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

Options

A. tanx
B. `x(x-pi)`
C. `(x-pi)(1-e^(x))`
D. Not possible

Answer

D

Explanation

Given that `(dy)/(dx)=1+y^(2)`
`implies(dy)/(1+y^(2))=dx`
On integrating both sides.
`int(dy)/(1+y^(2))=intdximpliestan^(-1)y=x+c`
At x=0, y=0, then c=0
At `x=pi,y=0`, then `tan^(-1)0=pi+c`
`impliesc=-pi`
`therefore tan^(-1)y=ximpliesy=tanx=phi(x)`
therefore, solution becomes y=tanx. but tan x is not continuous function in `(0,pi)`. so. `phi(x)` is not possible in (`0,pi`).