# I didn't understand this recurrence relation solution.

Recently, i was trying to solve this recurrence relation $$a_{n+4} = \frac{-\alpha(x)}{(n+4)} \cdot a_{n+3} +\frac{-\beta(x)}{(n+3)\cdot (n+4)} \cdot a_{n+2}$$

But i can't solve for $$a_n$$

I've tried to solve using Wolfram|Alpha and i got this result.
Actually i'm really good at recurrence relations, but i can't understand how do i get this result and how to solve it.
Edit: n = 0,1,2,3....

Hint Multiplying by $$(n+4)!$$ you get $$(n+4)!a_{n+4} = -\alpha(x) *(n+3)! a_{n+3} -\beta(x) *(n+2)!a_{n+2}$$
Let $$b_n=n! a_n$$ then, your recurrence is $$b_{n+4}= -\alpha(x) \cdot b_{n+3} -\beta(x) \cdot b_{n+2}$$ which is a standard second order recurrence.
Solve it, and then $$a_n=\frac{b_n}{n!}$$
• @MuratGüven That is irrelevant... $\frac{\sqrt{\alpha^2-\beta}}{\beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$. – N. S. Feb 16 at 23:24