Solution to ODE $\frac{d^2z(x)}{dx^2}=b\cdot e^{(-a\cdot z(x))}$

Question

I am looking for solution $z(x)$ ($z$ as function of $x$) to differential equation. This is the eikonal in geometric optics that describes the rays direction for hot air over a surface (mirage).

$$\frac{d^2z(x)}{dx^2}=b\cdot e^{(-a\cdot z(x))}$$

Variable $x$, and $a$ and $b$ are constants. How to solve it ?

I am trying to solve it using Laplace transformation. And would like to ask what is the Laplace transform of $\exp(-a\cdot z(x))$

$Ys^2-s\cdot y(0)-y\prime(0)=?$

Answer 1

Hint:

$\dfrac{d^2z}{dx^2}=be^{-az}$

$\dfrac{dz}{dx}\dfrac{d^2z}{dx^2}=be^{-az}\dfrac{dz}{dx}$

$\int\dfrac{dz}{dx}\dfrac{d^2z}{dx^2}~dx=\int be^{-az}\dfrac{dz}{dx}~dx$

$\int\dfrac{dz}{dx}~d\left(\dfrac{dz}{dx}\right)=\int be^{-az}~dz$

$\dfrac{1}{2}\left(\dfrac{dz}{dx}\right)^2=-\dfrac{be^{-az}}{a}+c$

$\left(\dfrac{dz}{dx}\right)^2=\dfrac{C_1-be^{-az}}{a}$

$\dfrac{dz}{dx}=\pm\dfrac{\sqrt{C_1-be^{-az}}}{\sqrt a}$

Answer 2

$$\int_{0}^\infty e^{-azx}e^{-sx}dx=\int_{0}^\infty e^{-x(az+s)}dx=-\frac{e^{-x(az+s)}}{az+s}|_{0}^\infty=\frac{1}{az+s}$$