JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution curve of the differential equation

(1 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0

y (0) = 0

. If the curve

y = y (x)

passes through the point

(α, \frac{- 1}{2})

, then a value of

α

is :

Answer: (D)

Solution

Separating variables,

\frac{d y}{y + 1} = - \frac{cos x}{1 + sin x} d x

, so

ln (y + 1) = - ln (1 + sin x) + C

. The condition

y (0) = 0

gives

ln 1 = - ln 1 + C

, hence

C = 0

and

y + 1 = \frac{1}{1 + sin x}

. At the point

(α, - \frac{1}{2})

\frac{1}{2} = \frac{1}{1 + sin α}

, so

1 + sin α = 2

and

sin α = 1

, giving

α = \frac{π}{2}

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