Differential EquationsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Differential Equations: Let Solution Curve Differential Equation Curve Passes Throug (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let y=y(x)y=y(x) be the solution curve of the differential equation (1+sin⁡x)dydx+(y+1)cos⁡x=0(1+\sin x)\dfrac{dy}{dx}+(y+1)\cos x=0, y(0)=0y(0)=0. If the curve y=y(x)y=y(x) passes through the point (α,−12)\left(\alpha,-\dfrac12\right), then a value of α\alpha is
Aπ6\dfrac{\pi}{6}
Bπ4\dfrac{\pi}{4}
Cπ3\dfrac{\pi}{3}
Dπ2\dfrac{\pi}{2}correct
Solution
Step 1: Separate variables:
dyy+1=−cos⁡x1+sin⁡x dx.\frac{dy}{y+1}=-\frac{\cos x}{1+\sin x}\,dx.
 Step 2: Integrate both sides (right side has numerator = derivative of denominator):
log⁡e(y+1)=−log⁡e(1+sin⁡x)+C.\log_e(y+1)=-\log_e(1+\sin x)+C.
 Step 3: Apply y(0)=0y(0)=0: log⁡e1=−log⁡e1+C⇒C=0\log_e1=-\log_e1+C\Rightarrow C=0. So
log⁡e(y+1)=−log⁡e(1+sin⁡x) ⇒ y+1=11+sin⁡x.\log_e(y+1)=-\log_e(1+\sin x)\ \Rightarrow\ y+1=\frac{1}{1+\sin x}.
 Step 4: Put y=−12y=-\dfrac12:
11+sin⁡x=12 ⇒ 1+sin⁡x=2 ⇒ sin⁡x=1 ⇒ x=π2.\frac{1}{1+\sin x}=\frac12\ \Rightarrow\ 1+\sin x=2\ \Rightarrow\ \sin x=1\ \Rightarrow\ x=\frac{\pi}{2}.
 Correct answer: (4)
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