# Partial Fraction Decomposition for Laplace Transform

As part of trying to solve a differential equation using Laplace transforms, I have the fraction $\frac{-10s}{(s^2+2)(s^2+1)}$ which I am trying to perform partial fraction decomposition on so that I can do a inverse Laplace transform. When I try to work out this fraction, I get that $-10=0(A+B)$ since $s^1$ does not show up in the fraction. How does partial fraction decomposition work for a fraction like this?

I think you should work on the following, instead:

$$\frac{As+B}{s^2+2}+\frac{Cs+D}{s^2+1}$$

Or use the convolution theorem.

• Thank you, I can't believe I forgot about that. It's been about a year since I last did partial fractions. – Chris Crutchfield Mar 6 '15 at 3:28
• @ChrisCrutchfield: Thanks! I am I could help you. :-) – mrs Mar 6 '15 at 6:24

$\dfrac{-10s}{(s^2+2)(s^2+1)}=\dfrac{10s}{s^2+2}-\dfrac{10s}{s^2+1}$

Beacuse

$$\dfrac{-10s}{(s^2+2)(s^2+1)}=\frac{As+B}{s^2+2}+\frac{Cs+D}{s^2+1}$$

then

$$(A+C)s^3+(B+D)s^2+(A+2C)s+(B+2D)=-10s$$

$A+C=0$

$B+D=0$

$A+2C=-10$

$B+2D=0$

imply $A=10$, $B=0$, $C=-10$ and $D=0$.