$dy/dx=\sqrt{x^2+y^2}$ 
$$\frac{dy}{dx}=\sqrt{x^2+y^2}$$

slope=distance from origin, should be simple and interesting. May have no solution! 
I have tried several approaches, best:
$(\frac{dy}{dx}-y)(\frac{dy}{dx}+y)=x^2$  multiply by $e(-x) * e(+x)$ as integrating factor. Substitute $\frac{1}{2}x^2=t$.
Second approach:
$y=x\sinh(u)$ and $x=e(t)$ yields $\frac{du}{dt} + \tanh(u)=e(t)$.
Sorry, I am not yet using the proper format.
 A: Similar to To solve $ \frac {dy}{dx}=\frac 1{\sqrt{x^2+y^2}}$:
Apply the Euler substitution:
Let $u=y+\sqrt{x^2+y^2}$ ,
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
