Solution of $ (dy/dx)^2 = (x^2+y^2) $ How do I find the solution of:
$$ (dy/dx)^2 = (x^2+y^2). $$ 
I tried $ y=tx$ and $x=r\cos(t) ,y=r\sin(t)$ substitutions but they did not help. I am not  able to change it into any variable separable form and the only meaning I can make out is that the slope at any point is equal to the distance of it from origin. 

$\left(\dfrac{dy}{dx}\right)^2=x^2+y^2$
$\dfrac{dy}{dx}=\pm\sqrt{x^2+y^2}$
Apply the Euler substitution:
Let $u=y\pm\sqrt{x^2+y^2}$ ,
Then $y=\dfrac{u}{2}-\dfrac{x^2}{2u}$
$\dfrac{dy}{dx}=\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}$
$\therefore\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}=u-\left(\dfrac{u}{2}-\dfrac{x^2}{2u}\right)$
$\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}=\dfrac{u}{2}+\dfrac{x^2}{2u}$
$\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}=\dfrac{u}{2}+\dfrac{x^2+2x}{2u}$
$(u^2+x^2)\dfrac{du}{dx}=u^3+(x^2+2x)u$
Let $v=u^2$ ,
Then $\dfrac{dv}{dx}=2u\dfrac{du}{dx}$
$\therefore\dfrac{u^2+x^2}{2u}\dfrac{dv}{dx}=u^3+(x^2+2x)u$
$(u^2+x^2)\dfrac{dv}{dx}=2u^4+(2x^2+4x)u^2$
$(v+x^2)\dfrac{dv}{dx}=2v^2+(2x^2+4x)v$
Let $w=v+x^2$ ,
Then $v=w-x^2$
$\dfrac{dv}{dx}=\dfrac{dw}{dx}-2x$
$\therefore w\left(\dfrac{dw}{dx}-2x\right)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$
$w\dfrac{dw}{dx}-2xw=2w^2+(4x-2x^2)w-4x^3$
$w\dfrac{dw}{dx}=2w^2+(6x-2x^2)w-4x^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{dx}=-\dfrac{1}{z^2}\dfrac{dz}{dx}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{dx}=\dfrac{2}{z^2}+\dfrac{6x-2x^2}{z}-4x^3$
$\dfrac{dz}{dx}=4x^3z^3+(2x^2-6x)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
