The largest value of c such that there exists a differential function $h\left(x\right)$ for$-c<x<c$ that $2$ is a solution of $y_1 = 1 + y^2$ with $h\left(0\right) = 0$ is

 Given :$h\left(0\right)=0,-c<x<c$
Consider, $y_1=1+y^2$
then, we have $y_1=\frac{dy}{dx}$
then, we have,
$\frac{dy}{dx}=1+y^2$
$\Rightarrow \frac{dy}{1+y^2}=dx$.
Integrating the equation, we have.
${\tan }^{-1}y=x+c \left\{\int \frac{1}{1+y^2}dy={\tan }^{-1}y\right\}$
As we know, there exists a differential function $h\left(x\right)$,
for $-c<x<c$. So,
${\tan }^{-1}\left(h\right(x\left)\right)=x+c$
$0=0+c \left\{h\right(0)=0$ is given $\}$.

$0=C$
 we have .
${\tan }^{-1}h\left(x\right)=x$.
i.e., $h\left(x\right)=\tan x$,
which is differential for $x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)$
$\therefore$ Largest value of $c=\frac{\pi }{2}$.