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JEE Main 1999 Mathematics question on Differential Equations

A solution of the differential equation (dydx)2−xdydx+y=0\left(\frac{dy}{dx}\right)^2 - x\frac{dy}{dx} + y = 0 is
  1. A.
    y=2y = 2
  2. B.
    y=2xy = 2x
  3. C.
    y=2x−4y = 2x - 4
  4. D.
    y=2x2−4y = 2x^2 - 4
Show answer and solution

Correct option: C

Solution

The equation is p² - xp + y = 0 where p = dy/dx, so y = xp - p².

This is Clairaut's form. The general solution is y = cx - c², obtained by replacing p with constant c.

For c = 2: y = 2x - 4, which matches option 3. Verify directly: dy/dx = 2, so (2)² - x(2) + (2x-4) = 4 - 2x + 2x - 4 = 0 ✓

[1999]

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