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(B) Given the area under the curve $y=f(x)$ from $x=0$ to $x=a$ is $\int_0^a f(x) dx = e^{-a}+4a^2+a-1$.By the Fundamental Theorem of Calculus,differe

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

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(B) Given the area under the curve $y=f(x)$ from $x=0$ to $x=a$ is $\int_0^a f(x) dx = e^{-a}+4a^2+a-1$.By the Fundamental Theorem of Calculus,differentiating both sides with respect to $a$ gives $f(a) = \frac{d}{da}(e^{-a}+4a^2+a-1) = -e^{-a}+8a+1$.Thus,$f(x) = -e^{-x}+8x+1$.Given the general solution $y = c_1 f(x) + c_2$,we differentiate with respect to $x$:$\frac{dy}{dx} = c_1 f'(x) = c_1(e^{-x}+8)$.$\frac{d^2y}{dx^2} = c_1 f''(x) = c_1(e^{-x})$.From the second derivative,we have $c_1 = e^x \frac{d^2y}{dx^2}$.Substituting $c_1$ into the first derivative equation:$\frac{dy}{dx} = (e^x \frac{d^2y}{dx^2})(e^{-x}+8) = \frac{d^2y}{dx^2}(1+8e^x)$.Rearranging gives $(8e^x+1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.

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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