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? 
 A: $$2xy\mathrm dx+(x^2+y^2)\mathrm dy = 0$$
$$2xy \mathrm dx=-(x^2+y^2)\mathrm dy$$
$$\frac{dx}{dy}=-\frac{(x^2+y^2)}{2xy}$$
Now put, $x=yu \Rightarrow \mathrm dx=u\mathrm dy+y\mathrm du$, i.e. 
$\frac {dx}{dy}=u+y\frac{du}{dy}$
Using these substitution and after simplification a you will get that
$$\frac{2u}{3u^2+1} \mathrm du=-\frac{1}{y}\mathrm dy$$
On integration you will get that,
$$\frac{1}{3}\ln(u^2+\frac{1}{3})+\ln y=\ln c$$
$$\ln {(u^2+\frac{1}{3})y^3=\ln c^3}$$
$$(u^2+\frac{1}{3})y^3=C$$ Now replace back $u=x/y$, you will get your answer.
A: The variables $y$ and $u$ can be separated from each other:
$$
2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0
$$
$$
2u\left( du + u \, \frac{dy} y \right) + (u^2+1 ) \, \frac{dy} y = 0
$$
$$
2u\left( \frac{du} u + \frac{dy} y \right) + \left( u + \frac 1 u \right) \, \frac{dy} y = 0
$$
$$
2 \left( \frac{du} u + \frac{dy} y \right) + \left( 1 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
2 \frac{du} u + 2 \frac{dy} y + \left( 1 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
2 \frac{du} u + \left( 3 + \frac 1 {u^2} \right) \, \frac{dy} y = 0
$$
$$
\left( 3 + \frac 1 {u^2} \right) \, \frac{dy} y = -2 \frac{du} u
$$
$$
\frac{dy} y = \frac{-2 \, du/u}{3 + \frac 1 {u^2}}
$$
$$
\frac{dy} y = \frac{-2 u\, du}{3u^2 + 1}
$$
$$
\log|y| = - \frac 1 3 \log|3u^2 + 1| + \text{constant}
$$
$$
y = (3u^2+1)^{-1/3}\cdot\text{constant}
$$
$$
y^3 = \frac{\text{constant}}{3u^2+1}
$$
A: Yes, you get $dy/y+ 2u/(3u^2+ 1)du= 0$. 
The integral of $dy/y$ is $ln(|y|)$ and the integral of $2u/(3u^2+ 1)du$, using the substitution $v= 3u^2+ 1$ so that $dv= 6udu$ or $2udu= dv/3$, is the integral of $1/v dv/3$ which is $$(1/3)ln(|v|)= (1/3)ln(3u^2+ 1)= (1/3)ln(3x^2/y^2+ 1)$$
So integrating $dy/y+ 2u/(3u^2+ 1)= 0$ becomes $$ln(|y|)+ (1/3)ln(3x^2/y^2+ 1)= C$$ or, equivalently, $-3ln(|y|)= ln(3x^2/y^2+ 1)+ C$.
Now, taking the exponential of both sides $y^{-3}= C(3x^2/y+ 1)$.
Finally, multiply on both sides by $y^3$ to get $1= C(3x^2y+ y^3)$ which is the same as $1/C= C'= 3x^2y+ y^3$.
