If $g$ is 2 times differentiable in $[a,b]$ and $g''+g'\,g=g$ and $g(a)=g(b)=0$, prove that $g=0$.

Hint: Consider what the equation says when $x$ is a global maximum or global minimum of $g$.

A full answer is hidden below.

Suppose $x\neq a,b$ is a global maximum of $g$. Then $g'(x)=0$ and $g''(x)\leq 0$. But the given equation then says that $g''(x)=g(x)$ so $g(x)\leq 0$. Since $g(a)=g(b)=0$, this means the maximum value of $g$ can only be $0$. Similarly, the minimum value of $g$ is $0$, so $g(x)=0$ for all $x$.