JEE Main 2023 · 1 February · morning · Single correct · +4 / −1

If y=y(x) is the solution curve of the differential equation
(d y)/(d x)+y tan x=x sec x, 0 ≤ x ≤ π/3, y(0)=1, then y(π/6) is equal to

  1. (a) π/12-√3/2 log ₑ((2 √3)/e)
  2. (b) π/12+√3/2 log ₑ((2 √3)/e)
  3. (c) π/12+√3/2 log ₑ(2/(e √3))
  4. (d) π/12-√3/2 log ₑ(2/(e √3))
Answer and solution

Answer (d) π/12-√3/2 log ₑ(2/(e √3))

Solution

Maths sir · 4 steps

The integrating factor is sec x. Then integrate x sec² x by parts and let y(0)=1 fix the constant.

  1. Integrating factor. The equation has the form dy/dx + Py = Q with P = tan x and Q = x sec x. The integrating factor is e^(∫ tan x dx) = eˡⁿˢᵉᶜˣ = sec x.

    y sec x = ∫ x sec² x dx = x tan x - ∫ tan x dx = x tan x + ln cos x + C

  2. Find the constant. Put x = 0, y = 1. Then 1 × 1 = 0 + ln 1 + C, so C = 1. The solution is y sec x = x tan x + ln cos x + 1.
  3. Evaluate at the point. At x = π/6, sec x = 2/√3, tan x = 1/√3 and cos x = √3/2. Multiplying both sides by √3/2 gives y = π/12 + √3/2(1 + ln√3/2) = π/12 + √3/2 ln e√3/2.
  4. So. Since ln e√3/2 = -ln 2/e√3, we get y(π/6) = π/12 - √3/2 ln(2/e√3), which is option (d).

The idea to keep. For dy/dx + y tan x = Q, the integrating factor is always sec x. Fix C from the initial condition before you substitute the new x.

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 2023.