The solution of the differential equation,
$\frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{x}-\mathrm{y})^{2}$, when $\mathrm{y}(1)=1$, is :-
Correct Option: , 4
$x-y=t \Rightarrow \frac{d y}{d x}=1-\frac{d t}{d x}$
$\Rightarrow 1-\frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{t}^{2} \Rightarrow \int \frac{\mathrm{dt}}{1-\mathrm{t}^{2}}=\int 1 \mathrm{dx}$
$\Rightarrow \frac{1}{2} \ln \left(\frac{1+\mathrm{t}}{1-\mathrm{t}}\right)=\mathrm{x}+\lambda$
$\Rightarrow \frac{1}{2} \ln \left(\frac{1+x-y}{1-x+y}\right)=x+\lambda \quad$ given $\quad y(1)=1$
$\Rightarrow \frac{1}{2} \ell n(1)=1+\lambda \Rightarrow \lambda=-1$
$\Rightarrow \ln \left(\frac{1+x-y}{1-x+y}\right)=2(x-1)$
$\Rightarrow-\ell n\left(\frac{1-x+y}{1+x-y}\right)=2(x-1)$
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