Any help on one ODE Some weeks ago I was asked to solve one ODE. I tried all methods I know, but couldn't crack this equation. Also I tried to use Matlab's dsolve function - without any result.
$y' + \frac{2x}{y} = x^2$
Does anyone have any suggestions on this?
P.S. This is not for homework or anything like this, just want to know if this equation is integrable?
 A: If you mean a (real) analytical solution with $y(0)=0$, then the answer is 'no'.  If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, you'll immediately see that the origin is an isolated singular point of elliptic type (i.e., the eigenvalues of the linearization at $(x,y)=(0,0)$ are pure imaginary).  In particular, there is no nontrivial integral curve of $\omega$ that passes through the origin.
All of the integral curves of $\omega$ near the origin spiral around it and converge inward as they go counterclockwise around the origin.  To see this, just consider the vector field
$$
Z = (2x - x^2y)\frac{\partial\ }{\partial y} - y\frac{\partial\ }{\partial x}
$$
which is tangent to the integral curves of $\omega$, and consider the convex function $H = 2x^2 + y^2$.  The derivative of $H$ with respect to $Z$ is $-2x^2y^2$, which is non-positive and vanishes only on the axes, which are not tangent to $Z$.  Since $H$ is strictly decreasing along each integral curve of $Z$, the integral curves of $Z$ spiral in to the origin.
Added note about complex solutions:  I neglected to mention that, over the complex numbers, there are, of course, two solutions satisfying $y(0)=0$.  I doubt that there is a closed form expression for them in elementary terms, but one easily obtains that they have convergent power series expansions of the form
$$
y_{\pm}(x) = {} \pm i\sqrt{2}\ x f_0(x^4) + x^3 f_1(x^4),
$$
where 
$$
f_0(t) = 1 - \frac{t}{3\cdot2^6} + \frac{13\ t^2}{5\cdot3^2\cdot2^{13}} - \cdots
$$
and 
$$
f_1(t) = \frac{1}{2^2} + \frac{t}{3\cdot2^9} - \frac{11\ t^2}{5\cdot3^3\cdot2^{15}} + \cdots.
$$
A: If it is any help, $ y = \pm \sqrt{2} (k-x^2)^{1/2}$ are homogeneous solutions, that is, they solve $$y^\prime + \dfrac{2x}{y} = 0$$
A: Maple classifies this as an Abel equation of the second kind, class A: see 
http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Abel2A
Maple does not find a closed-form solution, moreover the equation has no symmetries according to symgen.
