The solution of the differential equation, \frac{dy}{dx} = (x-y)^2 , when y (1) = 1 , is : [11 Jan 2019 Shift 2]

Correct Answer is : - loge|1−x+y1+x−y| = 2 (x - 1)

log_e |\frac{2-y}{2-x}| = 2 (y - 1)

log_e |\frac{2-x}{2-y}| = x - y

- log_e |\frac{1+x-y}{1-x+y}| = x + y - 2

- log_e |\frac{1-x+y}{1+x-y}| = 2 (x - 1)

JEE Main 11 Jan 2019 Shift 2 Solved Paper

Section: Mathematics

Question : 43 of 90

Marks: +1, -0

The solution of the differential equation,

= (x−y)2 , when y (1) = 1 , is :

[11 Jan 2019 Shift 2]

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