JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution curve of the differential equation

sec y \frac{d y}{d x} + 2 x sin y = x^{3} cos y

y (1) = 0

. Then

y (3)

is equal to :

Answer: (C)

Solution

sec y d y / d x + 2 x sin y = x^{3} cos y ⟹ sec^{2} y d y / d x + 2 x tan y = x^{3}

(dividing by

cos y

).
Let

t = tan y

d t / d x + 2 x t = x^{3}

, an integrating factor

e^{x^{2}}

gives

t e^{x^{2}} = \int x^{3} e^{x^{2}} d x = \frac{e ^{x^{2}}}{2} (x^{2} - 1) + C

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

y (1) = 0 ⟹ 0 = (1 - 1) + 2 C e^{- 1} ⟹ C = 0

, so

2 tan y = x^{2} - 1

.
At

x = 3

2 tan y = 3 - 1 = 2 ⟹ tan y = 1 ⟹ y = π /4

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