Question

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

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

Answer 1

Answer: (A)

$y=x{\tan }^{-1}\left(\ell n\frac{e}{x}\right)$

$\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$

$\frac{y}{\cos x}=m$

$\frac{dy}{dx}=\frac{dm}{dx}+m$

$\frac{xdm}{dx}+m=m-{\cos }^{2}m$

$\frac{xdm}{dx}=-{\cos }^{2}m$

${\sec }^{2}dm=\frac{-dx}{x}$

Integration

$\tan m=-\ln x+c$

$\tan m=\ln \left(\frac{e}{x}\right)$

$m={\tan }^{-1}\left[\ln \left|\frac{e}{x}\right|\right]$

$y=x{\tan }^{-1}\left[\ln \left(\frac{e}{x}\right)\right]$