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JEE Main 2019 · 11 Jan · Shift I

JEE Main 2019 Mathematics question on Differential Equations

If y(x)y(x) is the solution of the differential equation dydx+(2x+1x)y=e−2x, x>0\frac{dy}{dx} + \left(\frac{2x+1}{x}\right)y = e^{-2x},\ x > 0, where y(1)=12e−2y(1) = \frac{1}{2}e^{-2}, then
  1. A.
    y(x)y(x) is decreasing in (12,1)\left(\frac{1}{2}, 1\right)
  2. B.
    y(x)y(x) is increasing in (0,1)(0, 1)
  3. C.
    y(log⁡e2)=log⁡e4y(\log_e 2) = \log_e 4
  4. D.
    y(log⁡e2)=log⁡e24y(\log_e 2) = \dfrac{\log_e 2}{4}
Show answer

Correct option: A

[JEE Main 2019 · 11 Jan · Shift I]

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