Explanation

(d)
Given that, $\frac{dy}{dx} = 1 + y^{2}$
$\Rightarrow \ \frac{dy}{1 + y^{2}} = dx$
On integrating both sides, we get
$\int_{}^{}\frac{dy}{1 + y^{2}} = \int_{}^{}{dx}$
$\Rightarrow \tan^{- 1}{y = x + c}$
At $x = 0,\ y = 0,$ then $c = 0$
At $x = \pi,\ y = 0$, then $\tan^{- 1}{0 = \pi + c\ \Rightarrow c = - \pi}$
$\therefore\tan^{- 1}{y = x\ \Rightarrow y = \tan{x = \phi(x)}}$
Therefore, solution becomes $y = \tan x$
But $\tan x$ is not continuous function in $(0,\ \pi)$
So, $\phi(x)$ is not possible in $(0,\ \pi)$.