JEE Main 2021 · 16 March · evening · Single correct · +4 / −1

If y = y(x) is the solution of the differential equation
dy/dx + (tan x) y = sin x, 0 ≤ x ≤ π/3, with y(0) = 0, then y( π/4 ) equal to :

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

Answer (b) ( 1/2√2 ) logₑ 2

Solution

Maths sir · 4 steps

Here the integrating factor is sec x, and then the right side becomes ∫ tan x dx.

  1. Integrating factor. The equation has the form dy/dx + Py = Q with P = tan x. So the integrating factor is e^(∫ tan x dx) = eˡⁿˢᵉᶜˣ = sec x. On 0 ≤ x ≤ π/3, sec x > 0, so no modulus sign is needed.

    y sec x = ∫ sin x sec x dx = ∫ tan x dx = ln(sec x) + C

  2. Use y(0) = 0. At x = 0: 0 × 1 = ln 1 + C, so C = 0. That gives y = cos x ln(sec x).
  3. Put x = π/4. cos π/4 = 1/√2 and ln(sec π/4) = ln√2 = 1/2 ln 2. Multiplying gives 1/√2 × 1/2 ln 2.
  4. So. y(π/4) = 1/2√2 logₑ 2, which is option (b).

The idea to keep. For y' + Py = Q, multiply by e^(∫ P dx). Then y × IF = ∫ Q × IF dx. Divide by the IF at the end before you substitute the point.

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 2021.