TITLE: Differential equation in optics
QUESTION [6 upvotes]: While playing with the eikonal equation (classical optics) I stumbled this differential equation :
$$
y y'' = k\left((y')^2 + 1\right)
$$
where $y=y(x)$ should be defined for $x>0$, $y>0$ and where $k\in \mathbb Z$. I don't know if there is a general solution $y(x)$ to that problem for a given $k$. What I know is that solving this for :

$k=0$ is easy
$k=1$ gives something like $y(x)=\cosh(x)$

Question : Is there a known solution for $k\in \mathbb Z \backslash \{0,1\}$ ?

Remark 1 : The general solution for $k=1$ is, as asked, $y(x)=y_0 \cosh((x-x_0)/y_0)$ for some constants $x_0,y_0$.
Remark 2 : The case $k=-1$ is solved below.

REPLY [16 votes]: Rewrite it as $$\frac{2y'y''}{(y')^2+1} = \frac{2ky'}{y}$$
Integrate to get $$\ln \left((y')^2+1\right) = 2k \ln y + c$$
So $$(y')^2+1 = C_1y^{2k}$$
Rearrange to get $$1=\frac{y'}{\sqrt{C_1y^{2k}-1}}$$
Integrate to get $$x+c_2 = \int \frac{dy}{\sqrt{C_1y^{2k}-1}}$$
Substitute $y=\sec^{1/k}(t)/C_1^{1/2k}$ to get $$x+c_2 = \int \frac{1}{kC_1^{1/2k}}\frac{\sin(t)}{\cos^{1+1/k}(t)}\frac{1}{\tan(t)}{dt} = \frac{1}{kC_1^{1/2k}}\int\frac{dt}{\cos^{1/k}(t)}$$
This last integral can be expressed in terms of hypergeometric functions (via Wolfram Alpha), but this is about as far as you can go in general.  
You can easily recover some special cases, e.g. $k=\pm 1$, from the last expression.  The next simplest case, $k=2$, yields an elliptic integral .  Elliptic integrals are well behaved special functions, so you might consider  this a very satisfactory solution.  
You can't hope for an elementary solution for general k, because that would amount to finding an expression for special functions like elliptic integrals in terms of elementary functions.