{f(s)}{s^{2}}= \frac{f(0)}{s^{2}}+\frac{f'(0)}{s}+f''(0)\cdot s^{2}+\cdots$
But by simple fractions the function F(s) should look like this:
$F(s)=\frac{A}{s^{2}}+\frac{B}{s}+\frac{Cs+D}{(s^{2}+12s+37)}$
So $A=f(0)$ and $B=f'(0)$
So, for example with $a_{1}=25$ and $a_{2}=37$, by the "Taylor Method" I get $A= 1$ and $B=13/37$.
Is there a way using the Residues theorem or the Taylor expantion or another trick to obtain C and D getting rid of making distribution of the factors? With the distribution I got $C=\frac{-13}{37}$ and $D=\frac{-14}{37}$.
If this hint of the exercise helps: I'm trying to write that in this way to calculate de inverse of the Laplace transform. Thanks.
If you want to use residues, you could say $\frac {Cs + D}{s^2 + 12s + 37} = \frac {Q}{s+6+i} + \frac {R}{s+6-i}$
In which case $Q$ is the residue at $-6-i$ and $R$ is the residue at $-6+i$ and $D = i(Q-R)$ and $C = 6(Q+R)$
But ultimately I think it is easier to say
$Bs^3 + Cs^3 = 0\\ As^2 + 12Bs^2 + Ds^2 = 0$
And for the inverse Laplace transform, you are going to want to say $s^2 + 12s + 37 = (s+6)^2 + 1$