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Maths Question 20 – JEE-MAIN 2026

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Let y=y(x) be the solution curve of the differential equation (1+sin⁡x)dydx+(y+1)cos⁡x=0,y(0)=0. If the curve y=y(x) passes through the point (α,−12), then a value of α is :

Identify that the given differential equation is a first-order separable differential equation.

● Video Walkthrough
Step 1: Separate Variables and Integrate✦ Active

Rearrange the given differential equation to separate variables and integrate both sides:

(1+sin⁡x)dydx=−(y+1)cos⁡x dyy+1=−cos⁡x1+sin⁡xdx Integrating both sides yields: ∫dyy+1=−∫cos⁡x1+sin⁡xdx ln⁡|y+1|=−ln⁡|1+sin⁡x|+C ln⁡|(y+1)(1+sin⁡x)|=C (y+1)(1+sin⁡x)=A, where A=eC is an arbitrary positive constant.
Step 2: Apply Initial Condition○ Expand

Use the initial condition y(0)=0 to find the value of the constant A.

Substitute x=0 and y=0 into the general solution: (0+1)(1+sin⁡0)=A (1)(1+0)=A⟹A=1 The particular solution is (y+1)(1+sin⁡x)=1.
Step 3: Find α using the given point○ Expand

The curve passes through the point (α,−12). Substitute these values into the particular solution.

(−12+1)(1+sin⁡α)=1 (12)(1+sin⁡α)=1 1+sin⁡α=2 sin⁡α=1 From the given options, α=π2 satisfies sin⁡α=1.
💡 Teacher's Secret Hint

Ensure to check all options if multiple values could satisfy the condition, but in this case, only one option matches.

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