Let y = y(x) be the solution of the differential equation dy/dx + (√2 y)/(2cos ⁴x - cos ²x) = xe^(tan ⁻¹(√2 cot 2x)), 0 < x < π/2 with y( π/4 ) = (π ²)/32. If y( π/3 ) = (π ²)/18e^(- tan ⁻¹(α )), then the value of 3α² is equal to ___________.

Solution

Maths sir · 4 steps

Turn the denominator into cos 2x form, and the exponent on the right becomes your integrating factor.

1. Fix the denominator. The original JEE question has 2 cos⁴ x - cos 2x in the denominator, and the answer key is based on it. The given cos² x does not fit the exponent on the right, so it is a misprint for cos 2x. Then 2 cos⁴ x - (2 cos² x - 1) = 1 - 2 sin² x cos² x = 1 - (sin² 2x)/2 = (1+cos² 2x)/2, so P(x) = 2√2/(1+cos² 2x).

   d/dx tan⁻¹(√2 cot 2x) = (-2√2 cosec² 2x)/(1+2 cot² 2x) = -2√2/(sin² 2x + 2 cos² 2x) = -2√2/(1+cos² 2x)

2. Integrating factor. This derivative is exactly -P(x), so ∫ P dx = -tan⁻¹(√2 cot 2x) and the integrating factor is e^(-tan⁻¹(√2 cot 2x)). Multiplying the right side by it leaves only x, so y e^(-tan⁻¹(√2 cot 2x)) = x²/2 + C.
3. Use the condition. At x = π/4, cot π/2 = 0, so the exponential factor is 1. Then π²/32 = π²/32 + C, which gives C = 0 and y = x²/2 e^(tan⁻¹(√2 cot 2x)).
4. So. At x = π/3, cot 2π/3 = -1/√3, so y = π²/18 e^(tan⁻¹(-√(2/3))) = π²/18 e^(-tan⁻¹√(2/3)). Hence α = √(2/3) and 3α² = 3 × 2/3 = 2.

The idea to keep. When the right side of a linear equation carries an odd exponential, differentiate that exponent first. Very often it is ±∫ P dx, and the integrating factor cancels it completely.

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