Given $(x+y)(dx-dy)=dx+dy$
This can be written as
$xdx-xdy+ydx-ydy=dx+dy$
=> $ dx(x+y-1)=dy(x+y+1)$
(ie)$dx(x+y-1)=dy(x+y+1)$
=>$\large\frac{dy}{dx}=\large\frac{x+y-1}{x+y+1}$-----(1)
Let $x+y=z$ on differentiating w.r.t x we get
$1+\large\frac{dy}{dx}=\frac{dz}{dx}$
or $ \large\frac{dy}{dx}=\frac{dz}{dx}-1$
Now substituting in equ (1) we get,
$\large\frac{dz}{dx}$$-1=\large\frac{z-1}{z+1}$
=>$\large\frac{dz}{dx}=\frac{z-1}{z+1}$$+1$
$=\large\frac{z-1+z+1}{z-1}$
$\large\frac{dz}{dx}=\frac{2z}{z-1}$
On seperating the variables we get
$\bigg(\large\frac{z-1}{2z}\bigg)$$dz=dx$
=>$\large\frac{1}{2}\bigg(\frac{z-1}{z}\bigg)$$dz=dx$
On integrating we get
$\large\frac{1}{2}\bigg[\int \frac{z}{z} $$dz-\int \large\frac{1}{z}$$dz\bigg]=\int dx$
=>$\large\frac{1}{2}\bigg[\int $$dz-\int \large\frac{1}{z}$$dz\bigg]=\int dx$
$\large\frac{1}{2}\bigg[$$z-\log z\bigg]=x+c$
$z-\log z=2x+c$
Now substituting for $z=(x+y)$
$x+y-\log(x+y)=2x+c$
$y=x+\log (x+y)+c$
or $y-x=\log _e (x+y)$
$c\;e^{\large y-x}=x+y$