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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Differential Equations

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

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

$\large\frac{dy}{\sqrt {1-y^2}}=\frac{dx}{1+x^2}$
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.
answered Feb 7, 2014 by meena.p
 

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