second order difference equation question

Question

$y_{n+2} - 2y_{n+1} + 2y_n = 62^n$

sub $y_n= r^n$

then $y_{n+2}=r^{n+2}$, $y_{n+1}=r^{n+1}$

so $r^{n+2} - 2r^{n+1} + 2r^n = 0$

$r^n( r^2 - 2r + 2) = 0$

I got a problem here, I can solve for $r$, can anyone tell me what I can do to find $r$ in order to continue

Answer 1

I'm assuming the recursion is $$y_{n+2}-2y_{n+1}+2y_n={62^n}$$ If you want to go overkill you could use generating functions. I'll take $$y_0=y_1=0$$ for simplicity. Then we get $$G-2xG+2x^2G=6\sum_{n=0}^{\infty}{(2x)}^n={62x^2\over {1-2x}}$$ At this point I give up. If you really needed a closed form you could solve for $G$, use partial fractions to simplify, and then expand as power series to get $y_n$. According to WolframAlpha the answer isn't actually too bad...