JEE Main 2019 — Math solution | Fundamenthol

JEE Main2019Math

Q.

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

A
$lo g_{e} \frac{2 - y}{2 - x} = 2 (y - 1)$
B
$- lo g_{e} \frac{1 + x - y}{1 - x + y} = x + y - 2$
C
$- lo g_{e} \frac{1 - x + y}{1 + x - y} = 2 (x - 1)$
D
$lo g_{e} \frac{2 - x}{2 - y} = x - y$

Solution

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