Solve $2xyy^\prime=x^2+2y^2$ 
Solve $2xyy^\prime=x^2+2y^2$

I tried to solve with substitution $v=y^2$
then $\frac{dy}{dx}=\frac{1}{2}v^{-1/2}\frac{dv}{dx}$
and substituting the given equation I get $$\frac{x^2+v}{2x\sqrt{v}}=\frac{1}{2\sqrt{v}}\frac{dv}{dx}$$
$$\frac{x^2+v}{x}=\frac{dv}{dx}$$
$$\frac{dv}{dx}=x+\frac{v}{x}$$
But now it seems like I just need to make another substitution to solve this.

Note that you have a linear equation
\begin{eqnarray*}
\frac{dv}{dx} +P v =Q 
\end{eqnarray*}
where $P$ and $Q$ are functions of $x$ only. We multiply through by a function $R$ in order to make the LHS an exact diffeential of a product
\begin{eqnarray*}
R\frac{dv}{dx} +RP v =RQ \\
\frac{d}{dx} (Rv) = R \frac{dv}{dx} +R'v.
\end{eqnarray*}
So we need $R'=RP$ or $R=e^{\int P dx }$ ... This should shed some light on how we get the second line below.
You lost a factor of $2$ somewhere. We have
\begin{eqnarray*}
 x\frac{dv}{dx}=x^2+2v \\
 \frac{1}{ x^2}\frac{dv}{dx} - \frac{2v}{x^3}=\frac{1}{x} \\
\frac{d}{dx} \left( \frac{v}{x^2} \right) =\frac{1}{x} \\
 \frac{v}{x^2} =\ln(x)+\ln(A) \\
v=x^2 \ln(Ax).
\end{eqnarray*}
