A curve passes through the point ≤ft(1, π/6). Let the slope of the curve at each point (x, y) be y/x + ≤ft(y/x), x > 0. Then the equation of the curve is [JEE Adv 2013 P1]

Correct Answer is : sin (yx)≤ft(y/x)(xy) = log x + 121/221

Correct Answer is : sin (yx)≤ft(y/x)(xy) = log x + 121/221

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Differential Equations

Section: Mathematics

Question : 17 of 23

Marks: +1, -0

\left(1, \frac{\pi}{6}\right)

(x, y)

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

Solution:

Given slope at

(x, y)

\frac{dy}{dx} = \frac{y}{x}

+ sec

\left(\frac{y}{x}\right)

let

\frac{y}{x} = t

⇒ y = xt ⇒

\frac{dy}{dx} = t + x \frac{dt}{dx}

t + x

\frac{dt}{dx}

= t + sec(t)

\int

cot t dt =

\int\frac{1}{x}

sin t = lnx + c

sin

\left(\frac{y}{x}\right)

= lnx + c

This curve passes through

\left(1, \frac{\pi}{6}\right)

sin

\left(\frac{\pi}{6}\right)

= ln(1) + c ⇒ c =

\frac{1}{2}

sin

\left(\frac{y}{x}\right)

= lnx +

\frac{1}{2}

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