$y_{n+3} − 3 y_{n+1} + 2 y_n = (−2)^n$ I get the solution to be $y_n = A(-2)^n + Bn + C + \frac{1}{9}n(-2)^{n-1}$ but wolfram alpha gets $y(n) = c_1 (-2)^n+c_2+c_3 n+\frac{1}{27} 2^{n-1} e^{i \pi n} (4-3 n)$
Explanation required.
Help is much appreciated.