I'm trying to do a partial fraction expansion on $$Y = n \cdot \frac{e^{-pt_{0}}}{(p+n)(p^2+\omega^2)}$$
which gives $A(p^2+\omega^2)+(Bp+C)(p+n) = 1$ which implies
$$A = -B$$ $$Bpn=0\rightarrow B=0\rightarrow A=0$$
which is incorrect. Any help with what I'm missing would be greatly appreciated
The condition
$$ A(p^2 + \omega^2) + (B p + C)(p + n ) = 1 $$
leads to the equations
\begin{eqnarray} A + B &=& 0 \\ B n + C &=& 0 \\ A \omega^2 + C n &=& 1 \end{eqnarray}
whose solution is
\begin{eqnarray} A &=& \frac{1}{\omega^2 + n^2} = -B \\ C &=& \frac{n}{\omega^2 + n^2} \end{eqnarray}
So the fraction is
$$ Y = \frac{n e^{-pt_0}}{\omega^2 + n^2}\left[ \frac{1}{p + n} + \frac{n-p}{p^2 + \omega^2} \right] $$
