Concept:
Given a homogeneous 1st order differential equation of the form:
$$\frac{dy}{dx}=F(\frac{y}{x})$$
...where F is a function of y/x
...we can make a substitution via the following:
$$let\;v=\frac{y}{x}$$
$$y=vx $$
...and via the product rule:
$$\frac{dy}{dx}=v'x+x'v\;,\;x'=1 $$
therefore:
$$\frac{dy}{dx}=v'x+v $$
(Ex 1) Find the general solution of the following:
$$\frac{dy}{dx}=\frac{x^2+3y^2}{2xy} $$
First, attempt to make the right side of the equation a function solely of y/x. Upon inspection you may realize that it is possible to do this by dividing both the numerator and denominator by x-squared.
$$\frac{dy}{dx}=\frac{\frac{x^2}{x^2}+\frac{3y^2}{x^2}}{\frac{2xy}{x^2}} $$
...which becomes:
$$\frac{1+3(\frac{y}{x})^2}{2(\frac{y}{x})} $$
Now, let:
$$v=\frac{y}{x}\;,\;y=vx$$
$$\frac{dy}{dx}=v+v'x$$
...and make the appropriate substitutions:
$$v+x\frac{dv}{dx}=\frac{1+3v^2}{2v}$$
The above equation is now separable. Go ahead and solve:
$$2v^2+2vx\frac{dv}{dx}=1+3v^2$$
$$2vx\frac{dv}{dx}=1+v^2$$
$$\int_{}{}\frac{2v}{1+v^2}\;dv=\int_{}{}\frac{1}{x}\;dx$$
$$ln|1+v^2|=ln|x|+C$$
Since C is an arbitrary constant at this point, we can write:
$$ln(1+v^2)=ln\;x+ln\;C$$
...and recalling one of our log rules, the above can be written as:
$$ln(1+v^2)=ln(Cx)$$
...which becomes:
$$1+v^2=Cx$$
Remember that we are trying to find a general solution for y. So at this point we can go ahead and back-substitute for v in the above equation:
$$1+(\frac{y}{x})^2=Cx$$
$$1+\frac{y^2}{x^2}=Cx$$
$$\frac{y^2}{x^2}=Cx-1$$
$$y^2=Cx^3-x^2$$
...and finally, the general solution to our original differential equation is represented by:
$$y=\pm \sqrt{Cx^3-x^2}$$