Suppose that $q:\mathbb{R}\to\mathbb{R}$ is a continuous function such that $$\forall t\in\mathbb{R}: q(t)\leq-1$$ Let $a<b$ and consider the equation $x''(t)+q(t)x(t)=0$ along with conditions $$x(a)=0,\ \sqrt3 x'(b)=x(b)$$ Prove that the there is only one solution and it is the trivial one ($x\equiv0$).
I suppose that there is some generalization of Grönwall's inequality for this.
Clearly, there is no pair of independent solutions as their Wronskian vanishes, e.g. in $a$ and $b$. So, there is some solution $x_1(t)$ and all the others are multiplications of it by a constant.
Multiplying the equation by $x'$ we get $$ \bigl((x')^2\bigr)'+q(t)(x^2)'=0\implies\bigl((x')^2\bigr)'=-q(t)(x^2)'\ge(x^2)'\implies\bigl((x')^2-(x^2)'\bigr)'\ge0. $$ It follows that $(x')^2-(x^2)$ is increasing. If $x$ is a solution such that $x(a)=0$ with $x'(a)\ne0$, then $$ (x'(t))^2-(x(t)^2)\ge(x'(a))^2-(x(a)^2)=(x'(a))^2>0,\quad t>a, $$ and $$ |x'(t)|>|x(t)|,\quad t>a. $$