Show that every nontrivial solution to the differential equation has at most one zero

Question

Show that every nontrivial solution to the equation $y''-e^{x}y=0$ can have at most one zero on the interval $0<x<+\infty$.

My idea is to compare it to the eq $y''-y=0$, then use the Sturm Comparison Theorem, but I'm unsure of how to execute this.

Answer 1

Prompted by rafa11111, I formulate my comment as an answer.

Suppose that a nontrivial solution $\varphi$ of $y'' - e^{x}y = 0$ on $(-\infty, \infty)$ has more than one zero. (Remark: In my comment I used the fact that the zeros are isolated, but that is not needed). Since $\varphi$ is continuous, the set $\{x \in (-\infty, \infty): \varphi(x) \ne 0\}$ is an open subset of $(-\infty, \infty)$, hence an (at most) countable union of pairwise disjoint open intervals (a fact from topology). Since its complement contains at least two members, there must be $x_1 < x_2$ such that $\varphi(x_1) = \varphi(x_2) = 0$ and $\varphi(x) < 0$ (say) for all $x \in (x_1, x_2)$. The function $\varphi$ restricted to the compact interval $[x_1, x_2]$ attains its (negative) minimum somewhere in $(x_1, x_2)$, say at $x^*$. But $\varphi''(x^*) = e^{x^*} \varphi(x^*) < 0$, which is impossible.