Question

Which one of the following is the general solution of the differential equation \[ \frac{d^2 y}{dx^2} - 8 \frac{dy}{dx} + 16y = 2e^{4x} ? \]

• A. \( \alpha_1 e^{4x} + \alpha_2 x e^{4x} \), where \( \alpha_1, \alpha_2 \in \mathbb{R} \)
• B. \( \alpha_1 e^{4x} + \alpha_2 x e^{4x} + 2x e^{4x} \), where \( \alpha_1, \alpha_2 \in \mathbb{R} \)
• C. \( \alpha_1 e^{4x} + \alpha_2 e^{4x} + 2x e^{4x} \), where \( \alpha_1, \alpha_2 \in \mathbb{R} \)
• D. \( \alpha_1 x e^{-4x} + \alpha_2 x^2 e^{4x} \), where \( \alpha_1, \alpha_2 \in \mathbb{R} \)

Hint:

To solve a second-order linear differential equation with constant coefficients, first solve the homogeneous equation, then use an appropriate guess for the particular solution.

Answer 1

The Correct Option is A

Step 1: Solve the homogeneous equation.
The homogeneous equation is \[ \frac{d^2 y}{dx^2} - 8 \frac{dy}{dx} + 16y = 0 \] The characteristic equation is: \[ r^2 - 8r + 16 = 0 \] Solving for \( r \), we get: \[ r = 4 \] Thus, the general solution of the homogeneous equation is: \[ y_h = \alpha_1 e^{4x} + \alpha_2 x e^{4x} \] Step 2: Solve the non-homogeneous equation.
For the non-homogeneous term \( 2e^{4x} \), we guess a particular solution of the form: \[ y_p = A x^2 e^{4x} \] Substitute into the original equation to find \( A \). The result is: \[ y_p = 2x e^{4x} \] Step 3: Write the general solution.
The general solution is the sum of the homogeneous and particular solutions: \[ y = y_h + y_p = \alpha_1 e^{4x} + \alpha_2 x e^{4x} + 2x e^{4x} \] Final Answer: \[ \boxed{\alpha_1 e^{4x} + \alpha_2 x e^{4x}} \]