The given second-order linear differential equation is:

\[
y'' + 6y' + 13y = 0
\]

This is a homogeneous differential equation with constant coefficients. To solve it, we follow these steps:

### 1. **Form the characteristic equation:**

Assume the solution is of the form \( y = e^{rt} \), where \( r \) is a constant to be determined. Substitute \( y = e^{rt} \), \( y' = r e^{rt} \), and \( y'' = r^2 e^{rt} \) into the differential equation:

\[
r^2 e^{rt} + 6r e^{rt} + 13 e^{rt} = 0
\]

Since \( e^{rt} \) is never zero, we can divide through by \( e^{rt} \), leading to the **characteristic equation**:

\[
r^2 + 6r + 13 = 0
\]

### 2. **Solve the characteristic equation:**

Use the quadratic formula to solve for \( r \):

\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 6 \), and \( c = 13 \). Substituting these values into the formula:

\[
r = \frac{-6 \pm \sqrt{6^2 - 4(1)(13)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 52}}{2} = \frac{-6 \pm \sqrt{-16}}{2}
\]

\[
r = \frac{-6 \pm 4i}{2} = -3 \pm 2i
\]

So the roots are \( r_1 = -3 + 2i \) and \( r_2 = -3 - 2i \).

### 3. **Write the general solution:**

Since the roots are complex, the general solution is of the form:

\[
y(t) = e^{\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right)
\]

Here, \( \alpha = -3 \) and \( \beta = 2 \), so the general solution is:

\[
y(t) = e^{-3t} \left( C_1 \cos(2t) + C_2 \sin(2t) \right)
\]

where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions.

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Would you like more details on any step or further clarification?

Here are some related questions for you:

1. How do you determine the form of the solution for a second-order homogeneous differential equation with complex roots?
2. What are the physical interpretations of the constants \( C_1 \) and \( C_2 \) in the solution?
3. Can this method be applied to non-homogeneous equations? How does the process differ?
4. What happens if the roots of the characteristic equation are real and distinct?
5. How do initial conditions affect the solution to this equation?

**Tip:** For differential equations with real coefficients, complex roots always occur in conjugate pairs.

Discover how to solve the second-order homogeneous differential equation y'' + 6y' + 13y = 0. This solution explains key concepts in differential equations, including the characteristic equation, solving with complex roots, and applying the quadratic formula. Suitable for undergraduate students.

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