Solution of the differential equation 

$\frac{x+y-1}{x+y-2}\frac{dy}{dx}=\frac{x+y+1}{x+y+2}\text{ suchthat }y=1\text{ when }x=1$

Let $x+y=u.$ Then, $1+\frac{dy}{dx}=\frac{du}{dx}$

So, the given differential equation becomes 

$\begin{array}{l}\left(\frac{u-1}{u-2}\right)\left(\frac{du}{dx}-1\right)=\frac{u+1}{u+2}\\ \Rightarrow \left.\frac{du}{dx}-1=\frac{u+1}{u+2}\right)\left(\frac{u-2}{u-1}\right)\\ =\frac{du}{dx}=\frac{u^2-u-2}{u^2+u-2}+1\\ \Rightarrow \frac{du}{dx}=\frac{2u^2-4}{u^2+u-2}\\ =\frac{u^2+u-2}{u^2-2}du=2dx\\ \end{array}$

On integrating, we get 

$\begin{array}{l}u+\frac{1}{2}\log \left|u^2-2\right|=2x+C\\ \Rightarrow (x+y)+\frac{1}{2}\log \left|(x+y\mid {)}^2-2\right|=2x+C\\ \Rightarrow 2(y-x)+\log \left|(x+y{)}^2-2\right|=2C\end{array}$

When $x=1,$ we have $y=1$

$\therefore \log 2=2C$

Substituting the value of C in (i), we get

$\begin{array}{r}2(y-x)+\log \left|(x+y{)}^2-2\right|=\log 2\\ \Rightarrow 2(y-x)+\log \left|\frac{(x+y{)}^2-2}{2}\right|=0\end{array}$