The solution of the differential equation, d y ~d x=(x-y)^2, when y(1)=1, is:

The given differential equation $\frac{d y}{d x}=(x-y)^{2}$dxdy​=(x−y)2 Let $x-y=t \Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$x−y=t⇒1−dxdy​=dxdt​ $\Rightarrow \frac{d y}{d x}=1-\frac{d t}{d x}$⇒dxdy​=1−dxdt​ Now, from equation (1) $\left(1-\frac{d t}{d x}\right)=(t)^{2}$(1−dxdt​)=(t)2 $\displaystyle \Rightarrow \quad 1-t^{2}=\frac{d t}{d x} \Rightarrow \int d x=\int \frac{d t}{1-t^{2}}$⇒1−t2=dxdt​⇒∫dx=∫1−t2dt​ $\Rightarrow-x=\frac{1}{2 \times 1} \ln \left|\frac{t-1}{t+1}\right|+c$⇒−x=2×11​ln​t+1t−1​​+c $\Rightarrow-x=\frac{1}{2} \ln \left|\frac{x-y-1}{x-y+1}\right|+c$⇒−x=21​ln​x−y+1x−y−1​​+c $\because$∵ The given condition $y(1)=1$y(1)=1 $-1=\frac{1}{2} \ln \left|\frac{1-1-1}{1-1+1}\right|+c \Rightarrow c=-1$−1=21​ln​1−1+11−1−1​​+c⇒c=−1 Hence, $2(x-1)=-\ln \left|\frac{1-x+y}{1-y+x}\right|$2(x−1)=−ln​1−y+x1−x+y​​