<s>[INST] Solving a differential equation by power series. Here's a question I'm struggling with and I'd like help on it. Please, this is not a homework problem.  

I want to find the power series solutions about the origin of two linearly independent solutions of $$w''-zw=0.$$

Also, how do I show that these solutions are analytic?  
thanks.
 [/INST] $w(x)=a_0+a_1x+\frac{a_2x^{2}}{2!}+\frac{a_3x^{3}}{3!}+...$
$w''(x)=a_2+a_3x+\frac{a_4x^{2}}{2!}+\frac{a_5x^{3}}{3!}+...$
$w''-xw=0$  
$(a_2+a_3x+\frac{a_4x^{2}}{2!}+\frac{a_5x^{3}}{3!}+...)-x(a_0+a_1x+\frac{a_2x^{2}}{2!}+\frac{a_3x^{3}}{3!}+...)=0$
$a_2+(a_3-a_0)x+(\frac{a_4}{2!}-a_1)x^{2}+(\frac{a_5}{3!}-\frac{a_2}{2!})x^{3}+....=0$
$a_2=0$
$a_3-a_0=0$
$\frac{a_4}{2!}-a_1=0$
$\frac{a_5}{3!}-\frac{a_2}{2!}=0$
$[n>2]$
$\frac{a_n}{(n-2)!}-\frac{a_{n-3}}{(n-3)!}=0$
$a_n=(n-2)a_{n-3}$
$a_0=c_1$
$a_1=c_2$
$a_2=0$
$a_3=a_0=c_1$
$a_4=2a_1=2c_2$
$a_5=3a_2=0$
if $n>=0$ then $a_{3n+2}=0$
$w(x)=c_1+c_2x+\frac{c_1x^{3}}{3!}+\frac{2c_2x^{4}}{4!}+\frac{4c_1x^{6}}{6!}+\frac{2.5c_2x^{7}}{7!}+\frac{4.7c_1x^{9}}{9!}+\frac{2.5.8c_2x^{10}}{10!}+......$
$w(x)=c_1(1+\frac{x^{3}}{3!}+\frac{4x^{6}}{6!}+\frac{4.7x^{9}}{9!}+.....)+c_2(x+\frac{2x^{4}}{4!}+\frac{2.5x^{7}}{7!}+\frac{2.5.8x^{10}}{10!}+.....)$
$w(x)=c_1(1+\sum_{k=1}^\infty \frac{1.4.7...(3k-2)x^{3k}}{(3k)!})+c_2(x+\sum_{k=1}^\infty \frac{2.5.8...(3k-1)x^{3k+1}}{(3k+1)!})$ 
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