Why does the solution of $y'' = -(1+e^x)y$ decay like $e^{-x/4}$? I am trying to show that any solution of $y'' = -(1+e^x)y$ goes to $0$ as $x \to \infty$.
I was able to show that the solutions hit $0$ infinitely often.
It seems (empirically) that the solutions decay like $e^{-x/4}$.
More specifically, $e^{x/4}y(x)$ is bounded (away from both $0$ and $\pm \infty$) as $x \to \infty$.
How would one show this, or any similar bound?
The substitution $z = e^{x/4}y$ doesn't yield anything useful.
I also tried (unsuccessfully) comparing the solutions with solutions of $y'' = -ax$ with $a = (1+e^{x_0})$, for large $x_0$.
 A: Short answers were already given. So it should be redondant to continue on this way. On the other hand, the analytic solving of the ODE might interests some people :
$$y''=-(1+e^x)y$$
Let $t=e^x \quad ; \quad \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=t\frac{dy}{dt}\quad ; \quad \frac{d^2y}{dx^2}=\frac{d\left( t\frac{dy}{dt}  \right)}{dt}\frac{dt}{dx}=t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}$
$$t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}+(1+t)y=0$$
This is a Bessel ODE which general solution is :
$$y=c_1 J_{2i}(2\sqrt{t})+c_2J_{-2i}(2\sqrt{t})$$
The first term of the asymptotic expension of the Bessel function is :
$$y\:\sim\: C_1 \sin(\theta-\frac{\pi}{4})\cosh(\pi)\frac{1}{\sqrt{\pi\:\theta}} + \:C_2 \cos(\theta-\frac{\pi}{4})\sinh(\pi)\frac{1}{\sqrt{\pi\:\theta}} +O\left( \frac{1}{\theta^{3/2}} \right)$$
where $\theta = 2\sqrt{t}=2e^{x/2}$
$$y\:\sim\: \left( C_1 \sin(2e^{x/2} -\frac{\pi}{4})\cosh(\pi) \frac{1}{\sqrt{2\pi}} +\:C_2 \cos(2e^{x/2}-\frac{\pi}{4})\sinh(\pi) \frac{1}{\sqrt{2\pi}} \right)e^{-x/4} +O\left( e^{-3x/4} \right)$$
Thus, the factor of decay is $e^{-x/4}$
A: first $y^{''}/y = -(1 + e^x) $
integrate b  oth side by x
$ \int y^{''}/ydx = -(x +  e^x) +c$
we got
$ y^{'}/y + \int_0^x ( y^{'} )^2/y^2 ds = -(x + e^x) + c_0 $
integrate one more time by dx
$ln(y) + \int_0^x \int_0^t ( y^{'} )^2/y^2 dsdt = -(x^2/2 + e^x + c_0 x) + c_1 $
so
$ y = e^{-(x^2/2 + e^x + c_0 x) + c_1 - \int_0^x \int_0^t ( y^{'} )^2/y^2 dsdt} $
let $x \to \infty$
we have
$ y = O( e^{-x^2/2})$
