Question 75

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 $$e^{-a} + 4a^2 + a - 1$$. Then the differential equation, whose general solution is $$y = c_1 f(x) + c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is

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f(x) = d/dx[e⁻ˣ+4x²+x-1] = -e⁻ˣ+8x+1. f'(x) = e⁻ˣ+8. y = c₁f(x)+c₂ → y' = c₁f'(x) → y'' = c₁f''(x) = c₁(−e⁻ˣ). Also y' = c₁(e⁻ˣ+8). So c₁ = y'/(e⁻ˣ+8) and y'' = -c₁e⁻ˣ = -y'e⁻ˣ/(e⁻ˣ+8). (e⁻ˣ+8)y''+e⁻ˣy'=0 → multiply by eˣ: (1+8eˣ)y''+y'=0.

Option (4): (8eˣ+1)y''+y'=0.

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