JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Differential Equations

The solution of the differential equation

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

when

y (1) = 1

, is:

Answer: (B)

Solution

Substituting

u = x - y

\frac{d u}{d x} = 1 - \frac{d y}{d x} = 1 - u^{2}

, a separable equation with

\frac{d u}{1 - u ^{2}} = d x

. Integrating,

\frac{1}{2} lo g \frac{1 + u}{1 - u} = x + C

, and the condition

y (1) = 1

means

u = 0

x = 1

, so

C = - 1

. Hence

lo g \frac{1 + x - y}{1 - x + y} = 2 (x - 1)

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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