The solution of the differential equation,

The given differential equation

$\frac{d y}{d x}=(x-y)^{2}$ .......(1)

Let $x-y=t \Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$

$\Rightarrow \quad \frac{d y}{d x}=1-\frac{d t}{d x}$

Now, from equation (1)

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

$\Rightarrow \quad 1-t^{2}=\frac{d t}{d x} \Rightarrow \int d x=\int \frac{d t}{1-t^{2}}$

$\Rightarrow \quad-x=\frac{1}{2 \times 1} \ln \left|\frac{t-1}{t+1}\right|+c$

$\Rightarrow \quad-x=\frac{1}{2} \ln \left|\frac{x-y-1}{x-y+1}\right|+c$

$\because$ The given condition $y(1)=1$

$-1=\frac{1}{2} \ln \left|\frac{1-1-1}{1-1+1}\right|+c \Rightarrow c=-1$

Hence, $2(x-1)=-\ln \left|\frac{1-x+y}{1-y+x}\right|$