Let g be a differentiable function such that ∫₀ˣ g(t) d t=x-∫₀ˣ tg(t) d t, x ≥ 0 and let y=y(x) satisfy the differential equation (d y)/(d x)-y tan x=2(x+1) sec x g(x), x ∈[0, π/2). If y(0)=0, then y(π/3) is equal to

Solution

Maths sir · 5 steps

First pull g(x) out of the integral equation by differentiating, then the DE becomes an easy linear one.

1. Find g. Differentiate both sides with respect to x. This gives g(x) = 1 - x g(x), so g(x) = 1/(1+x).
2. Simplify the DE. Now 2(x+1)sec x g(x) = 2 sec x, so the equation becomes dy/dx - y tan x = 2 sec x. It is linear with P = -tan x.
3. Integrating factor. The integrating factor is e^(-∫ tan x dx) = eˡⁿᶜᵒˢˣ = cos x. Multiplying through by cos x makes the left side an exact derivative.

   d/dx(y cos x) = 2 sec x cos x = 2 ⇒ y cos x = 2x + C

4. Use y(0) = 0. Putting x = 0 gives C = 0, so y = 2x sec x.
5. So. y(π/3) = 2 × π/3 × 2 = 4π/3, which is option (a).

The idea to keep. When an unknown function sits inside an integral with variable limits, differentiate once to free it. Then check whether it cancels neatly into the DE.

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