Question

If `dy/dx + 2y tan x = sin x, 0 < x < π/2` and `y(π/6)` = 0, then the maximum value of y(x) is ______.

पर्याय

• `4/sqrt(3)`

• `sqrt(3)/2`

• 0

• `1/6`

Answer 1

If `dy/dx + 2y tan x = sin x, 0 < x < π/2` and `y(π/6)` = 0, then the maximum value of y(x) is `underlinebb(sqrt(3)/2)`.

Explanation:

Given, `dy/dx + 2y tan x = sin x, 0 < x < π/2`

Which is a linear differential equation.

Here, P = 2 tan x and Q = sin x

IF = `e^(intP  dx)`

= `e^(2 int tan x  dx)`

= `e^(2 log sec x)`

= sec² x

∴ Required solution of differential equation,

y.IF = `int(Q xx IF) dx + C`

`\implies` y sec² x = `int (sin x xx sec^2 x) dx + C`

= `int tan x sec dx + C`

∴ y sec² x = sec x + C

As, `y(π/6)` = 0

`\implies 0 . sec^2 (π/6) = sec  π/6 + C`

`\implies` C = `-2/sqrt(3)`

 ∴ y sec² x = `sec x - 2/sqrt(3)`  ...[From equation (i)]

`\implies` y = `cos x - 2/sqrt(3) cos^2 x`

= `-2/sqrt(3) (cos^2 x - sqrt(3)/2 cos x)`

= `- 2/sqrt(3)[cos^2 x - sqrt(3)/2 cos x + (sqrt(3)/4)^2 - (sqrt(3)/4)^2]`

= `- 2/sqrt(3) [(cos x - sqrt(3)/4)^2 - 3/4]`

= `3/(2sqrt(3)) - 2/sqrt(3) (cos x - sqrt(3)/4)^2`

Minimum vaIue of `(cos x - sqrt(3)/4)` is 0.

∴ Maximum value of y = `3/(2sqrt(3)) = sqrt(3)/2`