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Question:

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

loge√√√2−y2−x√√√=2(y−1)

loge√√√2−x2−y√√√=x−y

−loge√√√1+x−y1−x+y√√√=x+y−2

−loge√√√1−x+y1+x−y√√√=2(x−1)

Solution:

x−y=t⟹dydx=1−dtdx⟹1−dtdx=t2⟹∫dt1−t2=∫1dx⟹12ln(1+t1−t)=x+λ⟹12ln(1+x−y1−x+y)=x+λ