Differential Equations Question 134
Question: Differential equation of $ y=\sec ({{\tan }^{-1}}x) $ is
[UPSEAT 2002]
Options:
A) $ (1+x^{2})\frac{dy}{dx}=y+x $
B) $ (1+x^{2})\frac{dy}{dx}=y-x $
C) $ (1+x^{2})\frac{dy}{dx}=xy $
D) $ (1+x^{2})\frac{dy}{dx}=\frac{x}{y} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\sec ({{\tan }^{-1}}x) $
$ \frac{dy}{dx}=\sec ({{\tan }^{-1}}x)\tan ({{\tan }^{-1}}x).\frac{1}{1+x^{2}} $
$ =\frac{xy}{1+x^{2}} $
Therefore $ (1+x^{2})\frac{dy}{dx}=xy $ .
BETA
