If Wronskian is zero at some point in an interval, then the Wronskian is zero at all points in the interval

Question

Suppose that $y_1$ and $y_2$ are solutions to the homogeneous DE $$y^{\prime \prime}+p(x)y^{\prime}+q(x)y=0$$ on $I$ and assume that for some $x_0 \in I$, we have $$W(y_1,y_2)(x_0)>0~.$$ Show that $$W(y_1,y_2)(x)>0\qquad \forall ~~x \in I~$$ I have no idea on how to start. Anyone can help ?

Answer 1

Hint: show that $W$ satisfies

$$W' + p(x)W = 0$$

which would imply that

$$W(x) = W(x_0) \exp{\left [ - \int_{x_0}^x dt \: p(t) \right ]}$$

So if $W(x_0) > 0, \mathrm{then} \; W(x) > 0 \; \forall \, x \in I$.

Answer 2

Let $Y= \left( \begin{matrix} y' \\ y \end{matrix} \right)$. Then $y$ is a solution of $y''+p(x)y'+q(x)y=0$ $(*)$ iff $Y$ is a solution of $Y'=A(x)Y$ $(**)$, with $A(x)= \left( \begin{matrix} -p(x) & -q(x) \\ 1 & 0 \end{matrix} \right)$.

Now, let $y_1, y_2$ be two solutions of $(*)$ with $Y_1,Y_2$ the associated solutions of $(**)$. If $\det(Y_1(x_0),Y_2(x_0))=W(y_1,y_2)(x_0)=0$ for some $x_0 \in I$, then $Y_1(x_0)$ and $Y_2(x_0)$ are colinear so $Y_1(x_0)=aY_2(x_0)$ for some $a \in \mathbb{R}$. According to Cauchy-Lipschitz theorem, $Y_1=aY_2$ and $\det(Y_1,Y_2)=W(y_1,y_2) \equiv 0$.

Therefore, $W(y_1,y_2)$ is either positive or negative or identically equal to zero.