A continuously differentiable function `phi(x)` in `(0,pi)` satisfying `y'=1+y^(2),y(0)=0=y(pi)` is

Options

A. tan x
B. `x(x-pi)`
C. `(x-pi)(1-e^(x))`
D. not possible

Answer

d

Explanation

Given that, `(dy)/(dx) = 1+Y(2) rArr (dy)/(1+y^(2)) = dx `
On integratng both sides , we get
` int (dy)/(1+y^(2)) = int dx rArr tan^(-1) y = x +C` ,
At x = 0 , y = 0 then C = 0
At x = `pi` , y = 0 then `tan^(-1) 0 = pi + C rArr C = - pi`
` :. tan^(-1) y = x`
` rArr y = tan c = phi (x)`
Therefore , solution becpmes y = tan x
But tan x is not continous functin in `(0,pi)`
So , ` phi (x)` is not possible in `(0,pi)`