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JEE Main 2019 · 12 Jan · Shift II

JEE Main 2019 Mathematics question on Differential Equations

If a curve passes through the point (1,−2)(1, -2) and has slope of the tangent at any point (x,y)(x, y) on it as x2−2yx\frac{x^2 - 2y}{x}, then the curve also passes through the point
  1. A.
    (3,0)(\sqrt{3}, 0)
  2. B.
    (−1,2)(-1, 2)
  3. C.
    (−2,1)(-\sqrt{2}, 1)
  4. D.
    (3,0)(3, 0)
Show answer and solution

Correct option: A

Solution

The ODE is dy/dx = (x² - 2y)/x, rewritten as dy/dx + (2/x)y = x. This is linear with integrating factor e^(∫2/x dx) = x². Multiplying: d/dx(x²y) = x³, so x²y = x⁴/4 + C. Using point (1, -2): 1·(-2) = 1/4 + C, giving C = -9/4. The curve is x²y = x⁴/4 - 9/4. At x = √3: 3y = 9/4 - 9/4 = 0, so y = 0. Thus the curve passes through (√3, 0).

[JEE Main 2019 · 12 Jan · Shift II]

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