Question

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

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

Answer 1

Answer: (C)

Given : $y=\sec \left(\tan ^{-1} x\right)$
Put $t=\tan ^{-1} x $
$\Rightarrow \tan t=x $
$\Rightarrow \sec ^{2} t\, d t=d x$
$y=\sec t $ Differentate w.r.t. $x'$
$\therefore y=\sec t $ Differentate w.r.t. '$x$'
$\frac{d y}{d x}=\sec t \tan t \frac{d t}{d x} $
$\Rightarrow \frac{d y}{d x}=\frac{\sec t \tan t}{1+x^{2}} $
$\Rightarrow \frac{d y}{d x}=\frac{x y}{1+x^{2}} $
$ \therefore \left(1+x^{2}\right) \frac{d y}{d x}=x y$