Prove each non-trivial solution, $u(x)$ of a normal, second-order linear ordinary differential equation, has only simple zeroes.

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Prove each non-trivial solution, $u(x)$ of a normal, second-order linear ordinary differential equation,

$$a_2(x)y'' + a_1(x)y' + a_0(x)y = 0$$ $\forall\ x \in I$

has only simple zeroes.

Where point $x_0$ is a zero of a function $u$ if and only if $u(x0) = 0$. A zero, $x_0$, of $u$ is simple if and only if $u'(x_0) \neq 0$.

Can someone advise how to begin this please? Do I use the wronskian to show linear independence?

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Assuming $a_2(x_0) \ne 0$ and the coefficients are continuous, the Picard–Lindelöf Existence and Uniqueness Theorem applies, so the only solution $u$ with $u(x_0) = 0$ and $u'(x_0)=0$ is the trivial $u = 0$.

Without the requirement $a_2(x_0) \ne 0$ the statement is not true, e.g. $x y'' - 2 y' = 0$ has a nontrivial solution $u = x^2$ whose zero at $x=0$ is not simple.