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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
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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