### 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:

$$y=\pm \sqrt{Cx^3-x^2}$$