Question

Let y = y(x) be a solution curve of the differential equation (y + 1)tan²xdx + tanxdy + ydx = 0, `x∈(0, π/2)`. If `lim_(x→0^+)` xy(x) = 1, then the value of `y(π/2)` is ______.

Options

• `π/4 - 1`

• `π/4 + 1`

• `π/4`

• `-π/4`

Answer 1

Let y = y(x) be a solution curve of the differential equation (y + 1)tan²xdx + tanxdy + ydx = 0, `x∈(0, π/2)`. If `lim_(x→0^+)` xy(x) = 1, then the value of `y(π/2)` is `underlinebb(π/4)`.

Explanation:

(y + 1)tan²xdx + tanxdy + ydx = 0

So, `(dy)/(dx) + (1 + y)tanx` = –ycotx

⇒ `(dy)/(dx) + y(tanx + cotx)` = –tanx 

IF = `e^(int(tanx + cotx)dx` = `e^(int(tan^2x + 1)/tanxdx` = tanx

∴ ytanx = `int-tan^2xdx + C`

⇒ ytanx = `int(1 - sec^2x)dx + C`

⇒ ytanx = x – tanx + C

Now `lim_(x→0^+)xy` = 1

 ⇒ `lim_(x→0^+)(x/(tanx))(x - tanx + C)` = 1

⇒ 1(0 – 0 + C) = 1

⇒ C = 1

Then, the function y tan x = x – tanx + 1 at x = `π/4`

 ⇒ `y(π/4)tan(π/4) = π/4 - tan  π/4 + 1`

∴ `y(π/4) = π/4`