Question

The slope of the tangent at (x, y) to a curve passing through $\left( 1,\frac{\pi}{4} \right)$ as given by $\frac{y}{x} - \cos^{2}\left( \frac{y}{x} \right)$, then the equation of the curve is

• A. $y = \tan^{- 1}\left\lbrack \log\left( \frac{e}{x} \right) \right\rbrack$
• B. $y = x\tan^{- 1}\left\lbrack \log\left( \frac{x}{e} \right) \right\rbrack$
• C. $y = x\tan^{- 1}\left\lbrack \log\left( \frac{e}{x} \right) \right\rbrack$
• D. None of these

Answer 1

Answer: (C)

$y = x\tan^{- 1}\left\lbrack \log\left( \frac{e}{x} \right) \right\rbrack$

Answer 2

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

Let y = vx ⇒ $\frac{dy}{dx} = v + x\frac{dv}{dx}$ ⇒ $v + x\frac{dv}{dx} = v - \cos^{2}v$

⇒ $x\frac{dv}{dx} = - \cos^{2}v$ ⇒ $\sec^{2}vdv = - \frac{dx}{x}$ ⇒ tan v = – ln x + c

⇒ tan (y/x) = – ln x + c

For x = 1, y = π/4

⇒ $\tan\pi/4 = - \ln 1 + c$ ⇒ $1 = 0 + c$

∴ c = 1

∴ $\tan(y/x) = 1 - \ln x$

⇒ $y/x = \tan^{- 1}(1 - \ln x) = \tan^{- 1}(\ln e - \ln x) = \tan^{- 1}\left\lbrack \ln\left( \frac{e}{x} \right) \right\rbrack$

∴ $y = x\tan^{- 1}\left\lbrack \ln\left( \frac{e}{x} \right) \right\rbrack$