JEE Main 2025 · 29 January · morning · Single correct · +4 / −1

Let y = y(x) be the solution of the differential equation :
cos x(log ₑ(cos x))² d y+(sin x-3 y sin x log ₑ(cos x)) d x=0, x ∈ (0, π/2 ). If y(π/4) = -1/(logₑ 2), then y(π/6) is equal to :

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

Answer (d) 1/(logₑ(3)−logₑ(4))

Solution

Maths sir · 4 steps

Everything here is built from ln cos x, so make that your new variable and the equation turns linear.

  1. Make it linear. Divide by cos x (ln cos x)² dx and write L = ln cos x. You get dy/dx - (3 tan x)/L y = -(tan x)/L², a linear equation in y.
  2. Change variable. Put t = ln cos x, so dt/dx = -tan x. Dividing the equation by -tan x gives dy/dt + 3/t y = 1/t². The integrating factor is e^(∫ 3/t dt) = t³.

    t³ y = ∫ t³ · 1/t² dt = t²/2 + C ⇒ y = 1/2t + C/t³

  3. Find C. At x = π/4, t = ln 1/√2 = -1/2 ln 2, so 1/2t = -1/(ln 2). That already equals the given y, so C = 0 and y = 1/(2 ln cos x).
  4. So. At x = π/6, 2 ln cos x = 2 ln√3/2 = ln 3/4 = ln 3 - ln 4. Hence y(π/6) = 1/(ln 3 - ln 4), option (d).

The idea to keep. When one expression like ln cos x keeps showing up and its derivative (-tan x) is also sitting in the equation, substitute it as t. The messy equation usually becomes a clean linear one.

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 2025.