Recurrence formula for power series (I asked this question in an earlier post and when I had follow up question they went unanswered so I'm posting it again.)
The given is:
$$y^{''}-x^2y^{'}-3xy=0$$
Converting the given into sigma notation:
$$\sum^\infty_{n=2}n(n-1)C_{n}x^{n-2}+\sum^\infty_{n=1}nC_nx^{n+1}-\sum^\infty_{n=0}3C_nx^{n+1}=0$$
Then reindexing and putting them all into a single power series:
$$2C_2+6C_3x-3C_0x+\sum^\infty_{n=2}\left[(k+2)(k+1)C_{k+2}+(k-1)C_{k-1}-3C_{k-1}\right]x^k=0$$
Then the resulting expressions follow:
$$2C_2+6C_3x-3C_0x=0$$
$$(k+2)(k+1)C_{k+2}+(k-1)C_{k-1}-3C_{k-1}=0$$
How I got  $C_{k+2}$ is:
$$(k+2)(k+1)C_{k+2}+(k-1)C_{k-1}=3C_{k-1}$$
$\bullet$Then dividing by $C_{k-1}$ on both sides
$$(k+2)(k+1)C_{k+2}+(k-1)=3$$
$$(k+2)(k+1)C_{k+2}=3-(k-1)$$
$\bullet$ Now dividing again to get $C_{K+2}$ alone
$$C_{k+2}=\frac{3-(k-1)}{(k+2)(k+1)}$$
which is my recurrence formula.
In my previous post, one answer said I could separate the first expression into $2C_2=0$ and $ 6C_3-3C_0=0$ which I haven't been taught so I am confused as to how thats possible, also which $C$-term am I meant to solve for, $C_3, C_2$ or $C_0$? And is my recurrence formula correct? 
 A: $$\sum^\infty_{n=2}n(n-1)C_{n}x^{n-2}-\sum^\infty_{n=1}nC_nx^{n+1}-\sum^\infty_{n=0}3C_nx^{n+1}=0$$
Change indices:
$$\sum^\infty_{n=0}(n+2)(n+1)C_{n+2}x^{n}-\sum^\infty_{n=2}(n-1)C_{n-1}x^{n}-\sum^\infty_{n=1}3C_{n-1}x^{n}=0$$
Which gives us for $n \ge 2$:
$$$$
$$(n+2)(n+1)C_{n+2}=(n+2)C_{n-1}$$
$$\implies C_{n+2}=\dfrac {C_{n-1}}{n+1}$$
And
$$2C_2+3x(2C_3-C_0)=0$$
$$\implies C_2=0, 2C_3=C_0$$
You find the following recurrence formula:
$$C_{3n+1}=\dfrac {C_1}{n!3^n} \,\, n \in \mathbb{N}$$
Which gives the solution:
$$ y_1(x)=\sum_{n=0}^\infty {C_{3n+1}x^{3n+1}}$$
$$ \boxed {y_1(x)=x\sum_{n=0}^\infty \dfrac {C_1x^{3n}}{n!3^n}=C_1xe^{x^3/3}}$$

For the second solution it's hard to find a recurrence formula and a pattern. Maybe you can leave the series that way and calculate some of its terms:
$$y_2(x)=C_0x^0+C_3x^3+C_6x^6+.....$$
Where $C_3=\frac {C_0}2$. Use the recurrence formula for the other coeffcients:
$C_{n+2}=\dfrac {C_{n-1}}{n+1}$
For example $C_6=\dfrac {C_3}{5}=\dfrac {C_0}{2*5}$
