Solving $(xy)y'= x^2+3y^2$ I am having a very frustrating time with the back book that says my answer is way off but to me everything looks fine:
\begin{align*}
(xy)y'&= x^2+3y^2\\
y' &= \frac{x^2}{xy} + \frac{3y^2}{xy}\\
y' &= \frac{x}{y} + \frac{3y}{x}\\
y' &= \frac{1}{v} + 3v\\
y' &= \frac{1 + 3v^2}{v}\\
v+\frac{dv}{dx}x &= \frac{1+3v^2}{v}\\
\frac{dv}{dx}x&= \frac{1+3v^2-v^2}{v}\\
\frac{dv}{dx}x &= \frac{1+2v^2}{v}\\
\int \frac{v}{2v^2+1}\,dv &= \int\frac{1}{x}\,dx\\
u &= 2v^2+1\\
du &= 4v\,dv\\
dv &= \frac{1}{4v}\,du\\
\int \frac{v}{u} \frac{1}{4v}\,du &= \int \frac{1}{x} \,dx\\
\int \frac{1}{4u}\,du &= \ln|x| + c\\
\frac{1}{4} \int \frac{1}{u}\,du &= \ln|x| +c\\
\frac{1}{4} \ln|2v^2 + 1| &= \ln |x| + c\\
\ln|2v^2 + 1|&= 4\ln|x|+c\\
2v^2 + 1 &= e^{4\ln|x|}e^c\\
2v^2 + 1 &= Cx^4\\
2v^2 &= Cx^4\\
v^2 &= Cx^4\\
\frac{y}{x} &= \sqrt{Cx^4}\\
y &= x\sqrt{Cx^4}
\end{align*}
However the book says the answer is $x^2 + 2y^2 = Cx^6.$ I am fairly sure there are no mistakes.
 A: Where did the $+1$ go?
$$2v^2+1=Cx^4$$
$$\frac{2y^2}{x^2}+1=Cx^4$$
$$2y^2+x^2=Cx^6$$
$$x^2+2y^2=Cx^6$$
A: Hint.
A way to solve this DE is by making $z = y^2$ and then
$$
\frac 12 x z' - 3z = x^2
$$
which is a linear DE. Solving for $z$ we have
$$
z = y^2 = C_1 x^6-\frac 12x^2
$$
NOTE
The homogeneous part
$$
\frac 12 x z'_h -3z_h = 0
$$
is separable giving
$$
z_h = C_0 x^6
$$
A: As you set 
$$v=\frac{y}{x}$$
you need to substitute it back into 
$$2v^2+1=Cx^4$$ 
which forms
$$2\Big(\frac{y}{x}\Big)^2 + 1 = Cx^4 \implies x^2+2y^2 = Cx^6$$
A: Your solution simplifies to the cubical parabola $$y=kx^3.$$ Substituting this into the original equation with $k=1$ for simplicity shows something must be wrong despite your confidence to the contrary, for we have on the one hand $(xy)y'=x^4(3x^2)=3x^6,$ and on the other side $x^2+3y^2=x^2+3x^6,$ which are clearly not equal, off in fact by $x^2.$
The most suspicious operation is where you subsumed the $+1$ into the constant -- wrongly, it appears. There's no legitimate way you could have managed that! 
