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?



Any help on one ODE
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 
$$
f\_1(t) = \frac{1}{2^2} + \frac{t}{3\cdot2^9} - \frac{11\ t^2}{5\cdot3^3\cdot2^{15}} + \cdots.
$$


