Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e–a + 4a2 + a –1. Then the differential equation, whose general solution is y = c1f(x) + c2, where C1 and C2 are arbitrary constants, is

Engineering

Mathematics

Introduction to Determinants

Question

Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e^–a + 4a² + a –1. Then the differential equation, whose general solution is y = c₁f(x) + c₂, where C₁ and C₂ are arbitrary constants, is

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

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

$(8 e^{x} - 1) \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} = 10$

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

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$\int_{0}^{a} f (x) dx = e^{- a} + 4 a^{2} + a - 1$

f(a) = –e^–a + 8a + 1

$y = C_{1} (1 + 8 x - \frac{1}{e^{x}}) + C_{2}$

$\frac{dy}{dx} = C_{1} (8 + \frac{1}{e^{x}}) \Rightarrow \frac{(\frac{dy}{dx})}{(8 + \frac{1}{e^{x}})} = C_{1}$

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