Q .     The solution of the differential equation `\ frac {dy}{dx} + \ frac {y}{2} \sec x = \ frac {\tan x}{2y}`, where `0\le x < \ frac {\pi}{2}` and `y(0) = 1`, is given by:

JEE 2016 Mains 10 April
A

`y^{2} = 1 + \ frac {x}{\sec x + \tan x}`

B

`y = 1 + \ frac {x}{\sec x + \tan x}`

C

`y = 1 - \ frac {x}{\sec x + \tan x}`

D

` y^{2} = 1 - \ frac {x}{\sec x + \tan x}`

HINT

Put `y^{2} = t`

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