Question

If $dx+dy=(x+y)(dx-dy)\Rightarrow y+\log (x+y)=x+C$ is the solution of differential equation

Answer 1

Given,

$dx+dy=(x+y)(dx-dy)$

$-(x+y-1)dx=-(x+y+1)dy$

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

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

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

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

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

integrating on both sides, we get,

$\int \frac{v+1}{2v}dv=\int dx$

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

$x+y+\log (x+y)=2x+C$

$y+\log (x+y)=x+C$