(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?
$$\sum^\infty_{n=2}n(n-1)C_{n}x^{n-2}-\sum^\infty_{n=1}nC_nx^{n+1}-\sum^\infty_{n=0}3C_nx^{n+1}=0$$ 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 $C_6=\dfrac {C_3}{5}=\dfrac {C_0}{2*5}$