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

The Correct Option is C

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:

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