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JEE Main 2013

JEE Main 2013 Mathematics question on Differential Equations

A curve passes through the point (1,π6)\left(1, \dfrac{\pi}{6}\right). Let the slope of the curve at each point (x,y)(x, y) be yx+sec⁡(yx)\dfrac{y}{x} + \sec\left(\dfrac{y}{x}\right), x>0x > 0. Then, the equation of the curve is
  1. A.
    sin⁡(yx)=log⁡x+12\sin\left(\dfrac{y}{x}\right) = \log x + \dfrac{1}{2}
  2. B.
    cosec(yx)=log⁡x+2\text{cosec}\left(\dfrac{y}{x}\right) = \log x + 2
  3. C.
    sec⁡(yx)=log⁡x+2\sec\left(\dfrac{y}{x}\right) = \log x + 2
  4. D.
    cos⁡(yx)=log⁡x+12\cos\left(\dfrac{y}{x}\right) = \log x + \dfrac{1}{2}
Show answer and solution

Correct option: A

Solution

The ODE is dy/dx = y/x + sec(y/x), a homogeneous equation. Substitute v = y/x, so y = vx and dy/dx = v + x·dv/dx.

Then: v + x·dv/dx = v + sec(v), giving x·dv/dx = sec(v).

Separating: cos(v)dv = dx/x. Integrating: sin(v) = log x + C.

So sin(y/x) = log x + C. Using the point (1, π/6): sin(π/6) = log 1 + C → 1/2 = 0 + C, so C = 1/2.

The equation of the curve is sin(y/x) = log x + 1/2.

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