Prove some properties of the Wronskian

The Wronskian $W(x)$ is defined by

$W(x) = u_1(x)u_2'(x) - u_2(x)u_1'(x)$

for given functions $u_1$ and $u_2$ .

If the Wronskian $W(x)$ of two functions $u_1(x)$ and $u_2(x)$ is zero for all $x$ in an open interval $I$ , prove that $\frac{u_2(x)}{u_1(x)}$ is constant for all $x \in I$ . Equivalently, if $\frac{u_2}{u_1}$ is not constant on $I$ then there is some $c \in I$ such that $W(c) \neq 0$ .
Prove that the derivative of the Wronskian is given by
$W'(x) = u_1(x) u_2''(x) - u_2(x)u_1''(x).$

Proof. (Note: I think we need the additional assumption that $u_1(x) \neq 0$ for any $x \in I$ .) With our additional assumption we have,
$d \left( \frac{u_2}{u_1} \right) = \frac{u_1 u_2' - u_1' u_2}{(u_1)^2} = 0$

since $W = u_1 u_2' - u_1' u_2 = 0$ by assumption. Thus, $\frac{u_2}{u_1}$ is constant by the zero derivative theorem $. \qquad \blacksquare$
Proof. This is a direct computation of the derivative of the Wronskian,
$\begin{align*} &&W &= u_1(x) u_2'(x) - u_1'(x) u_2(x) \\ \implies && W' &= u_1(x)u_2''(x) + u_1'(x) u_2'(x) - u_1''(x) u_2(x) - u_1'(x) u_2'(x) \\ \implies && W' &= u_1(x)u_2''(x) - u_1''(x) u_2(x). \qquad \blacksquare \end{align*}$

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