Solving a Young Laplace equation for a meniscus against a flat plate This is more of a math question and one, furthermore, that I know the final answer to.  What I am asking is more of a "how do I get there" question as this question was generated during a self study situation.
So, for a flat plate with a meniscus on it at some contact angle the form is:
a y''$/(1+(y')^2)^{3/2}$ + b g y = 0
where a is surface tension, b is density, and g is the usual 9.8 m$/s^2$.   
Now, I know that when you "A first integration, together with the boundary condition dy/dx=y=0 as x goes to infinity yields:
1$/(1+y'^2)^{1/2}$=1-(bg/2a)$y^2$    "
--From de Gennes on Menisci
My problem is with the integration.  How do I integrate something like the 1st equation?  I find myself running in circles.  Worse, I know that this type of integration is something I've run across numerous times, but I can't seem to find the methodology in the current stack of books.
So I beg, can someone either tell me where to look to find the appropriate methodology to attack this problem or take me through the process?
I would be grateful for as long as it would be useful for you.
Thanks,
Sam
 A: I think it's a simple substitution: recognizing the derivative of some part of the expression in another part of the expression... in a useful way. :-D
The derivative of $1 + (y')^2$ almost appear in the numerator, but the $y'$ is lacking. Let's try multiplying the equation by $y'$, as the second term looks easier:
$$\frac{a y'' y'}{(1+(y')^2)^\frac{3}{2}} + b g yy' = 0$$
Now we can integrate the first term using $u = 1+(y')^2$:
$$\int dx \frac{a y'' y'}{(1+(y')^2)^\frac{3}{2}} = \int du \frac{a}{2}u^{-\frac{3}{2}} = -au^{-\frac{1}{2}} = \frac{a}{(1+(y')^2)^\frac{1}{2}}$$
Going for the second term (here using just $dy = y'\,dx$):
$$\int dx\,b g y y' = b g \int dy\,y = \frac{b g}{2}y^2$$
Putting it all together with the constant:
$$\frac{a}{(1+(y')^2)^\frac{1}{2}} + \frac{b g}{2}y^2 = C$$
Only the boundary conditions remain:
$$\lim_{x \rightarrow +\infty} y = lim_{x \rightarrow +\infty} y' = 0$$
$$\frac{a}{(1+(0)^2)^\frac{1}{2}} + \frac{b g}{2}0^2 = C$$
$$a = C$$
Replacing this value for $C$:
$$\frac{a}{(1+(y')^2)^\frac{1}{2}} + \frac{b g}{2}y^2 = a$$
Multiplying by $1/a$ and reordering terms:
$$\frac{1}{(1+(y')^2)^\frac{1}{2}} + \frac{b g}{2a}y^2 = 1$$
$$\frac{1}{(1+(y')^2)^\frac{1}{2}} = 1 - \frac{b g}{2 a}y^2$$
A: Another solution (hopefully, the answer given by mmc is alright, but I followed another procedure):
Using this change of variable $p=\displaystyle \frac{dy}{dx}$, we have $\displaystyle p\dot p = \frac{d^{2}y}{dx^{2}}$ (keep in mind that $y$ depends on $x$ and $p$ depends on $y$). Substituting in the original equation
$$\frac{a p \dot p}{(1+p^{2})^{\frac{3}{2}}}=-bgy$$
We notice that the left-hand side can be written as
$$\frac{d}{dy}\left(-\frac{a}{(1+p^{2})^{\frac{1}{2}}}\right)=-bgy$$
After integration we have
$$-\frac{a}{(1+y'^{2})^{\frac{1}{2}}}=-\frac{bgy^{2}}{2}+C$$
Using your boundary conditions, we find that $C=-a$. Therefore, we obtain the equation you were looking for
$$\frac{1}{(1+y'^{2})^\frac{1}{2}}=1-\frac{bgy^{2}}{2a}$$
