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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?

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    $\begingroup$ 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. $\endgroup$ Dec 26 '18 at 20:53
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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?

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  • $\begingroup$ Thanks! This helped me solve it $\endgroup$
    – Gabi G
    Dec 26 '18 at 21:47

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