Question

Solve:
$\frac{dy}{dx}=\frac{x+y+1}{x+y-1}$ when $y=\frac{1}{3}$ at $x=\frac{2}{3}$

Answer 1

Given,

$\frac{dy}{dx}=\frac{x+y+1}{x+y-1}$

substitute $v=x+y\to \frac{dv}{dx}=1+\frac{dy}{dx}$

$\frac{dv}{dx}-1=\frac{v+1}{v-1}$

$\frac{dv}{dx}=\frac{v+1}{v-1}+1$

$\frac{dv}{dx}=\frac{2v}{v-1}$

$\frac{v-1}{2v}dv=dx$

integrating on both sides, we get,

$\frac{1}{2}(v-\log v)=x+c$

$\frac{1}{2}(x+y-\log (x+y\left)\right)=x+c$

given, $y=\frac{1}{3},x=\frac{2}{3}$

$\frac{1}{2}(\frac{2}{3}+\frac{1}{3}-\log (\frac{2}{3}+\frac{1}{3}))=\frac{2}{3}+c$

$\frac{1}{2}(1-\log 1)=\frac{2}{3}+c$

$\frac{1}{2}-\frac{2}{3}=c\Rightarrow c=-\frac{1}{6}$

$\therefore \frac{1}{2}(x+y-\log (x+y\left)\right)=x-\frac{1}{6}$

$6x+6y-6\log (x+y)=2x-2$

$4x+6y-6\log (x+y)=-2$