A continuously differentiable function $y =f\left(x\right), x\in \left(0,\mathrm{\pi }\right)$satisfying $y\text{'} = 1 + y^2,y\left(0\right) = 0 = y\left(\mathrm{\pi }\right)$ is

 Given the differential equation,
Consider, $y^{\text{'}}=1+y^2,y\left(0\right)=0=y\left(\pi \right)$
$\frac{dy}{dx}=1+y^2\frac{dy}{dx}$, we have.
Then, $\frac{dy}{1+y^2}=dx$.
Integrating the equation, we have.

${\tan }^{-1}y=x+c ......\left(1\right)$$\left\{\int \frac{1}{1+y^2}dy={\tan }^{-1}y\right\}$
since $y\left(0\right)=0$ are $x=0$ and $y=0$  .
${\tan }^{-1}\left(0\right)=0+c$

$0=C$
Now, substitute the obtained value of $C$ in $\left(1\right)$ 
we have,
$$\begin{gathered} {\tan }^{-1}y=x+0. \\ {\tan }^{-1}y=x. \\ \Rightarrow y=\tan x. \end{gathered}$$