MCQMediumJEE Main 2024 · 8 April, Shift 1Linear Differential Equations

Mathematics Question from JEE Main 2024 · 8 April, Shift 1

Let f(x)f(x) be a positive function such that the area bounded by y=f(x)y = f(x), y=0y = 0, from x=0x = 0 to x=a>0x = a > 0 is ea+4a2+a1e^{-a} + 4a^2 + a - 1. Then the differential equation, whose general solution is y=c1f(x)+c2y = c_1 f(x) + c_2, where c1c_1 and c2c_2 are arbitrary constants, is:

  • A

    (8ex1)d2ydx2+dydx=0(8e^x - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0

  • B

    (8ex+1)d2ydx2dydx=0(8e^x + 1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0

  • C

    (8ex+1)d2ydx2+dydx=0(8e^x + 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0

  • D

    (8ex1)d2ydx2dydx=0(8e^x - 1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0

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