Answer
The solution of the differential equation is given by
$$
| (y+2x)|^{3}.| (y-2x)|= c
$$
where c is an arbitrary constant.
Work Step by Step
$$
y^{'}=\frac{dy}{dx}=\frac{y-4x}{x-y} \quad (*)
$$
Divide each term in the numerator and denominator by the highest power of x, namely x. We obtain
$$
y^{'}=\frac{dy}{dx}=\frac{\frac{y}{x}-4}{1-\frac{y}{x}} =f(x,y)
$$
where $f(x,y)$ can be expressed as a function of the ratio $\frac{y}{x}$ only .
Thus eq. (*) is homogeneous.
Now we assume a new dependent variable $ v = \frac{y}{x }$
and we can express
$$
y^{'}=\frac{dy}{dx}= x \frac{dv}{dx}+v = \frac{v-4}{1-v}
$$
$$
x \frac{dv}{dx} = \frac{v-4}{1-v} -v= \frac{v^{2}-4}{1-v} \quad (**)
$$
The differential equation (**) is separable and can be written as
$$
[\frac{1-v}{{v^{2}-4}}[ dv= \frac{dx}{x}
$$
by using partial fractions and some simplifications the previous differential equation can be written as
$$
[\frac{\frac{1}{4}}{{v-2}}+\frac{\frac{-1}{4}}{{v+2}}-\frac{1}{2}(\frac{2v}{v^{2}-4})] dv= \frac{dx}{x}
$$
integrating the left side with respect to $v$ and the right side with respect to $x$ gives
$$
\frac{1}{4} \int{\frac{dv}{v-2}}-\frac{1}{4} \int{\frac{dv}{v+2}}-\frac{1}{2} \int {\frac{2v dv}{v^{2}-4}}= \int{\frac{dx}{x}}
$$
we obtain
$$
\frac{1}{4} ln | (v-2)|-\frac{1}{4} ln | (v+2)|- \frac{1}{2} ln | (v^{2}-4)|= ln | x |+ ln | c |
$$
where c is an arbitrary constant.
simplification
$$
ln | \frac{1}{(v+2)^{3} (v-2)}|= 4 ln | xc |
$$
we can obtain $v$ implicitly in terms of $ x$ as follows
$$
| \frac{1}{(v+2)^{3} (v-2)}|= c| x |^{4}
$$
we find the solution of Eq. (*) by replacing $v$ by $ y/x $ in the previous expression. Hence the solution of eq. (*) is given by
$$
| (y+2x)|^{3}.| (y-2x)|= c
$$
where c is an arbitrary constant.
