Solving differential equation - how to find inhomogeneous solution Given the following
$\ y'= \frac{3y^2-x^2}{2xy}$
I need to tell if the equation is linear, which I think it is because:
$\ y'= \frac{3y^2-x^2}{2xy} = \frac{3y}{2x}-\frac{x}{2y}$
Now I need to solve the equation with separation of variables which is only possible with substitution. So I substitute with $\ u(x)= \frac{y}{x}$ and do the following
$\ y'= \frac{3y}{2x}-\frac{x}{2y} => y'=\frac{3y}{2x}-\frac{1}{2u(x)} = \frac{3}{2x}y-\frac{1}{2u(x)}$
so I can use separation of variables and get the homogeneous solution and the inhomogeneous. For the homogeneous I got
$\ y_h=e^cx^{3/2} = Cx^{3/2} $
but I really don't know how to get the inhomogeneous solution because I don't exactly know what to do with the substitution, I am glad for help.
 A: A start:   The DE is not linear. But it is homogeneous. A standard way to begin is to let $y=ux$. Then $\frac{dy}{dx}=\frac{du}{dx}+u$. The right-hand side simplifies to $\frac{3u^2-1}{2u}$. After some further simplification the equation can be written as $2u\frac{du}{dx}=u^2-1$.  This can be solved by separation of variables.
A: $$y'(x)=\frac{3y(x)^2-x^2}{2xy(x)}\Longleftrightarrow$$
$$y'(x)-\frac{3y(x)}{2x}=-\frac{x}{2y(x)}\Longleftrightarrow$$
$$2y(x)y'(x)-\frac{3y(x)^2}{2x}=-x\Longleftrightarrow$$

Let $r(x)=y(x)^2$, which gives $r'(x)=2y(x)y'(x)$:

$$r'(x)-\frac{3r(x)}{x}=-x\Longleftrightarrow$$

Let $v(x)=\exp\left[\int-\frac{3}{x}\space\text{d}x\right]=\frac{1}{x^3}$:

$$\frac{r'(x)}{x^3}-\frac{3r(x)}{x^4}=-\frac{1}{x^2}\Longleftrightarrow$$

Substitute $-\frac{3}{x^4}=\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^3}\right)$:

$$\frac{r'(x)}{x^3}-\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^3}\right)r(x)=-\frac{1}{x^2}\Longleftrightarrow$$

Apply the reverse product rule:

$$\frac{\text{d}}{\text{d}x}\left(\frac{r(x)}{x^3}\right)=-\frac{1}{x^2}\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}x}\left(\frac{r(x)}{x^3}\right)\space\text{d}x=\int-\frac{1}{x^2}\space\text{d}x\Longleftrightarrow$$
$$\frac{r(x)}{x^3}=\frac{1}{x}+\text{C}\Longleftrightarrow$$
$$r(x)=x^2\left(\text{C}x+1\right)\Longleftrightarrow$$
$$y(x)^2=x^2\left(\text{C}x+1\right)\Longleftrightarrow$$
$$y(x)=\pm\sqrt{x^2\left(\text{C}x+1\right)}\Longleftrightarrow$$
$$y(x)=\pm x\sqrt{\text{C}x+1}$$
