Let y=y(x) be the solution curve of the differential equation
sin ( 2x² )log ₑ( tan x² )dy + ( 4xy - 4√2 x sin ( x² - π/4 ) )dx = 0, 0 < x < √(π/2), which passes through the point (√(π/6), 1). Then |y(√(π/3))| is equal to ______________.

Solution

Maths sir · 4 steps

Put t = x² and the integrating factor turns out to be ln(tan t) itself.

1. Substitute t. Let t = x², so 4x dx = 2 dt. The equation becomes sin 2t ln(tan t) dy/dt + 2y = 2√2 sin(t - π/4), which is linear in y.
2. Integrating factor. The coefficient of y is P = 2/(sin 2t ln(tan t)). Since d/dt ln(tan t) = (sec² t)/(tan t) = 2/(sin 2t), P has the form u'/u with u = ln(tan t). So the integrating factor is ln(tan t), and the right side becomes 2(sin t - cos t)/(2 sin t cos t) = sec t - csc t.

   y ln(tan t) = ln(sec t + tan t) - ln(csc t - cot t) + C

3. Find C. At t = π/6, y = 1: ln(tan t) = -1/2 ln 3, sec t + tan t = √3 and csc t - cot t = 2 - √3. So -1/2 ln 3 = 1/2 ln 3 - ln(2-√3) + C, giving C = ln(2-√3) - ln 3.
4. So. At t = π/3: sec t + tan t = 2+√3 and csc t - cot t = 1/√3. Then 1/2 ln 3 · y = ln((2+√3)(2-√3)) + 1/2 ln 3 - ln 3 = -1/2 ln 3. So y = -1, and |y| = 1.

The common slip: -1. Dropping the modulus at the end reports y = -1 instead of |y| = 1.

The idea to keep. When the coefficient of y has the form f'(t)/f(t), the integrating factor is just f(t). Differentiate the ugly factor first, it is usually there for a reason.

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