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 $\int_0^a f(x) \, dx = e^{-a} + 4a^2 + a - 1.$ 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:

• A. $(8e^x - 1)\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$
• B. $(8e^x + 1)\frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$
• C. $(8e^x + 1)\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$
• D. $(8e^x - 1)\frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$

Answer 1

Answer: (C)

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

Answer 2

The given integral is:

$\int_0^a f(x) dx = e^{-a} + 4a^2 + a - 1.$

Step 1: Differentiate with respect to $a$:

$f(a) = -e^{-a} + 8a + 1.$

Step 2: Differentiate again:

$f'(a) = e^{-a} + 8.$

Step 3: General solution:

The general solution for $y$ is: $y = c_1 f(x) + c_2 \implies \frac{dy}{dx} = c_1 f'(x), \quad \frac{d^2y}{dx^2} = c_1 f''(x).$

Substitute values:

$f''(x) = -e^{-x}, \quad f'(x) = e^{-x} + 8.$

The differential equation becomes:

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

Final Answer:

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