How can I find inverse z transform of $$X(z)=\frac{5}{z^{2}-z-6}$$
What I did:
first i factored denominator and i got (z+2)(z-3), now we get A(-2^{n}) + b(3^{n}). To get A and B i used Partial Fraction Decomposition and got A=-1 and B=1. But Wolframalpha gave me another answer! $$=\frac{1}{6}(3(-2^{n}+2*3^{n}))$$ what i did wrong?
Doing partial fractions as per usual, $$\frac{5}{(z+2)(z-3)}= \frac{A}{z+2} + \frac{B}{z-3} \implies 5 = A(z-3)+ B(z+2).$$ Equating coefficients, we see that $$A+B=0 \mbox{ and } 2B-3A=5.$$ This linear system has the unique solution $A=-1$, $B=1$. Therefore, we see that $$\frac{5}{(z+2)(z-3)}=\frac{1}{z-3} - \frac{1}{z+2}.$$
So the poles of $X(z)$ are encircled, choose $C$ to be $\{z\in\mathbb{C}:|z|=4\}$ with standard orientation.
Therefore, by definition, the inverse $Z$-transform of $X(z)$ is \begin{align*} \frac{1}{2\pi j} \int_C X(z)z^{n-1} dz &= \frac{1}{2\pi j} \int_C \left(\frac{1}{z-3} - \frac{1}{z+2}\right)z^{n-1} dz \\ &= \frac{1}{2\pi j} \left(\int_C \frac{z^{n-1}}{z-3}dz - \int_C\frac{z^{n-1}}{z+2} dz\right). \\ \end{align*} We then apply Cauchy's Integral Formula to get $$\frac{1}{2\pi j} \left(\int_C \frac{z^{n-1}}{z-3}dz - \int_C\frac{z^{n-1}}{z+2} dz\right)=3^{n-1} - (-2)^{n-1}.$$