Q:

Integrate the full Laplace-Young equation

I have been given the question:
Recall the Laplace–Young equation governing the shape z = h(x) of a meniscus near a vertical,
planar wall
$l_{c}^2 \frac{h_{xx}}{(1+h{x}^2)^{3/2}} = h$
Integrating the full Laplace–Young equation once (without making an assumption of shallow slopes), show that the rise height of the meniscus on the wall, $h_0$, is given in terms of the contact
angle of the liquid $ \theta$ by
$h_0 = ±l_c [2(1 − sin \theta)]^{1/2}$
and discuss when each of the ± branches of the result are appropriate.
I have no idea how to start to integrate the equation. Any help would be appreciated.

A:

To solve
$$\tag{1}l_c^2\frac{h_{xx}}{(1 + h_x^2)^{3/2}} = h,$$
note that
$$\tag{2}\frac{d}{dx} (1 + h_x^2)^{-1/2} =  - \textstyle\frac{1}{2}(1+ h_x^2)^{-3/2}2h_xh_{xx} = -h_x\frac{h_{xx}}{(1+h_x^2)^{-3/2}}.$$
Multiplying both sides of (1) by $-h_x$ and using (2) we obtain
$$\tag{3}l_c^2\frac{d}{dx} (1 + h_x^2)^{-1/2} = -h_xh = - \frac{d}{dx}(\textstyle\frac{1}{2} h^2) $$
Integrating both sides of (3) yields
 $$l_c^2(1 + h_x^2)^{-1/2} = C - \textstyle\frac{1}{2}h^2,$$
where $C$ is a constant of integration.
It remains to apply boundary conditions.  Far from the wall ($x \to \pm \infty$ depending on how the coordinate system is specified) we have $h, h_x \to 0$ which fixes $C = l_c^2$.
Thus,
$$h^2 = 2l_c^2 ( 1 - (1 + h_x^2)^{-1/2})$$ 
You should be able to solve for $h(0)$, by specifying the  contact angle at the wall, which should yield $(1 + h_x^2(0))^{-1/2} = \sin \theta$.

