JEE Main 2020 · 5 September · evening · Single correct · +4 / −1

Let y = y(x) be the solution of the differential equation
cosxdy/dx + 2ysinx = sin2x, x ∈ ( 0,π/2 ).
If y( π/3 ) = 0, then y( π/4 ) is equal to :

  1. (a) 1/√2 - 1
  2. (b) √2 - 2
  3. (c) 2 - √2
  4. (d) 2 + √2
Answer and solution

Answer (b) √2 - 2

Solution

Maths sir · 4 steps

Divide by cos x first, and the standard linear form shows its integrating factor sec² x.

  1. Standard linear form. Divide every term by cos x, which is nonzero on (0, π/2). Since sin 2x = 2 sin x cos x, the right side becomes 2 sin x, so dy/dx + (2 tan x) y = 2 sin x.
  2. Integrating factor. With P = 2 tan x, ∫ P dx = 2 ln sec x, so the integrating factor is e²ˡⁿˢᵉᶜˣ = sec² x.

    y sec² x = ∫ 2 sin x sec² x dx = ∫ 2 sec x tan x dx = 2 sec x + C

  3. Use the condition. At x = π/3, y = 0 and sec π/3 = 2, so 0 = 4 + C and C = -4. Multiplying through by cos² x gives y = 2 cos x - 4 cos² x.
  4. So. At x = π/4, y = 2 × 1/√2 - 4 × 1/2 = √2 - 2, which is option (b).

The idea to keep. Always get the coefficient of dy/dx to 1 before finding the integrating factor. Then ∫ sec x tan x dx = sec x handles most such right-hand sides.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Differential Equations

Asked about 200 times in JEE Main since 2020. This question is from 2020.