Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a differentiable function with \( f'(x) = f(x) \) for all \( x. \) Suppose that \( f(\alpha x) \) and \( f(\beta x) \) are two non-zero solutions of the differential equation \[ 4 \frac{d^2 y}{dx^2} - p \frac{dy}{dx} + 3y = 0 \] satisfying \( f(\alpha x)f(\beta x) = f(2x) \) and \( f(\alpha x)f(-\beta x) = f(x). \) Then, the value of \( p \) is ...........

Hint:

For exponential-type solutions, compare coefficients of exponents to relate constants systematically.

Answer 1

Correct Answer: 8

Step 1: Given \( f'(x) = f(x) \), so \( f(x) = Ce^x. \) 
Thus, \[ f(\alpha x) = Ce^{\alpha x}, f(\beta x) = Ce^{\beta x}. \]

Step 2: Substitute into given conditions. 
\[ f(\alpha x)f(\beta x) = C^2 e^{(\alpha + \beta)x} = f(2x) = Ce^{2x}. \] \[ \Rightarrow e^{(\alpha + \beta)x} = \frac{1}{C} e^{2x}. \] Ignoring constants, \( \alpha + \beta = 2. \) Next, \[ f(\alpha x)f(-\beta x) = C^2 e^{(\alpha - \beta)x} = f(x) = Ce^x \Rightarrow \alpha - \beta = 1. \]

Step 3: Solve for \( \alpha, \beta. \) 
Adding and subtracting: \[ \alpha = \frac{3}{2}, \beta = \frac{1}{2}. \]

Step 4: Substitute into the differential equation. 
Let \( y = f(\alpha x) = e^{\alpha x}. \) \[ \frac{dy}{dx} = \alpha e^{\alpha x}, \frac{d^2y}{dx^2} = \alpha^2 e^{\alpha x}. \] Substitute into equation: \[ 4\alpha^2 e^{\alpha x} - p\alpha e^{\alpha x} + 3e^{\alpha x} = 0. \] \[ \Rightarrow 4\alpha^2 - p\alpha + 3 = 0. \]

Step 5: Use \( \alpha = \frac{3}{2}, \beta = \frac{1}{2}. \) 
Both satisfy the same equation: \[ 4\alpha^2 - p\alpha + 3 = 0, 4\beta^2 - p\beta + 3 = 0. \] Subtract second from first: \[ 4(\alpha^2 - \beta^2) - p(\alpha - \beta) = 0. \] \[ 4(\alpha + \beta)(\alpha - \beta) = p(\alpha - \beta). \] Since \( \alpha - \beta = 1, \, \alpha + \beta = 2, \) \[ p = 8. \]

Final Answer: \[ \boxed{8} \]