Key Concepts and Formulas

• Fundamental Theorem of Calculus (Part 1) / Leibniz Integral Rule: If $G(a) = \int_{k}^{a} f(x) dx$, where $k$ is a constant, then $G'(a) = f(a)$.
• Formation of Differential Equations: Given a general solution $y = c_1 f(x) + c_2$, differentiate successively to eliminate the arbitrary constants $c_1$ and $c_2$.
• Chain Rule: $\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$

Step-by-Step Solution

Step 1: Find $f(a)$ using the Fundamental Theorem of Calculus.

We are given that the area bounded by $y = f(x)$, $y = 0$ from $x = 0$ to $x = a$ is $e^{-a} + 4a^2 + a - 1$. This can be written as: $\int_0^a f(x) dx = e^{-a} + 4a^2 + a - 1$ To find $f(a)$, we differentiate both sides of the equation with respect to $a$ using the Fundamental Theorem of Calculus: $\frac{d}{da} \left( \int_0^a f(x) dx \right) = \frac{d}{da} \left( e^{-a} + 4a^2 + a - 1 \right)$ $f(a) = -e^{-a} + 8a + 1$ Since $a$ is just a variable, we can replace it with $x$ to get $f(x)$: $f(x) = -e^{-x} + 8x + 1$

Step 2: Obtain the general solution.

We are given that the general solution is $y = c_1 f(x) + c_2$, where $f(x) = -e^{-x} + 8x + 1$. Substituting this into the general solution, we get: $y = c_1(-e^{-x} + 8x + 1) + c_2$

Step 3: Differentiate the general solution with respect to $x$.

Differentiating the general solution with respect to $x$, we get: $\frac{dy}{dx} = c_1(e^{-x} + 8)$ Notice that we have eliminated $c_2$.

Step 4: Differentiate again with respect to $x$.

Differentiating the equation from Step 3 again with respect to $x$, we get: $\frac{d^2y}{dx^2} = c_1(-e^{-x})$

Step 5: Eliminate $c_1$ from the equations.

From the second derivative, we can express $c_1$ as: $c_1 = -\frac{1}{e^{-x}} \frac{d^2y}{dx^2} = -e^x \frac{d^2y}{dx^2}$ Substitute this expression for $c_1$ into the first derivative equation: $\frac{dy}{dx} = \left( -e^x \frac{d^2y}{dx^2} \right) (e^{-x} + 8)$ $\frac{dy}{dx} = -e^x \frac{d^2y}{dx^2} e^{-x} - 8e^x \frac{d^2y}{dx^2}$ $\frac{dy}{dx} = -\frac{d^2y}{dx^2} - 8e^x \frac{d^2y}{dx^2}$ $0 = \frac{d^2y}{dx^2} + 8e^x \frac{d^2y}{dx^2} + \frac{dy}{dx}$ $0 = (1 + 8e^x) \frac{d^2y}{dx^2} + \frac{dy}{dx}$ $(8e^x + 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$

Common Mistakes & Tips

• Careless Differentiation: Pay close attention to signs when differentiating exponential functions and polynomials. A small mistake can lead to a completely different answer.
• Elimination Strategy: Choose the easiest way to eliminate the arbitrary constants. Sometimes solving for $c_1$ and $c_2$ and substituting back is more complex than necessary.
• Understanding the Fundamental Theorem of Calculus: Make sure you correctly apply the Fundamental Theorem.

Summary

We started by finding $f(x)$ using the Fundamental Theorem of Calculus. Then, using the given general solution, we differentiated it twice to eliminate the arbitrary constants $c_1$ and $c_2$. This gave us the required differential equation. The differential equation whose general solution is $y=c_1 f(x)+c_2$ is $(8e^x + 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$.

Final Answer The final answer is $\boxed{(8e^x + 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0}$, which corresponds to option (B).