Give the correct answer and solve the given equation displaystyle{left({x}-{y}right)}{left.{d}{x}right.}+{left({3}{x}+{y}right)}{left.{d}{y}right.}={0},text{when} {x}={3},{y}=-{2}

The given diferential equation is
${\displaystyle (x-y)dx+(3x+y)dy=0,}$
with ${\displaystyle x=3,y=2}$ Now
${\displaystyle (x-y)dx+(3x+y)dy=0,\Rightarrow \frac{dy}{dx}=\frac{y-x}{3x=y}}$
since the given differential equation is a homogenuous equation, putting ${\displaystyle y=vx}$, we have
${\displaystyle \frac{dy}{dx}=v+xd\frac{v}{dx}}$
So the equation becomes.
${\displaystyle v+xd\frac{v}{dx}=\frac{v-1}{v+3}}$
${\displaystyle \Rightarrow xd\frac{v}{dx}=\frac{v-1}{v+3}-v=-\frac{{(v+1)}^2}{v+3}}$
${\displaystyle \Rightarrow \frac{v+3}{{(v+1)}^2}dv+\frac{dx}{x}=0}$
${\displaystyle \Rightarrow \frac{v+1+2}{{(v+1)}^2}dv+\frac{dx}{x}=0}$
${\displaystyle \Rightarrow d\frac{v}{(v+1)}+\frac{2dv}{{(v+1)}^2}+\frac{dx}{x}=0}$
${\displaystyle \Rightarrow \int \frac{dx}{(v+1)}+\int \frac{2dv}{{(v+1)}^2}+\int \frac{dx}{x}=C}$ where C is a integrating constant
${\displaystyle \Rightarrow \ln (v+1)=\frac{2}{(v+1)}+\ln \left(x\right)=C}$
${\displaystyle \Rightarrow \ln \left(x(v+1)\right)-\frac{2}{(v+1)}=C}$
putting ${\displaystyle v=\frac{y}{x}}$ we have
${\displaystyle \Rightarrow \ln (y+x)-2\frac{x}{(x+y)}=C}$
Since ${\displaystyle x+3,y+2}$, we have ${\displaystyle C=\ln 5-\frac{6}{5}}$
Hence the solution of the equation is
${\displaystyle \Rightarrow \ln (y+x)-2\frac{x}{(x+y)}=\ln 5-\frac{6}{5}}$