# 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}

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