I am trying to find the solution to $y''-2xy'+10y=$ given $y(0)=0, y'(0)=3$.
My work so far: $$ \sum_{n=2}^\infty n(n-1)a_nx^{n-2} -2 \sum_{n=1}^\infty na_nx^{n-1} +10\sum_{n=0}^\infty a_nx^{n}=0$$ $$ \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n} -2\sum_{n=0}^\infty na_nx^{n} +10\sum_{n=0}^\infty a_nx^{n}=0 $$ $$ \sum_{n=0}^\infty [(n+2)(n+1)a_{n+2} -2 na_n+10 a_n]x^n=0 $$ $$ \sum_{n=0}^\infty [(n+2)(n+1)a_{n+2} +8 a_n]x^n=0 $$ $$(n+2)(n+1)a_{n+2} +8a_n=0, n=0,1,2,...$$ $$a_{n+2} = \frac{-8a_n}{(n+1)(n+2)}, n=0,1,2,...$$ Next step is finding the recurrence relation and the generalized form for $a_n$, but I am having trouble with this. I know to plug in $n=0$, $n=1$, and so on until the relation can be reduced to $a_0$ and $a_1$ terms, but I cannot seem to successfully find the general form. Any hints or suggestions?
hint
After writing that the coefficient of $x^n $ is zero , you will get
$$a_{n+2}=\frac {2n-10}{(n+1)(n+2)} a_n$$
and treat the two cases : $n $ even to use $a_0$ and $n $ odd to use $a_1$.
Since $a_0=y (0)=0$, we conclude that $$a_{2p}=0$$
and since $a_1=y'(0)=3$, we find
$$a_3=\frac {-8}{2\times 3}3 $$ $$a_5=\frac {4}{4\times 5}4$$ $$a_7=0=a_9=a_{11}=.. $$
finally $$\boxed {y=3x-4x^3+\frac {4}{5}x^5}$$