A continuously differentiable function (x) , in ,(0, , ) satisfyi… - Vidyadip

MCQ

A continuously differentiable function $\phi (x)\,{\rm{in}}\,(0,\,\pi )$ satisfying $y' = 1 + {y^2},\,\,y(0) = 0 = y(\pi )$ is

A
$\tan x$
B
$x(x - \pi )$
C
$(x - \pi )$ $(1 - {e^x})$
✓
Not possible

✓

Answer

Correct option: D.

Not possible

d
(d) $\frac{{dy}}{{dx}} = 1 + {y^2}$ ==> $\frac{{dy}}{{1 + {y^2}}} = dx$

Integrating both sides,

$\int {\frac{{dy}}{{1 + {y^2}}} = \int {dx} } $ ==> ${\tan ^{ - 1}}y = x + c$

At $x = 0,$$y = 0,$ then $c = 0$

At $x = \pi ,$$y = 0,$ then ${\tan ^{ - 1}}0 = \pi + c$ ==> $c = - \pi $

$\therefore {\tan ^{ - 1}}y = x$==>$y = \tan x = \phi (x)$

Therefore, solution is $y = \tan x$

But $\tan x$ is not continuous function in $(0,\,\pi )$

Hence, $\phi \,(x)$ is not possible in $(0,\,\pi )$.

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