I am trying to answer the following question:
Let $y''+p(x)y"+q(x)y'=0 $ be a second order linear homogenous ODE where $p(x), q(x)$ are continuous. i.e The Existence and Uniqueness Theorem holds.
Prove that if $y_1$ and $y_2$ are solutions of the ODE such that both $y_1$ and $y_2$ have a local maximum at some point $x_0$, and neither $y_1$ nor $y_2$ are identically $0$ and $q \neq 0 $ at some point in every neighborhood of $x_0$ then $y_1(x)=ky_2(x)$ for some constant $k>0$.
Work so far:
Because at $x_0$ we have $y_1'(x_0)=y_2'(x_0)=0$ Then $W(y_1,y_2) = 0 \Rightarrow y_1 $ and $y_2$ are linearly depedendent $\Rightarrow$ $y_1(x)=ky_2(x)$ for some constant $k$. I just can't figure out how to show that $k>0$.
Clearly i have not used all of the information given about $y_1$ and $y_2$ and $q(x)$. Also what i have shown thus far could have been proved given only that both $y_1$ and $y_2$ have local extremum at some point $x_0$ (not necessarily both achieving maximum at $x_0$).
Edit:
Still unsolved. Realized that $y_1'(x_0)=0 \Rightarrow y_1(x_0) \neq 0$ because otherwise $y_1$ would be the trivial solution. Likewise $y_2(x_0) \neq 0$.
$\Rightarrow y_i''=-q(x_0)y_i(x_0)$ So that under the assumption that $q(x_0) \neq 0$ we have $y_i''(x_0) \neq 0$ and since for $i=1,2$ we have $y_i(x_0)$ is a local maximum then $y_i''(x_0) < 0 $ but because $y_1(x)=ky_2(x) \Rightarrow y_1''(x)=ky_2''(x)$ we have $k >0$ but this is all under the assumption that $q(x_0) \neq 0$
Also, with no extra assumptions, $k \neq 0 $ since otherwise $y_1$ would be the trivial solution which we have assumed it not to be.
All help is welcome
Thanks