Solution of $y''+e^xy=0$ is unbounded as $x\to\infty$ Consider the differential equation $y''+e^xy=0$. Can we say something about the behaviour of $y$ as $x\to\infty$? In particular, is it unbounded?
I think, to solve the equation, we need to use the power series method. By is there a way to understand the behaviour beforehand, say by using the sturm-picone or similar theorems?Thanks beforehand.
 A: Your equation can be solved explicitly in terms of Bessel functions:
$$y(x)=C_1 J_0\left(2 e^{x/2}\right)+ C_2 Y_0\left(2 e^{x/2}\right). $$
Using the asymptotic behaviors of the Bessel functions $J_0(t),Y_0(t)$ as $t \to 0^+$ and as $t\to +\infty$ we can deduce that the solution is bounded if and only if $C_2=0$.
Edit:
If you prefer the answer in terms of initial value problems, the solution defined by the initial conditions
$$y(x_0)=y_0, \\
y'(x_0)=y_1, $$
is bounded if and only if
$$ y_1 J_0\left(2 e^{x_0/2}\right)+e^{x_0/2} y_0 J_1\left(2 e^{x_0/2}\right)=0. $$

Further Edit:
Changing the independent variable according to 
$$t=2 e^{x/2}, \quad y(x)=Y(t), $$
results in the Bessel equation of order $0$
$$ t^2 \ddot{Y}+t \dot{Y} + t^2 Y=0. $$
The solution is thus
$$ Y(t)=C_1 J_0(t)+C_2 Y_0(t), $$
and therefore $y(x)$ is given as above.
A: In order to prove that the solutions of (the transformed equation through a change of variable)
$$ x u''(x) + u'(x) + x u(x) = 0 \tag{1}$$
are bounded on $[1,+\infty)$ we do not need to invoke the asymptotics of Bessel functions, we may directly exploit the structure of the differential equation. On short intervals $[a,b]$ it is reasonable to claim that the solution of $(1)$ with $u(a)=u_0, u'(a)=u_1$ is close to the solution of 
$$ a v''(x) + v'(x) + a v(x) = 0 \tag{2}$$
with boundary conditions $v(a)=u_0, v'(a)=u_1$. On the other hand $(2)$ is a ODE with constant coefficients and characteristic polynomial $az^2+z+a$, hence the solutions of $(2)$ are bounded by $\sqrt{u_0^2+u_1^2+\frac{u_0 u_1}{a}}$ on $[a,b]$. Let us consider the sequence of intervals $[H_1,H_2],[H_2,H_3],\ldots$ By denoting as $\sigma_m$ the following supremum
$$ \sigma_m = \sup_{x\in[H_m,H_{m+1}]}\left|u(x)-v(x)\right| $$
(where the boundary conditions for $v$ on $[H_m,H_{m+1}]$ are given by the values of $u$ and $u'$ at the left endpoint) we have that
$$ \{\sigma_m\}_{m\geq 1}\in \ell^1 \tag{3} $$
allows to state that the boundedness of the solutions of $(1)$ is a consequence of the boundedness of the solutions of $(2)$. By Frobenius power series method both the solutions of $(1)$ and $(2)$ are analytic functions: by considering $d(x)=u(x)-v(x)$ on $[H_m,H_{m+1}]$ we have that both $d$ and $d'$ vanish at the left endpoint, so by considering the termwise difference between $(1)$ and $(2)$ it is not difficult to derive
$$ \sigma_m \ll |[H_m,H_{m+1}]|^2 H_m^2 \ll \frac{\log^2 m}{m^2}$$
having $\{\sigma_m\}_{m\geq 1}\in\ell^1$ as a straightfoward consequence.

Now that the boundedness of the solutions of $(1)$ is proved, a "bootstrap" argument allows to prove that $u(x)^2+u'(x)^2 \ll \frac{1}{x}$ as $x\to +\infty$, hence the solutions of $(1)$ are not only bounded but convergent to zero and with a bounded derivative. Indeed, by defining the energy of a solution as $E(u)=u(x)^2+u'(x)^2$
and by multiplying both sides of $(1)$ by $2u'(x)$, we have
$$ 0 = x E'(u) + 2u'(x)^2 \leq x E'(u) + 2 E(u)$$
and any non-trivial bound for the energy immediately gives a simultaneous bound for $u$ and $u'$.

Summarizing, Tricomi's approximation
$$ J_0(x) \approx \frac{\sin(x)+\cos(x)}{\sqrt{\pi x}} \text{ for large values of }x$$
can be essentially derived by standard manipulations of Bessel's differential equation $(1)$. 
A recommended book is Watson's A treatise on the theory of Bessel functions.
