If y(x) satisfies the differential equation y’ - y tan x = 2x sec x and y(0) = 0, then: (a) y(π/4) = π2/8√2 (b) y’(π/4) = π2/18 (c) y(π/3) = π2/9 (d) y’(π/3) = (4π/3) + (2π2/3√3)

 Solution:

Given y’ – y tan x = 2x sec x

(dy/dx) – y tan x = 2x sec x

Divide both sides by sec x

=> (dy/dx) cos x – y sin x = 2x

(d/dx)(y cos x) = 2x

Integrating both sides

We get y cos x = x2 + c

Given y(0) = 0

=> c = 0

=> y cos x = x2 

When x = π/4, y(π/4) = π2/8√2

When x = π/3, y(π/3) = 2π2/9

When x = π/4, y’(π/4) = π2/8√2 + π/√2

When x = π/3, y’(π/3) = (2π2/3√3) + (4π/3)

Hence option A and D are correct.