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)

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