Separating variables by substitution in a homogenous ODE I am brand new to ODE's, and have been having difficulties with this practice problem. Find a 1-parameter solution to the homogenous ODE:$$2xy \, dx+(x^2+y^2) \, dy = 0$$assuming the coefficient of $dy \ne 0$
The textbook would like me to use the subsitution $x = yu$ and $dx=y \, du + u \, dy,\  y \ne 0$
Rewriting the equation with the subsititution:
$$2uy^2(y \, du + u \, dy)+(x^2+y^2) \, dy = 0$$
divide by $y^2$
$$2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0$$
but after further simplification I end up getting ${dy \over y}$ which would mean I would get a logarithm after integrating, and the answer is given as $$3x^2y+y^3 = c$$
Could I get some help/hints as to how this answer was obtained? 

The variables $y$ and $u$ can be separated from each other:
$$
2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0
$$
$$
2u\left( du + u \, \frac{dy} y \right) + (u^2+1 ) \, \frac{dy} y = 0
$$
$$
2u\left( \frac{du} u + \frac{dy} y \right) + \left( u + \frac 1 u \right) \, \frac{dy} y = 0
$$
$$
2 \left( \frac{du} u + \frac{dy} y \right) + \left( 1 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
2 \frac{du} u + 2 \frac{dy} y + \left( 1 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
2 \frac{du} u + \left( 3 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
\left( 3 + \frac 1 {u^2} \right) \, \frac{dy} y = -2 \frac{du} u
$$
$$
\frac{dy} y = \frac{-2 \, du/u}{3 + \frac 1 {u^2}}
$$
$$
\frac{dy} y = \frac{-2 u\, du}{3u^2 + 1}
$$
$$
\log|y| = - \frac 1 3 \log|3u^2 + 1| + \text{constant}
$$
$$
y = (3u^2+1)^{-1/3}\cdot\text{constant}
$$
$$
y^3 = \frac{\text{constant}}{3u^2+1}
$$
