JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

sec x d y + {2 (1 - x) tan x + x (2 - x)} d x = 0

such that

y (0) = 2

. Then

y (2)

is equal to :

Answer: (A)

Solution

sec x d y = - {2 (1 - x) tan x + x (2 - x)} d x \Rightarrow \frac{d y}{d x} = 2 (x - 1) sin x + x (x - 2) cos x

, which is exactly

\frac{d}{d x} [(x^{2} - 2 x) sin x]

.

Integrating:

y = (x^{2} - 2 x) sin x + C

; using

y (0) = 2

gives

C = 2

y (2) = (4 - 4) sin 2 + 2 = 2

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option

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