The solution of the differential equation, dydx=(x−y)2, when y(1)=1 is

A
$- \log_{e} |\frac{1 + x - y}{1 - x + y}| = x + y - 2$
B
$\log_{e} |\frac{2 - x}{2 - y}| = x - y$
C
$- \log_{c} |\frac{1 - x + y}{1 + x - y}| = 2 (x - 1)$
D
$\log_{c} |\frac{2 - y}{2 - x}| = 2 (y - 1)$

Solution:

Given, $\frac{d y}{d x} = (x - y)^{2}$

Let, $x - y = t \Rightarrow \frac{d y}{d x} = 1 - \frac{d t}{d x} ∴ 1 - \frac{d t}{d x} = t^{2}$

$\begin{array}{l} \int \frac{d t}{1 - t^{2}} = \int d x \Rightarrow \frac{1}{2} \log_{e} |\frac{1 + t}{1 - t}| = x + c \\ \frac{1}{2} \log_{e} |\frac{1 + x - y}{1 - x + y}| = x + C \end{array}$

Given, $y (1) = 1$

$\begin{array}{l} ∴ \frac{1}{2} \log_{e} (1) = C + 1 \Rightarrow C = - 1 \\ ∴ \log_{e} |\frac{1 + x - y}{1 - x + y}| = 2 (x - 1) \Rightarrow - \log_{e} |\frac{1 - x + y}{1 + x - y}| = 2 (x - 1) \end{array}$

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

Solution:

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