Solution of the differential equation $\left(\frac{x+y-1}{x+y-2}\right) \frac{d y}{d x}=\left(\frac{x+y+1}{x+y+2}\right)$, given that $y=1$ when $x=1$, is

none of these

Let $x+y=u$. Then, $1+\frac{d y}{d x}=\frac{d u}{d x}$. So, the given differential equation becomes

$\left(\frac{u-1}{u-2}\right)\left(\frac{d u}{d x}-1\right)=\frac{u+1}{u+2}$

$\Rightarrow \frac{d u}{d x}-1=\left(\frac{u+1}{u+2}\right)\left(\frac{u-2}{u-1}\right)$

$\Rightarrow \frac{d u}{d x}=\frac{u^2-u-2}{u^2+u-2}+1$

$\Rightarrow \frac{d u}{d x}=\frac{2 u^2-4}{u^2+u-2}$

$\Rightarrow \frac{u^2+u-2}{u^2-2} d u=2 d x$

$\Rightarrow \left(1+\frac{u}{u^2-2}\right) d u=2 d x$

On integrating, we get

$u+\frac{1}{2} \log \left|u^2-2\right|=2 x+C$

$\Rightarrow (x+y)+\frac{1}{2} \log \left|(x+y)^2-2\right|=2 x+C$

$\Rightarrow 2(y-x)+\log \left|(x+y)^2-2\right|=2 C$       ....(i)

When $x=1$, we have $y=1$

∴  $\log 2=2 C$

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

$2(y-x)+\log \left|(x+y)^2-2\right|=\log 2$

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