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. 
 A: I made some elaborations to your problem: $$
(y')^2=x^2+y^2 \Leftrightarrow (y')^2-y^2=x^2 \Leftrightarrow (y'-y)(y'+y)=x^2 
$$
To find a solution we assume to have a decomposition for $x^2$: $$
\left\{\begin{align*}
y'-y&=p\\ 
y'+y&=x^2p^{-1}
\end{align*}\right.\Leftrightarrow
$$$$
\left\{\begin{align*}
y'&=(p+x^2p^{-1})/2\\ 
y&=(x^2p^{-1}-p)/2
\end{align*}\right.$$
At this point, deriving the second expression:$$
p+x^2p^{-1}=2xp^{-1}-x^2p^{-2}p'-p'
$$$$
(x^2p^{-2}+1)p'=(2x-x^2)p^{-1}-p
$$$$
p'=\frac{(2x-x^2)p-p^3}{x^2+p^2}
$$$$
p'=-p+\frac{2xp}{x^2+p^2}
$$
I don't know if you can find a closed expression for this one (I couldn't), but, at least, I managed to get rid of all the powers for the derivatives.
A: $\left(\dfrac{dy}{dx}\right)^2=x^2+y^2$
$\dfrac{dy}{dx}=\pm\sqrt{x^2+y^2}$
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
A: Most likely there is no closed form for the solution. Nevertheless one can find the first terms of a series to approximate a solution. 
The form of the equation shows that $y(x)$ is an even function. So, we are looking for a solution on the form $y(x)=\sum_{k=0}^{n}a_k (x^2)^k +O(x^{2n+2})$
$$\left(\sum_{k=0}^{n}a_k 2k x^{2k-1}\right)^2=x^2+\left(  \sum_{k=0}^{n}a_k x^{2k} \right)^2+O(x^{2k+2})$$
The expansion and identification of the powers of $x^2$ leads to :
$$y(x)=\frac{1}{2}x^2+\frac{3}{4}\frac{x^4}{4!}+\frac{15}{2^4}\frac{x^6}{6!}+\frac{105}{2^7}\frac{x^8}{8!}+\frac{17325}{2^{11}}\frac{x^{12}}{12!}-\frac{225225}{2^{13}}\frac{x^{14}}{14!}+O(x^{16})$$
Note that the coefficient of $x^{10}$ is $0$.
A: Well, it looks a little like the Ricatti Equation of which the substitution
$$y(x)=\frac{1}{u}\frac{du}{dx}$$
helps in the solution.
Here though, I could work the above ODE down to
$$(u')^{2}-2u(u')^{2}u''+u^{2}(u'')^{2}=u^{4}x^{2}$$
Not sure how helpful this avenue is; maybe an analytic solution isn't easy to come by.
