Differential Equation Solutions and Analytic Restrictions: A Proof

[NeoDevin]

Homework Statement

Show that if two (analytic) solutions $y_1$, $y_2$ to the differential equation for $y(z) : y'' + p_1(z)y' + p_0(z)y = 0$ exist in some domain and $y_1y_2' \neq y_1'y_2$ in that domain then $p_0(z)$ and $p_1(z)$ are restricted to be analytic.


2. The attempt at a solution
Expanding $y_1$ and $y_2$ in power series

$$y_1(z) = \sum_n a_n(x-x_0)^n$$

$$y_2(z) = \sum_n b_n(x-x_0)^n$$

and using the inequality gives that there exists an $n$ such that

$$\sum_{i=0}^nib_ia_{n-i+1} \neq \sum_{i=0}^n(n-i+1)b_ia_{n-i+1}$$

That's all I've gotten so far, I don't even know for sure if I'm on the right track, or way out to lunch. Any help would be appreciated.

 

[Dick]

Put the two solutions into the differential equation giving you two equations for the functions p0 and p1. Can you solve for p0 and p1 in terms of y's and their derivatives? I think you can since that's what y1*y2'!=y2*y1' is telling you.

 

[NeoDevin]

Thanks, got it. I was going in the totally wrong direction on that one.