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

none of these

The correct answer is option (4) : none of these

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

So, the given differential equation becomes

$\left(\frac{u-1}{u-2}\right) \left(\frac{du}{dx} - 1\right) = \frac{u+1}{u+2}$

$⇒\frac{du}{dx} - 1= \left(\frac{u+1}{u+2}\right)\left(\frac{u-2}{u-1}\right)$

$⇒\frac{du}{dx} = \frac{u^2-u-2}{u^2+u-2}+1$

$⇒\frac{du}{dx} = \frac{2u^2-4}{u^2+u-2}$

$⇒\frac{u^2+u-2}{u^2-2} du = 2dx$

$⇒\left(1+\frac{u}{u^2-2}\right) du = 2dx$

On integrating , we get

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

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

$⇒2(y-x) + log |(x+y)^2 -2|= 2C$ .........(i)

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

$∴log 2 = 2C$

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

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

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