# ODE of second order, proving that polynomials at $t_0$ are zero

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?

• Use the two equations to eliminate one of $p(t_0)$ or $q(t_0)$ and consider the remaining terms in view of the Wronskian. Dec 26 '18 at 20:53

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?