Let ` y= y (x) ` be the solution of the differential equation `( dy )/(dx) + y tan x = 2x + x ^2 tan x,x in (-(pi)/(2) , (pi)/(2)),` such that y(0)=1 . Then :

Options

A. `y'((pi)/(4)) -y'(-(pi)/(4)) = pi - sqrt2`
B. `y((pi)/(4) -y (-(pi)/(4)) = sqrt2`
C. `y((pi)/(4)) + y(-(pi)/(4)) = (pi^(2))/(2) + 2`
D. `y'((pi)/(4)) + y'(-(pi)/(4)) = -sqrt2`

Answer

A

Explanation

`(dy)/(dx) + y.tan x = 2x + x^(2) tan x`
`I.F. = e^(int tan x dx) = e^(log|sec x|) = sec x`
`y(sec x) = int (2x + x^(2) tan x) sec x dx + c = int 2x sec x dx + int x^(2) tan x sec x dx = x^(2) sec x + c`
`y = x^(2) + c. cosx rArr y(0) = 1 rArr c =1 rArr y = x^(2) + cos x rArr y' = 2x - sin x`
`rArr y'((pi)/(4)) = (pi)/(2) - (1)/(sqrt2) rArr y'(-(pi)/(4)) = (-pi)/(2) + (1)/(sqrt2)`