Question

A continuously differentiable function φ(x) in (0, π) satisfying $y^{'} = 1 + y^{2}$, $y(0) = 0 = y(\pi)$ is

• A. tanx
• B. x(x – π)
• C. $(x - \pi)(1 - e^{x})$
• D. Not possible

Answer 1

Answer: (D)

Not possible

Answer 2

For φ(x) = y, $y^{'} = 1 + y^{2}$ ⇒ $\frac{dy}{dx} = 1 + y^{2}$

⇒ $\int_{}^{}{\frac{dy}{1 + y^{2}} = \int_{}^{}{dx}}$

⇒ $\tan^{- 1}y = x + c$

∴ y = tan (x + c)

i.e., φ(x) = tan (x + c)

As y(0) = 0, 0 = tan c and y(π) = 0 ⇒ 0 = tan (π + c) = tan c

∴ c = 0

∴ φ(x) = y = tan x.

But tan x is not continuous in (0, π)

Since $\tan\frac{\pi}{2}$ is not defined.

Hence there exists not a function satisfying the given condition.