To solve $ \frac {dy}{dx}=\frac 1{\sqrt{x^2+y^2}}$ How do we solve the differential equation $ \dfrac {dy}{dx}=\dfrac 1{\sqrt{x^2+y^2}}$ ?
 A: Since this problem is not amenable to the standard repertoire of 1st order techniques, we might use some asymptotic analysis to understand the qualitative features of this ODE. Firstly, the derivative is always positive and bounded as $x$ or $y$ or both approach $\pm\infty$. The limiting differential equation in all of these cases is $y'=0$. And so in the asymptotic behavior is constant. On the other hand, the origin is a singularity and so we might ask what is the behavior near $(x,y)=(0,0)$. Suppose we investigate $y=c\sqrt{x}$ for $c\in\mathbb{R}$ near the origin. Then the differential equation near the origin becomes $y'=\frac{1}{\sqrt{x^2+c^2x}}\approx\frac{1}{c\sqrt{x}}$ and so $y(x)\approx \sqrt{x}/c$ and we see this matches our supposition up to a multiplicative constant. If we suppose that $y = cx^r$ near the origin and go through the same argument we find that $r=1/2$ necessarily. So the asymptotic behavior near the origin goes as the square root. 
A: $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x^2+y^2}}$
$\dfrac{dx}{dy}=\sqrt{x^2+y^2}$
Apply the Euler substitution:
Let $u=x+\sqrt{x^2+y^2}$ ,
Then $x=\dfrac{u}{2}-\dfrac{y^2}{2u}$
$\dfrac{dx}{dy}=\left(\dfrac{1}{2}+\dfrac{y^2}{2u^2}\right)\dfrac{du}{dy}-\dfrac{y}{u}$
$\therefore\left(\dfrac{1}{2}+\dfrac{y^2}{2u^2}\right)\dfrac{du}{dy}-\dfrac{y}{u}=u-\left(\dfrac{u}{2}-\dfrac{y^2}{2u}\right)$
$\left(\dfrac{1}{2}+\dfrac{y^2}{2u^2}\right)\dfrac{du}{dy}-\dfrac{y}{u}=\dfrac{u}{2}+\dfrac{y^2}{2u}$
$\left(\dfrac{1}{2}+\dfrac{y^2}{2u^2}\right)\dfrac{du}{dy}=\dfrac{u}{2}+\dfrac{y^2+2y}{2u}$
$(u^2+y^2)\dfrac{du}{dy}=u^3+(y^2+2y)u$
Let $v=u^2$ ,
Then $\dfrac{dv}{dy}=2u\dfrac{du}{dy}$
$\therefore\dfrac{u^2+y^2}{2u}\dfrac{dv}{dy}=u^3+(y^2+2y)u$
$(u^2+y^2)\dfrac{dv}{dy}=2u^4+(2y^2+4y)u^2$
$(v+y^2)\dfrac{dv}{dy}=2v^2+(2y^2+4y)v$
Let $w=v+y^2$ ,
Then $v=w-y^2$
$\dfrac{dv}{dy}=\dfrac{dw}{dy}-2y$
$\therefore w\left(\dfrac{dw}{dy}-2y\right)=2(w-y^2)^2+(2y^2+4y)(w-y^2)$
$w\dfrac{dw}{dy}-2yw=2w^2+(4y-2y^2)w-4y^3$
$w\dfrac{dw}{dy}=2w^2+(6y-2y^2)w-4y^3$
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 $\dfrac{dw}{dy}=-\dfrac{1}{z^2}\dfrac{dz}{dy}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{dy}=\dfrac{2}{z^2}+\dfrac{6y-2y^2}{z}-4y^3$
$\dfrac{dz}{dy}=4y^3z^3+(2y^2-6y)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
