A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$. Then the equation of the curve is

(A) $\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}$

(B) $\operatorname{cosec}\left(\frac{y}{x}\right)=\log x+2$

(C) $\sec \left(\frac{2 y}{x}\right)=\log x+2$

(D) $\cos \left(\frac{2 y}{x}\right)=\log x+\frac{1}{2}$

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Differential Equations

A curve passes through the point

(1, \frac{π}{6})

. Let the slope of the curve at each point

(x, y)

\frac{y}{x} + sec (\frac{y}{x}), x > 0

. Then the equation of the curve is

Answer: (A)

Given

\frac{d y}{d x} = \frac{y}{x} + sec (\frac{y}{x})

. Let

v = \frac{y}{x}

, so

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

. Substituting:

v + x \frac{d v}{d x} = v + sec v

, thus

x \frac{d v}{d x} = sec v

. Separating variables:

cos v d v = \frac{d x}{x}

. Integrating:

sin v = ln ∣ x ∣ + C

. Using the initial condition

(1, π /6)

sin (π /6) = ln 1 + C

, so

C = \frac{1}{2}

. Therefore,

sin (\frac{y}{x}) = lo g x + \frac{1}{2}

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option