JEE Main 2021 · 17 March · evening · Single correct · +4 / −1

Let y = y(x) be the solution of the differential equation
cos x(3 sin x + cos x + 3)dy = (1 + y sin x(3 sin x + cos x + 3))dx,0 ≤ x ≤ π/2,y(0) = 0. Then, y( π/3 ) is equal to :

  1. (a) 2log ₑ( (√3 + 7)/2 )
  2. (b) 2log ₑ( (3√3 - 8)/4 )
  3. (c) 2log ₑ( (2√3 + 10)/11 )
  4. (d) 2log ₑ( (2√3 + 9)/6 )
Answer and solution

Answer (c) 2log ₑ( (2√3 + 10)/11 )

Solution

Maths sir · 5 steps

Divide through, and it is linear in y with integrating factor cos x; then t = tan(x/2) finishes it.

  1. Make it linear. Write P = 3 sin x + cos x + 3 and divide both sides by cos x · P dx. This gives dy/dx - y tan x = 1/(cos x · P), a linear equation in y.
  2. Integrating factor. The factor is e^(-∫ tan x dx) = cos x. Multiplying by it cancels the cos x on the right, so d/dx(y cos x) = 1/(3 sin x + cos x + 3).
  3. Substitute t. Put t = tan x/2, so sin x = 2t/(1+t²), cos x = (1-t²)/(1+t²) and dx = (2 dt)/(1+t²). The denominator becomes (2t² + 6t + 4)/(1+t²), and the 1+t² cancels.

    y cos x = ∫ dt/(t+1)(t+2) = ln(t+1)/(t+2) + C

  4. Use y(0) = 0. At x = 0, t = 0, so 0 = ln 1/2 + C and C = ln 2. Hence y cos x = ln 2(t+1)/(t+2).
  5. So. At x = π/3, t = 1/√3 and cos x = 1/2, so y = 2 ln 2(1+√3)/(1 + 2√3). Multiplying top and bottom by 2√3 - 1 gives y(π/3) = 2 ln(2√3 + 10)/11, which is option (c).

The idea to keep. When a messy equation has y appearing only to the first power, divide it into the form dy/dx + Py = Q. For an integral over a sin x + b cos x + c, the substitution t = tan(x/2) always works.

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.