JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Differential Equations

y = y (x)

is the solution of the differential equation

(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0

and

y (0) = 0

, then

6 (y^{'} (0) + (y (lo g_{e} 3))^{2})

is equal to:

Answer: (C)

Solution

Separate variables:

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

. Integrate:

tan^{- 1} y = - 2 tan^{- 1} (e^{x}) + C

. Using

y (0) = 0

0 = - 2 tan^{- 1} (1) + C

, so

C = \frac{π}{2}

. Thus

tan^{- 1} y + 2 tan^{- 1} (e^{x}) = \frac{π}{2}

. Differentiate:

\frac{y ^{'}}{1 + y ^{2}} + \frac{2 e ^{x}}{1 + e ^{2 x}} = 0

. At

x = 0

y^{'} (0) = - \frac{2 ( 1 )}{1 + 0} = - 1

. At

x = ln 3

e^{x} = 3

, so

tan^{- 1} y = \frac{π}{2} - 2 tan^{- 1} (3) = \frac{π}{2} - \frac{2 π}{3} = - \frac{π}{6}

, giving

y = - \frac{1}{3}

. Therefore

6 (- 1 + \frac{1}{3}) = - 4

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