Differential equation in optics 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.
 A: I didn't came with a full solution, but I can address the case $k=-1$. Our equation is
$$yy^{\prime\prime}=k\left(\left(y^{\prime}\right)^2+1\right)$$
$$\left(yy^{\prime}\right)^{\prime}-\left(y^{\prime}\right)^{2}=k\left(\left(y^{\prime}\right)^2+1\right)$$
$$\left(yy^{\prime}\right)^{\prime}=\left(k+1\right)\left(\left(y^{\prime}\right)^2+1\right)-1$$
Now if $k=-1$ the above equation simplifies tremendously to
$$\left(yy^{\prime}\right)^{\prime}=-1$$
$$yy^{\prime}=-x+\frac{A}{2}$$
$$\int y{\rm d}y=\int\left(-x+\frac{A}{2}\right){\rm d}x$$
$$\frac{y^2}{2}=-\frac{x^2}{2}+\frac{A}{2}x+\frac{B}{2}$$
$$y=\pm\sqrt{-x^{2}+Ax+B}$$
EDIT: If I continue with @stressed-out line of thought, for the general case we have to solve
$$yu\frac{{\rm d}u}{{\rm d}y}=k\left(u^{2}+1\right)$$
with $u=y^{\prime}$. This is far more manageable, as one can use seperation of variables
$$\int\frac{u{\rm d}u}{u^{2}+1}=\int\frac{k{\rm d}y}{y}$$
$$\frac{1}{2}\ln\left(u^{2}+1\right)=k\ln y+\tilde{A}$$
$$u^{2}+1=Ay^{2k}$$
with $A\equiv e^{2\tilde{A}}$. Further algebra yields
$$y^{\prime}=\pm\sqrt{Ay^{2k}-1}$$
$$\int\frac{{\rm d}y}{\sqrt{Ay^{2k}-1}}=\pm\int{\rm d}x$$
$$\int\frac{{\rm d}y}{\sqrt{Ay^{2k}-1}}=\pm x+B$$
The problem now is in solving this integral, which turns out to be quite complicated for most $k$'s. However, there is now another case that is easily solvable - $k=\frac{1}{2}$ (though not mentioned by the OP). Indeed for $k=\frac{1}{2}$
$$\int\frac{{\rm d}y}{\sqrt{Ay-1}}=\pm x+B$$
$$\frac{2}{A}\sqrt{Ay-1}=\pm x+B$$
$$Ay-1=\left(\pm\frac{A}{2}x+\frac{AB}{2}\right)^{2}$$
$$y=\frac{1}{A}\left(\left(\frac{A}{2}x+C\right)^{2}+1\right)$$
with $C\equiv \frac{AB}{2}$. Also the case $k=-\frac{1}{2}$ can be solved like this, but it is slightly more complicated to calculate the integral.
A: Again, just an observation: 
$$yy''-k(y')^2+k=0$$
Take $y'=u$
then $$y''=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=\frac{du}{dy}u$$
We will therefore obtain:
$$(y u)\frac{du}{dy}-k(u^2+1)=0$$
This idea was successfully followed and beautifully done by @eranreches. He/She has written it in his/her post and therefore, this post will be a community-wiki.
A: 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. 
