Let y = y(x) be a solution curve of the differential equation (y + 1)tan ²x dx + tan x dy + y dx = 0, x ∈ ( 0,π/2 ). If lim _(x → 0 +) xy(x) = 1, then the value of y( π/4 ) is :

Solution

Maths sir · 4 steps

Pair y tan² x with y to get y sec² x, and the integrating factor comes out as plain tan x.

1. Regroup the terms. Collect the y terms: y tan² x + y = y sec² x. The equation becomes tan x dy + (y sec² x + tan² x) dx = 0. Divide by tan x dx to get dy/dx + (sec² x)/(tan x) y = -tan x.
2. Integrating factor. ∫ (sec² x)/(tan x) dx = ln(tan x), so the integrating factor is tan x, which is positive on (0, π/2). Multiplying through by it gives an exact derivative on the left.

   d/dx(y tan x) = -tan² x = 1 - sec² x ⇒ y tan x = x - tan x + C

3. Use the limit. Multiply by x/(tan x) to get xy = x/(tan x)(x - tan x + C). As x → 0⁺, x/(tan x) → 1 and x - tan x → 0, so xy → C. The condition gives C = 1.
4. So. The curve is y tan x = x - tan x + 1. At x = π/4, tan x = 1, so y(π/4) = π/4 - 1 + 1 = π/4, which is option (d).

The idea to keep. When a linear equation has sec² x/tan x as the coefficient of y, the integrating factor is just tan x. Always rewrite tan² x as sec² x - 1 before integrating.

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