Given a second order linear differential equation: $$x''+xe^t=0$$ It is clear that every its solution $x(t)$ oscillates and it can be proven that the distance between closest zeros of the function decreases. Though I still don't get how to prove that: $$\text{ If }b_n=\max\limits_{t\in[t_{n}; \ t_{n+1}]}|x(t)|, \text{ then }\ b_1>b_2>b_3>\dots$$ where $\{t_n\}$ is a sequence of zeros of the function $x(t)$.
Thanks in forward and sorry if dublicating (I couldn't find anything).
This not an answer to the question, but a comment too long to be put in the comment section.
Also, this is a discussion about the validity of the question if no boundary condition is specified.
$$\frac{d^2x}{dt^2}+x(t)e^t=0 \tag 1$$ Change of variable : $\theta=ae^{bt}\quad\to\quad d\theta=b\theta\:dt$
$\frac{dx}{dt}=\frac{dx}{d\theta}\frac{d\theta}{dt}=b\theta\frac{dx}{d\theta}$
$\frac{d^2x}{dt^2}=\frac{d}{d\theta}\left(b\theta\frac{dx}{d\theta}\right)\frac{d\theta}{dt}=\left(b\theta\frac{d^2x}{d\theta^2}+b\frac{dx}{d\theta} \right) b\theta=b^2\theta^2\frac{d^2x}{d\theta^2}+b^2\theta\frac{dx}{d\theta}$ $$b^2\theta^2\frac{d^2x}{d\theta^2}+b^2\theta\frac{dx}{d\theta}+x\frac{\theta^{1/b}}{a^{1/b}}=0$$ $$\frac{d^2x}{d\theta^2}+\frac{1}{\theta}\frac{dx}{d\theta}+\frac{\theta^{(1/b)-2}}{b^2a^{1/b}}x(\theta)=0$$ To transform Eq.$(1)$ to a Bessel equation, we just have to set $b=\frac12$ and $b^2a^{1/b}=1$, thus $a=2$ : $$\frac{d^2x}{d\theta^2}+\frac{1}{\theta}\frac{dx}{d\theta}+x(\theta)=0 \tag 2$$ The general solution is : $\quad x(\theta)=c_1J_0(\theta)+c_2Y_0(\theta)$
where $J_0$ and $Y_0$ are the Bessel functions of first and second kind and order $0$. $$x(t)=c_1J_0(2e^{t/2})+c_2Y_0(2e^{t/2})$$ Of course, if we consider independently $J_0(X)$ and $Y_0(X)$, the roots are well defined. There is a lot of studies about the roots of the Bessel functions, where one could find pertinent material to answer to the question which, then, is a well-asked question.
But if no boundary condition is specified, $c_1$ and $c_2$ can be any constants and the roots of $x(t)$ can be any values (depending on $c_1$ and $c_2$). I wonder if the question raised could be answered on a general manner in those non-specific conditions.