I am studying Laplace transforms right now and got stuck at this step that involves a weird partial fraction decomposition. It looks like the instructor skipped a bunch of steps and assigned numerators to a bunch of the fractions without assigning them dummy variables. Any idea how he got to this step so quickly?
I tried doing this the standard way (i.e. breaking each polynomial in the denominator into its own fraction, assigning variables for each numerator, and trying to solve), but it got messy very quickly.
laplace transform up to the part where I got stuck
$$ X=\frac{2}{s^4-1}\left(e^{-s}\cdot\frac{1}{s}-e^{-2s}\cdot\frac{1}{s}\right)=\frac{2}{\left(s^4-1\right)s}\left(e^{-s}-e^{-2s}\right)=2\cdot\frac{1}{\left(s^2+1\right)\left(s+1\right)\left(s-1\right)s}\left(e^{-s}-e^{-2s}\right)=2\left(\frac{As+B}{s^2+1}+\frac{\left(\frac{1}{4}\right)}{\left(s+1\right)}+\frac{\left(\frac{1}{4}\right)}{s-1}+\frac{\left(-1\right)}{s}\right)\left(e^{-s}-e^{-2s}\right) $$
Let's start from the general form of the decomposition of
$${1\over s(s-1)(s+1)(s^2+1)}={a\over s}+{b\over s-1}+{c\over s+1}+{ds+e\over s^2+1}$$
Now multiply both sides by $s$ then set $s=0$ you get $a=-1$
Multiply both sides by $s-1$ then set $s=1$ you get $b={1\over 4}$
Multiply both sides by $s+1$ then set $s=-1$ you get $c={1\over 4}$
Finally multiply by $s^2+1$ then set $s=i$ you get
$${1\over i(i^2-1)}=di+e={i\over 2}$$
And this leads to $d=1/2$ and $e=0$