Answer is (d)
Explanation:
The given differential equation is
y' – y tan x = 2x sec x
(dy/dx) – tan x.y = 2x sec x
which is a linear differential equation
Thus, IF = e–∫tan x dx = elog cos x = cos x
Hence, the solution is
y.cos x = ∫2x.sec x.cos x dx + c
y.cos x = ∫2x.dx + c
y.cos x = x2 + c
when x = 0, y = 0, then c = 0
Thus, the equation of the curve is
y.cos x = x2
When x = π/4, then y.1/√2 = π2/16
