The solution of the differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^2\) when \(y(1)=1\) is

Given differential equation is
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^2\)
Let \(x-y=\mathrm{t}\)
\(\begin{aligned}
\therefore \quad & 1-\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{dt}}{\mathrm{~d} x} \\
& \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=1-\frac{\mathrm{dt}}{\mathrm{~d} x}
\end{aligned}\)
From (i)
\(\begin{aligned}
& 1-\frac{\mathrm{dt}}{\mathrm{~d} x}=\mathrm{t}^2 \\
& 1-\mathrm{t}^2=\frac{\mathrm{dt}}{\mathrm{dx}}
\end{aligned}\)
\(\therefore \quad \mathrm{d} x=\frac{1}{1-\mathrm{t}^2} \mathrm{dt}\)
Integrating on both sides, we get
\(\begin{aligned}
& \int \mathrm{d} x=\int \frac{1}{1-\mathrm{t}^2} \mathrm{dt} \\
& x=\frac{1}{2} \log \left|\frac{1+\mathrm{t}}{1-\mathrm{t}}\right|+\mathrm{c} \\
& x=\frac{1}{2} \log \left|\frac{1+x-y}{1-(x-y)}\right|+\mathrm{c} \\
& x=\frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right|+\mathrm{c}
\end{aligned}\)
\(\text { But } y(1)=1\)
\(\begin{aligned}
& \therefore \quad c=1 \\
& x=\frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right|+1 \\
& x-1=\frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| \\
& 2(x-1)=\log \left|\frac{1+x-y}{1-x+y}\right| \\
& \Rightarrow-\log \left|\frac{1-x+y}{1+x-y}\right|=2(x-1)
\end{aligned}\)