The largest internal for which solution of differential equation $\large\frac{dy}{dx}=\sqrt {\large\frac{1-y^2}{(1+x^2)^2}}$ where $y(0)= \large\frac{1}{\sqrt 2}$ holds good is

Question

$(a)\;x \in [0, \infty] \\ (b)\;x \in (-\infty, 1] \\ (c)\;x \in (-\infty, \infty) \\ (d)\;x \in [-1,1] $

Answer 1

$\large\frac{dy}{\sqrt {1-y^2}}=\frac{dx}{1+x^2}$

It satisfied $(0, \large\frac{1}{\sqrt 2})$

$\sin ^{-1}y =\tan^{-1} x +c$

$\sin ^{-1}y =\tan ^{-1} x +\large\frac{\pi}{4}$

$\large\frac{-\pi}{2}$$ \leq \tan^{-1}x +\large\frac{\pi}{4} \leq$$ \large\frac{\pi}{2}$

$\large\frac{-3 \pi}{4}$$ \leq \tan^{-1}x \leq$$ \large\frac{\pi}{2}$

Variation of $\tan ^{-1}x$ is $\bigg(\large\frac{-\pi}{2},\frac{\pi}{2}\bigg)$

So variation of x is

$-\infty < x \leq 1$

Hence b is the correct answer.