Find two linearly independent power series solutions of $y''-xy'+y=0$

Question

I am trying to find two linearly independent power series solutions of $y''-xy'+y=0$. Upon solving for $C_{n+2}$ I got the following recursion: $$C_{n+2} = \frac{nC_n-C_n}{(n+2)(n+1)}$$ After that, I found a series of $C$s for $n=0,1,2,3,...$

Theses are some of them: $$C_2 = \frac{-1*C_0}{2!}$$ $$C_3 = 0$$ $$C_4 = \frac{-1*C_0}{4!}$$ $$C_5 = 0$$ $$C_6 = \frac{-1*3*C_0}{6!}$$ $$C_7 = 0$$ $$C_6 = \frac{-1*3*5*C_0}{8!}$$

Now I know that one of the power series solution is $n$ because the odd terms are just $0$. Now, I have difficulty finding the solution for the even terms because the pattern is not obvious. Please someone help me.

Answer 1

As you say, one solution is $y_1(x)=c_1 x$. The other solution has no nice expression, as far as I can tell. According to Wolfram Alpha, it is $$ y_2(x)=c_2 \sqrt{-x^2}\left( 2i \sqrt π\left( \text{Erfi}\left(\sqrt{\frac{x^2}2}\right) + 1\right) - \frac{2i \sqrt2 \,e^{x^2/2}}{\sqrt{x^2}} - 2 \sqrt π\right) $$

Answer 2

Hint

In numerators, you face terms looking like $$P_n=1\cdot 3 \cdot 5 \cdots (2n-1)$$ Insert the even numbers to rewrite $$P_n=1\cdot 3 \cdot 5 \cdots (2n-1) = \frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n-1)\cdot (2n) }{2\cdot 4 \cdot 6 \cdots (2n)}$$ Now take out the factor $2$ from each term in the denominator to get $$P_n=\frac{(2n)!}{2^n\, n!}$$

I am sure that you can take it from here (not missing the fact that $y=kx$ is a solution).