Question

The general solution of $\frac{d^2 y}{dx^2} - \frac{dy}{dx} - 6y = 0$ is given by}

• A. $y(x) = C_1 e^{-3x} + C_2 e^{3x}$
• B. $y(x) = C_1 e^{-3x} + C_2 e^{-2x}$
• C. $y(x) = C_1 e^{-2x} + C_2 e^{3x}$
• D. $y(x) = C_1 e^{-2x} + C_2 e^{2x}$

Hint:

Always double-check the factoring of the auxiliary quadratic equation to ensure the signs of the roots ($m = 3, -2$) are correct before writing down the exponents.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A second-order homogeneous linear differential equation with constant coefficients can be solved using its auxiliary (characteristic) equation.
Key Formula or Approach:
For the differential equation $a y'' + b y' + c y = 0$, the auxiliary equation is:
\[ a m^2 + b m + c = 0 \]
If the roots $m_1, m_2$ are real and distinct, the general solution is:
\[ y(x) = C_1 e^{m_1 x} + C_2 e^{m_2 x} \]

Step 2: Detailed Explanation:
Given the differential equation:
\[ \frac{d^2 y}{dx^2} - \frac{dy}{dx} - 6y = 0 \]
Write down the auxiliary equation:
\[ m^2 - m - 6 = 0 \]
Factor this quadratic equation:
\[ (m - 3)(m + 2) = 0 \]
This gives two real and distinct roots:
\[ m_1 = -2 \quad \text{and} \quad m_2 = 3 \]
Substitute these roots into the general solution template:
\[ y(x) = C_1 e^{-2x} + C_2 e^{3x} \]
Therefore, the general solution is $y(x) = C_1 e^{-2x} + C_2 e^{3x}$.

Step 3: Final Answer
The correct option is (C).