Question

Let \( m \) and \( n \) be fixed real numbers. If the function
\[ y(t) = C_1 e^t + C_2 e^{-t} \]
is a solution of
\[ \frac{d^2y}{dt^2} + m \frac{dy}{dt} + n y = 0 \]
for any constants \( C_1 \) and \( C_2 \), then \( m + n \) is equal to

• A. -2
• B. -1
• C. 0
• D. 1

Hint:

When solving second-order differential equations with exponential solutions, ensure to equate the coefficients of \( e^t \) and \( e^{-t} \) to zero to find relationships between the parameters.

Answer 1

The Correct Option is B

Step 1: Calculate the first and second derivatives of \( y(t) \). 
We are given the function: \[ y(t) = C_1 e^t + C_2 e^{-t}. \] The first derivative of \( y(t) \) is: \[ \frac{dy}{dt} = C_1 e^t - C_2 e^{-t}. \] The second derivative of \( y(t) \) is: \[ \frac{d^2y}{dt^2} = C_1 e^t + C_2 e^{-t}. \] Step 2: Substitute into the differential equation. 
Substitute \( y(t) \), \( \frac{dy}{dt} \), and \( \frac{d^2y}{dt^2} \) into the given differential equation: \[ \frac{d^2y}{dt^2} + m \frac{dy}{dt} + n y = 0, \] \[ (C_1 e^t + C_2 e^{-t}) + m(C_1 e^t - C_2 e^{-t}) + n(C_1 e^t + C_2 e^{-t}) = 0. \] Now group terms involving \( e^t \) and \( e^{-t} \): \[ (C_1 + m C_1 + n C_1)e^t + (C_2 - m C_2 + n C_2)e^{-t} = 0. \] This gives two separate equations: 1. \( C_1 (1 + m + n) = 0 \), 2. \( C_2 (1 - m + n) = 0 \). 
Since this equation must hold for any constants \( C_1 \) and \( C_2 \), the coefficients of \( e^t \) and \( e^{-t} \) must be zero. Therefore: \[ 1 + m + n = 0 \quad \Rightarrow \quad m + n = -1. \] Thus, the correct answer is \( \boxed{(B)} \).