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JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 20

Let y=y(x)y=y(x) be the solution curve of the differential equation
(1+sin⁡x)dydx+(y+1)cos⁡x=0,y(0)=0.(1+\sin x)\frac{dy}{dx}+(y+1)\cos x=0,\quad y(0)=0.
If the curve y=y(x)y=y(x) passes through the point (α,−12)\left(\alpha,\dfrac{-1}2\right), then a value of α\alpha is :

Choose one option
Answer and worked solution

Official answer: D

Exact derivative. (1+sin⁡x)y′+(y+1)cos⁡x=ddx[(1+sin⁡x)(y+1)]=0(1+\sin x)y'+(y+1)\cos x=\dfrac d{dx}\left[(1+\sin x)(y+1)\right]=0, so (1+sin⁡x)(y+1)=C(1+\sin x)(y+1)=C.

Constant. y(0)=0y(0)=0 gives C=1C=1.

At y=−12y=-\frac12: 12(1+sin⁡α)=1\frac12(1+\sin\alpha)=1, so sin⁡α=1\sin\alpha=1 and α=π2\alpha=\frac\pi2.

(D)

Topics:Linear first-order equations
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