JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0

y (0) = 1

. Then

y (\frac{π}{4})

is equal to :

Answer: (C)

Solution

Separating,

\int \frac{sec ^{2} x e ^{t a n x}}{1 + e ^{2 t a n x}} d x + \int \frac{d y}{1 + y ^{2}} = C

, i.e.

tan^{- 1} (e^{t a n x}) + tan^{- 1} y = C

. At

x = 0, y = 1

\frac{π}{4} + \frac{π}{4} = \frac{π}{2} = C

. At

x = \frac{π}{4}

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

. Option (3).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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