I'm working through an example of an inverse Laplace transform, starting with:
$$\hat{F}(s)=\frac{1}{(s^2+1)(s^2+2)}$$
From this, I know to rewrite it using partial fractions, to get $\frac{1}{(s^2+1)(s^2+2)}=\frac{A}{s^2+1}+\frac{B}{s^2+2}$, which I then simplify as:
$$1=A(s^2+2)+B(s^2+1)$$
I know from the example that $A=1$ and $B=-1$, but I cannot for the life of me find the process behind this step. If somebody could help me understand it, I'd be very grateful.
$$\newcommand{recp}[1]{\frac{1}{#1}}$$ As Paul said,we need four unknowns instead of two, but the linear coefficients will be $0$ as we will see shortly $$\hat{F}(s)=\recp{(s^2+1)(s^2+2)}=\frac{As+B}{s^2+1}+\frac{Cs+D}{s^2+2}$$ Multiplying both sides by the left denominator: $$1=(As+B)(s^2+2)+(Cs+D)(s^2+1)=As^3+2As+Bs^2+2B+Cs^3+Cs+Ds^2+D$$ Since this is for all $s$, we can separate the coefficients: $$\begin{cases} (1)A+C=0\\(2)B+D=0\\(3)2A+C=0\\(4)2B+D=1 \end{cases}$$ Subtracting the equations and plugging in: $$(4)-(2)=B=1\\ 1+D=0\\D=-1\\(3)-(1)=A=0\\C=0$$ Therefore $$\hat{F}(s)=\recp{s^2+1}-\recp{s^2+2}$$