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(D) Given the differential equation $\frac{dy}{dx} = 1 + y^2$. Separating the variables,we get $\frac{dy}{1 + y^2} = dx$. Integrating both sides,we ha

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Not possible

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(D) Given the differential equation $\frac{dy}{dx} = 1 + y^2$. Separating the variables,we get $\frac{dy}{1 + y^2} = dx$. Integrating both sides,we have $\int \frac{dy}{1 + y^2} = \int dx$,which gives $	an^{-1} y = x + c$. Applying the condition $y(0) = 0$,we get $	an^{-1}(0) = 0 + c$,so $c = 0$. Thus,the solution is $	an^{-1} y = x$,or $y = 	an x$. However,we are given the condition $y(\pi) = 0$. Substituting $x = \pi$ into $y = 	an x$,we get $y = 	an(\pi) = 0$,which satisfies the boundary condition. But the function $y = 	an x$ is not continuous in the interval $(0, \pi)$ because it has a vertical asymptote at $x = \frac{\pi}{2}$. Therefore,no such continuously differentiable function $\phi(x)$ exists in $(0, \pi)$.

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

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