JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
Let y = y(x) be the solution of the differential equation cos ec²xdy + 2dx = (1 + y cos 2x)cos ec²xdx, with y( π/4 ) = 0. Then, the value of (y(0) + 1)² is equal to :
- (a) e 1/2
- (b) e -1/2
- (c) e⁻¹
- (d) e
Answer and solution
Answer (c) e⁻¹
Solution
Maths sir · 4 steps
Divide by csc² x first: 1 - 2 sin² x turns into cos 2x and the equation separates.
- Clear the cosec. Divide every term by csc² x. Since 2/(csc² x) = 2 sin² x, you get dy + 2 sin² x dx = (1 + y cos 2x) dx.
- Tidy up. So dy/dx = 1 - 2 sin² x + y cos 2x. Use 1 - 2 sin² x = cos 2x to get dy/dx = (1 + y)cos 2x, which is variable separable.
∫ dy/(1+y) = ∫ cos 2x dx ⇒ ln|1+y| = (sin 2x)/2 + C
- Find C. At x = π/4, y = 0, so ln 1 = 1/2 sin π/2 + C. That gives 0 = 1/2 + C, so C = -1/2.
- So. At x = 0, ln|1 + y(0)| = 0 - 1/2, so |1 + y(0)| = e^(-1/2). Squaring gives (y(0)+1)² = e⁻¹, which is option (c).
The idea to keep. Before you reach for an integrating factor, simplify the trig. Often 1 - 2 sin² x or a similar identity collapses the equation into a simple separable form.
Answer from the NTA final answer key. Solution by PrepWiser.