Question

Differential equation of $y=\sec \left({\tan }^{-1}x\right)$ is:

• A. $\left(1+x^2\right)\frac{dy}{dx}=y+x$
• B. $\left(1+x^2\right)\frac{dy}{dx}=y-x$
• C. $\left(1+x^2\right)\frac{dy}{dx}=xy$
• D. $\left(1+x^2\right)\frac{dy}{dx}=\frac{x}{y}$

Answer 1

Answer: (C)

Given : $y=\sec \left({\tan }^{-1}x\right)$
Put $t={\tan }^{-1}x$
$\Rightarrow \tan t=x$
$\Rightarrow {\sec }^{2}tdt=dx$
$y=\sec t$ Differentate w.r.t. $x^{\prime }$
$\therefore y=\sec t$ Differentate w.r.t. '$x$'
$\frac{dy}{dx}=\sec t\tan t\frac{dt}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{\sec t\tan t}{1+x^2}$
$\Rightarrow \frac{dy}{dx}=\frac{xy}{1+x^2}$
$\therefore \left(1+x^2\right)\frac{dy}{dx}=xy$