Question

A curve passes through the point $\left(1, \frac{\pi}{4}\right) \&$ its slope at any point is given by $\frac{y}{x}-\cos ^2\left(\frac{y}{x}\right)$. Then the curve has the equation

• A. $ y=x \tan ^{-1}\left(\ln \frac{e}{x}\right)$
• B. $y=x \tan ^{-1}(\ln +2)$
• C. $y=\frac{1}{x} \tan ^{-1}\left(\ln \frac{e}{x}\right)$
• D. none

Answer 1

Answer: (A)

$\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x} $
$y=v x $
$V+x \frac{d v}{d x}=v-\cos ^2 v$
$\int \frac{d v}{\cos ^2 v}+\int \frac{d x}{x}=C$
$\tan v+\ln x=C $
$\tan \frac{y}{x}+\ln x=C$
if $x=1, y=\frac{\pi}{4} \Rightarrow C=1 $
$\tan \frac{y}{x}=1-\ln x=\ln \frac{e}{x} $
$y=x \tan ^{-1}\left(\ln \frac{y}{x}\right) \Rightarrow A$