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JEE Main 2019 · 9 Apr · Shift I

JEE Main 2019 Mathematics question on Differential Equations

The solution of the differential equation xdydx+2y=x2(x≠0)x\frac{dy}{dx} + 2y = x^2(x \neq 0) with y(1)=1y(1) = 1, is
  1. A.
    y=x24+34x2y = \frac{x^2}{4} + \frac{3}{4x^2}
  2. B.
    y=x35+15x2y = \frac{x^3}{5} + \frac{1}{5x^2}
  3. C.
    y=3x24+14x2y = \frac{3x^2}{4} + \frac{1}{4x^2}
  4. D.
    y=4x35+15x2y = \frac{4x^3}{5} + \frac{1}{5x^2}
Show answer and solution

Correct option: A

Solution

The ODE is dy/dx + (2/x)y = x, a linear equation with integrating factor e^(∫2/x dx) = x².

Multiplying: d/dx(x²y) = x³, so x²y = x⁴/4 + C.

Applying y(1) = 1: 1 = 1/4 + C, giving C = 3/4.

Therefore y = x²/4 + 3/(4x²).

[JEE Main 2019 · 9 Apr · Shift I]

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