How to solve the following differential equation We have the following DE: $$ \dfrac{dy}{dx} = \dfrac{x^2 + 3y^2}{2xy}$$
I don't know how to solve this. I know we need to write it as $y/x$ but I don't know how to in this case. 
 A: $y=vx$ so $y'=xv'+v$. Your equation is $xv'+v={{1 \over{2v}}+{{3v} \over {2}}}$.
Now clean up to get $xv'={{v^2+1}\over{2v}}$. Now separate ${2v dv \over {v^2+1}} = {dx \over x}$.
Edit
${{x^2+3y^2} \over {2xy}} = {{{x^2}\over{2xy}}+{{3y^2}\over{2xy}}}={ x \over {2y}}+{{3y}\over{2x}}$
A: You are asked to solve, in terms of $\frac{y}{x}$, the following ODE:
$$\frac{\operatorname{d}\!y}{\operatorname{d}\!x} = \frac{x^2+3y^2}{2xy}$$
The way to do this is to make a substitution. Solving differential equations is like grown-up integration. Just as we used integration by substitution to solve ordinary integrals, we can do the same to solve ODEs. 
The dependent variable in your ODE is $y$ and so we aim to replace it. Let us call $\frac{y}{x}=u$. We want to replace all of the $y$s, and so we use $y=ux$. As with integration by substitution, $u$ becomes our new independent variable. Let us see what we can do:
$$\frac{\operatorname{d}\!y}{\operatorname{d}\!x}= \frac{\operatorname{d}(ux)\!}{\operatorname{d}\!x}=x\frac{\operatorname{d}\!u}{\operatorname{d}\!x}+u$$
$$\frac{x^2+3y^2}{2xy} = \frac{x^2+3u^2x^2}{2x\,ux} = \frac{1+3u^2}{2u}$$
Putting these two substitutions together gives:
$$\frac{\operatorname{d}\!y}{\operatorname{d}\!x} = \frac{x^2+3y^2}{2xy} \implies x\frac{\operatorname{d}\!u}{\operatorname{d}\!x}+u=\frac{1+3u^2}{2u} \implies x\frac{\operatorname{d}\!u}{\operatorname{d}\!x}=\frac{1+u^2}{2u}$$
Taking the reciprocal of both sides we see that:
$$\frac{1}{x}\frac{\operatorname{d}\!x}{\operatorname{d}\!u}=\frac{2u}{1+u^2} \implies \int \frac{\operatorname{d}\!x}{x} = \int \frac{2u\operatorname{d}\!u}{1+u^2} \implies \ln|x|=\ln|k(1+u^2)|$$
It follows that $x=k(1+u^2)$ for some constant of integration. Given that $u=\frac{y}{x}$ we have:
$$x=k(1+u^2) \implies x=k\left(1+\frac{y^2}{x^2}\right) \implies y = \pm \frac{x}{k}\sqrt{k(x-k)}$$
