Question

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

• A. tan x
• B. $x(x-\pi )$
• C. $(x-\pi )(1-e^x)$
• D. Not possible

Answer 1

The correct option is D Not possible

$\frac{dy}{dx}=1+y^2\Rightarrow \frac{dy}{1+y^2}=dx$
Integrating both sides,
$\int \frac{dy}{1+y^2}=\int dx\Rightarrow ta{n}^{-1}y=x+c$
At x =0, y =0, then c =0
AT $x=\pi ,y=0$, then $ta{n}^{-1}0=\pi +c\Rightarrow c=-\pi$
$\therefore ta{n}^{-1}y=x\Rightarrow y=tanx=\varphi \left(x\right)$
Therefore, solution is y =tan x
But x is not continuous function in $(0,\pi )$
Hence, $\varphi \left(x\right)$ is not possible in $(0,\pi )$.