TITLE: ODE of second order, proving that polynomials at $t_0$ are zero
QUESTION [5 upvotes]: The following ODE is given: $$y''(t) + p(t)y'(t) + q(t)y(t)=0$$
When $p(t), q(t)$ are continuous functions.
We are given two linear independent solutions $y_1(t), y_2(t)$ and also $y_1''(t_0) = y_2''(t_0) = 0$.
I need to prove that $p(t_0) = q(t_0) = 0$.
What I've tried is just placing zero in the second derivative for each function in the ODE, and working with the Wronskian. However I end up with $$p(t)(y_1'(t_0) - y_2'(t_0)) + q(t)(y_1(t_0) - y_2(t_0))$$ which is not the Wronskian.
Any help?

REPLY [3 votes]: From the assumptions you get
$$
\begin{aligned}
0 + p(t_0) y_1'(t_0) + q(t_0) y_1(t_0) &= 0
,
\\
0 + p(t_0) y_2'(t_0) + q(t_0) y_2(t_0) &= 0
.
\end{aligned}
$$
That's a $2 \times 2$ linear system for $p(t_0)$ and $q(t_0)$. Can you take it from here?