If dy/dx + 2y tan x = sin x, 0 < x < π/2 and y( π/3 ) = 0, then the maximum value of y(x) is :

Solution

Maths sir · 4 steps

This is a linear equation: multiply by sec² x, integrate, and you get a quadratic in cos x.

1. Integrating factor. The equation has the form dy/dx + P y = Q with P = 2 tan x. So the integrating factor is e^(∫ 2 tan x dx) = e²ˡⁿˢᵉᶜˣ = sec² x.

   d/dx(y sec² x) = sin x sec² x = sec x tan x

2. Integrate and fix C. Integrating gives y sec² x = sec x + C. At x = π/3 we have sec x = 2 and y = 0, so 0 = 2 + C and C = -2. Multiplying by cos² x gives y = cos x - 2 cos² x.
3. Maximise the quadratic. Put t = cos x, with 0 < t < 1. Then y = t - 2t², a downward parabola with its vertex at t = 1/4, which lies inside the range.
4. So. The maximum is y = 1/4 - 2 × 1/16 = 1/8, which is option (a).

The idea to keep. For y' + Py = Q, the integrating factor e^(∫ P dx) turns the left side into one derivative. If the answer is a polynomial in cos x, substitute t = cos x and find the vertex.

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