I'm solving the problem as part of a homework in my second applied analysis course, and I'm having a hard time solving the recurrence. I got
$$ \begin{cases} \begin{align} a_1-2a_2&=0\\ (n+3)(n+2)a_{n+3}-(n+2)a_{n+2}+a_{n}&=0 \end{align} \end{cases} $$
From which I got
$$ \begin{align} a_{3}=-\frac{1}{3!}a_{0}+\frac{1}{3!}a_{1}\\ a_{4}=-\frac{1}{4!}a_{0}-\frac{1}{4!}a_{1}\\ a_{5}=-\frac{1}{5!}a_{0}-\frac{4}{5!}a_{1}\\ a_{6}=\frac{3}{6!}a_{0}-\frac{8}{6!}a_{1}\\ a_{7}=\frac{8}{7!}a_{0}-\frac{3}{7!}a_{1}\\ a_{8}=\frac{14}{8!}a_{0}+\frac{2}{8!}a_{1}\\ \end{align} $$.
Any ideas on how to solve it?
$$y(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+a_7x^7+a_8x^8+...$$. $$y(x)=a_0[1-\frac{1}{3!}x^3-\frac{1}{4}x^4-\frac{1}{5!}x^5+\frac{3}{6!}x^6+\frac{8}{17}x^7+\frac{4}{8!}x^8+...]+a_1[x-\frac{1}{2}x^2-\frac{1}{3!}x^3+\frac{1}{4!}x^4-\frac{5}{5!}x^5-\frac{8}{6!}x^6-\frac{3}{7!}x^7+\frac{2}{8!}x^8+,,,]$$