logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

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] $
Can you answer this question?
 
 

1 Answer

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

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...