Solutions with same points of inflection

Question

Given $y'' + p(t)y'+ q(t)y = 0$, where the functions $p(t)$ and $q(t)$ are continuous on the whole real line and suppose that $p(t) > 0$ and $q(t) > 0$ for all values of $t$. Is it possible that its two linearly independent solutions have a point of inflection for the same value of $t$?

Answer 1

We have for all $t_0 \in\mathbb{R}$,

$y_1 ''(t_0) + p(t_0)y'_1 (t_0) + q(t_0)y_1(t_0) = 0$

$y_2 ''(t_0) + p(t_0)y'_2 (t_0) + q(t_0)y_2(t_0) = 0$

We let $t_0$ be the common point of inflection for $y_1(t)$ and $y_2(t)$. Then we have $y_1 ''(t_0)= 0 $ and $y_2''(t_0)=0$. Substituting this into the equations and multiplying the first by $y_2(t_0)$ and the second by $y_1(t_0)$, we obtain

$p(t_0)y'_1 (t_0)y_2(t_0) + q(t_0)y_1(t_0)y_2(t_0) = 0$

$p(t_0)y'_2 (t_0)y_1(t_0) + q(t_0)y_2(t_0)y_1(t_0) = 0$

Subtracting the first from the second, we obtain,

$p(t_0)(y'_2 (t_0)y_1(t_0) - y'_1 (t_0)y_2(t_0))= 0$

We get $p(t_0)W(t_0) = 0$. Since $p(t_0) > 0$ , we conclude that $W(t_0)=0$, which is a contradiction to the fact that $y_1$ and $y_2$ are linearly independent. Hence, it is not possible for two linearly independent solutions have a point of inflection for the same value of $t$.