Answer
the solution of differential equation is given by
$$
| (y-x)|=| (y+3x)|^{5} c
$$
where c is an arbitrary constant.
Work Step by Step
$$
y^{'}=\frac{dy}{dx}=\frac{4y-3x}{2x-y} \quad (*)
$$
Divide each term in the numerator and denominator by the highest power of $ x$ , namely $x$. We obtain
$$
y^{'}=\frac{dy}{dx}=\frac{4(\frac{y}{x})-3}{2-(\frac{y}{x})} =f(x,y)
$$
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
$$
y^{'}=\frac{dy}{dx}= x \frac{dv}{dx}+v = \frac{4v-3}{2-v}
$$
$$
x \frac{dv}{dx} = \frac{4v-3}{2-v}-v= \frac{v^{2}+2v-3}{2-v} \quad (**)
$$
The differential equation (**) is separable and can be written as
$$
[\frac{2-v}{{{v^{2}+2v-3}}}] dv= \frac{dx}{x}
$$
by using partial fractions and some simplifications the provisos differential equation can be written as
$$
[\frac{\frac{-5}{4}}{{v+3}}+\frac{\frac{1}{4}}{{v-1}}] dv= \frac{dx}{x}
$$
integrating the left side with respect to $v$ and the right side with respect to $x$ gives
$$
\frac{-5}{4} \int{\frac{dv}{v+3}}+\frac{1}{4} \int{\frac{dv}{v-1}}= \int{\frac{dx}{x}}
$$
we obtain
$$
\frac{-5}{4} ln | (v+3)|+\frac{1}{4} ln | (v-1)|= ln | x |+ ln | c |
$$
where c is an arbitrary constant.
simplification
$$
ln | \frac{(v-1)}{(v+3)^{5} }|= 4 ln | xc |
$$
we can obtain $v$ implicitly in terms of $ x$ as the follows
$$
\frac{ |(v-1)|} {|(v+3)^{5} |}= | x |^{4}c
$$
we find the solution of Eq. (*) by replacing $v$ by $ y/x $ in the previous expression. Hence the solution of eq. (*) is given by
$$
\frac{ |(\frac{y}{x}-1)|} {|(\frac{y}{x}+3)^{5} |}= | x |^{4}c
$$
this implies that
$$
\frac{ |(\frac{y-x}{x})|} {|(\frac{y+3x}{x})^{5} |}= | x |^{4}c
$$
Hence the solution of eq. (*) given by
$$
| (y-x)|=| (y+3x)|^{5} c
$$
where c is an arbitrary constant.
