$${dy\over dx}=-{4x+3y\over 2x+y}$$
If we divide right side of the given equation by $x$, we get:
$${dy\over dx}=-{4+3{y\over x} \over 2+{y\over x}}$$
Hence, given equation is a Homogeneous.$\\$
Let $v={y\over x}$, $v$ is a function of $x$.$\\$
Thus, $${dy\over dx}=-{4+3v\over 2+v}$$
Since $v={y\over x} \rightarrow y=vx$ $\\$
By differentiating both sides with respect to $x$, we get:
$${dy\over dx}=v+x{dv\over dx}=-{4+3v\over 2+v}$$
$$x{dv\over dx}={-(4+3v)\over 2+v}-v$$
$$x{dv\over dx}={-(4+3v)\over 2+v}-{v(2+v)\over 2+v}$$
$$x{dv\over dx}={-(v^2+5v+4)\over 2+v}$$
$${dv\over dx}={-(v+1)(v+4)\over 2+v}\cdot {1\over x}$$
$${2+v\over -(v+1)(v+4)}\cdot {dv\over dx}={1\over x}$$
$$-{1\over x}+{2+v\over -(v+1)(v+4)}\cdot {dv\over dx}=0$$
$${1\over x}+{2+v\over (v+1)(v+4)}\cdot {dv\over dx}=0$$
Notice that above equation has the form of $M(x)+N(y){dy\over dx}=0$ $\\$
Hence the equation is Seperable. $\\$
Rewrite the equation, we get:
$${2+v\over (v+1)(v+4)}dv=-{1\over x}dx$$
Integrating both sides, we get:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{-{1\over x}dx}$$
For $\int{{2+v\over (v+1)(v+4)}dv}$, we use Partial Fraction:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{({A\over v+1}+{B\over v+4})dv}$$
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{A(v+4)+B(v+1)\over (v+1)(v+4)}$$
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{(A+B)v+(4A+B)\over (v+1)(v+4)}$$
Hence, $$\begin{cases} A+B=1 \\4A+B=2\\ \end{cases} \implies
\begin{cases} A={1\over3}\\ B={2\over3} \end{cases}$$
Thus, rewrite gets us:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{{1\over 3(v+1)}+{2\over 3(v+4)}dv}=-\int{1\over x}dx$$
$${1\over 3}ln|v+1|+{2\over3}ln|v+4|=-ln|x|+c$$
where c is arbitrary constant. $\\$
Multiply both sides by $3$, we get:
$$ln|v+1|+ln|v+4|^2=-3ln|x|+3c$$
Now substitute $v={y\over x}$ back in:
$$ln|{y\over x}+1|+ln|{y\over x}+4|^2=-3ln|x|+3c$$
$$ln|{y+x\over x}|+ln|{y+4x\over x}|^2=-3ln|x|+3c$$
$$ln|y+x|-ln|x|+2ln|y+4x|-2ln|x|=-3ln|x|+3c$$
$$ln|y+x|+ln|y+4x|^2=3c$$
$$ln|y+x||y+4x|^2=3c$$
$$exp^{ln|y+x||y+4x|^2}=exp^{3c}$$
$$|y+x||y+4x|^2=c$$
where c is another arbitrary constant.
Therefore, we get the solution to the given differential equation.
