Using the power series method, find the general solution of $y^{''} − x^2 y^{'}−3xy=0$

Question

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: $$2C_2+6C_3x-3C_0x=0$$ $$(k+2)(k+1)C_{k+2}+(k-1)C_{k-1}-3C_{k-1}=0$$ Am I right so far? Also should it be $-3C_0x$ or $3C_0x$?

Answer 1

Yes, $-3C_0$ is correct, as you split this term off of the third series, it is not a result of a compensation for added terms.

In comparing coefficients, you get separate equations for the constant and linear terms, thus $$ 2C_2=0,\\ 6C_3-3C_0=0. $$ This tells you that $C_2=C_5=C_8=...=0$ and that $C_0$ and $C_1$ are the free variables that determine all the other coefficients.