The solution of the differential equation, \[\dfrac{dy}{dx}={{\left( x-y \right)}^{2}},\] when y(1) = 1, is:
\[\left( a \right){{\log }_{e}}\left| \dfrac{2-y}{2-x} \right|=2\left( y-1 \right)\]
\[\left( b \right){{\log }_{e}}\left| \dfrac{2-x}{2-y} \right|=x-y\]
\[\left( c \right)-{{\log }_{e}}\left| \dfrac{1+x-y}{1-x+y} \right|=x+y-2\]
\[\left( d \right)-{{\log }_{e}}\left| \dfrac{1-x+y}{1+x-y} \right|=2\left( x-1 \right)\]

Hint: First we will simplify our term by assuming t = x – y, then we change the coordinate in terms of t, then we separate the variable to interpret the given differential equation, we use \[\int{\dfrac{1}{1-{{t}^{2}}}dt}=\dfrac{1}{2}\ln \left| \dfrac{1+t}{1-t} \right|\] and \[\int{dn=x+e}\] to get the solution. To find the value of c, we will use y(1) = 1 and get our final answer.

Complete step by step answer:
We are asked to find the solution of \[\dfrac{dy}{dx}={{\left( x-y \right)}^{2}}.\] We will start by letting x – y as t, so that we will get simplified terms. So,
\[x-y=t\]
Differentiating with respect to x, we get,
\[\dfrac{d\left( x-y \right)}{dx}=\dfrac{d\left( t \right)}{dx}\]
\[\Rightarrow 1-\dfrac{dy}{dx}=\dfrac{dt}{dx}\]
\[\Rightarrow 1-\dfrac{dt}{dx}=\dfrac{dy}{dx}\]
Now putting this in \[\dfrac{dy}{dx}={{\left( x-y \right)}^{2}},\] we get,
\[1-\dfrac{dt}{dx}={{t}^{2}}\]
As x – y = t and \[\dfrac{dy}{dx}=1-\dfrac{dt}{dx}.\]
Now, simplifying, we get,
\[\Rightarrow \dfrac{dt}{dx}=1-{{t}^{2}}\]
\[\Rightarrow dt=\left( 1-{{t}^{2}} \right)dx\]
Shifting the terms, we get,
\[\dfrac{dt}{1-{{t}^{2}}}=dx\]
Now, integrating both the sides, we get,
\[\int{\dfrac{dt}{1-{{t}^{2}}}}=\int{dx}\]
We know that, \[\int{dx}=x+c\] while \[\int{\dfrac{dt}{1-{{t}^{2}}}}=\dfrac{1}{2}\ln \left| \dfrac{1+t}{1-t} \right|.\]
So, putting this value, we get,
\[\Rightarrow \dfrac{1}{2}\ln \left| \dfrac{1+t}{1-t} \right|=x+c\]
Putting, t = x – y, so we get,
\[\Rightarrow \dfrac{1}{2}\ln \left| \dfrac{1+x-y}{1-x+y} \right|=x+c\]
Now, to calculate the value of c, we know that y (1) = 1. This means that at x= 1, we have y = 1. Putting x = 1, y = 1 in
\[\Rightarrow \dfrac{1}{2}\ln \left| \dfrac{1+x-y}{1-x+y} \right|=x+c\]
We get,
\[\Rightarrow \dfrac{1}{2}\ln \left| \dfrac{1+1-1}{1-1+1} \right|=1+c\]
\[\Rightarrow 0=1+c\]
So, we get,
\[\Rightarrow c=-1\]
So, our solution is
\[\Rightarrow \dfrac{1}{2}\ln \left| \dfrac{1+x-y}{1-x+y} \right|=x-1\]
Taking 2 on the other side, we get,
\[\Rightarrow \ln \left| \dfrac{1+x-y}{1-x+y} \right|=2\left( x-1 \right)\]
Hence, the right option is (d).

Note:
Remember that log base e is written as ln and ln (1) = 0, that’s why \[\ln \left( \dfrac{1}{1} \right)=\ln \left( 1 \right)=0\] and the derivative of x – y = t gives us \[1-\dfrac{dy}{dx}=\dfrac{dt}{dx}.\] Because the derivative of x with respect to x is 1 while for other it is given as \[\dfrac{d}{dx}.\] So,
\[\dfrac{d\left( x-y \right)}{dx}=\dfrac{dt}{dx}\]
\[\Rightarrow \dfrac{dx}{dx}-\dfrac{dy}{dx}=\dfrac{dt}{dx}\]
\[\Rightarrow 1-\dfrac{dy}{dx}=\dfrac{dt}{dx}\]