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The equation provided is a second-order linear homogeneous differential equation:

\[
9y'' + 18y' + 19y = 0
\]

### Step 1: Write the characteristic equation
For a second-order linear homogeneous differential equation, the corresponding characteristic equation is obtained by assuming a solution of the form \( y = e^{rt} \). Substituting this into the differential equation gives:

\[
9r^2 + 18r + 19 = 0
\]

### Step 2: Solve the characteristic equation
The characteristic equation is quadratic:

\[
9r^2 + 18r + 19 = 0
\]

We can solve this using the quadratic formula:

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

where \( a = 9 \), \( b = 18 \), and \( c = 19 \). Substituting these values:

\[
r = \frac{-18 \pm \sqrt{18^2 - 4(9)(19)}}{2(9)}
\]
\[
r = \frac{-18 \pm \sqrt{324 - 684}}{18}
\]
\[
r = \frac{-18 \pm \sqrt{-360}}{18}
\]
\[
r = \frac{-18 \pm 6i\sqrt{10}}{18}
\]
\[
r = -1 \pm \frac{i\sqrt{10}}{3}
\]

So the roots are complex: \( r = -1 \pm \frac{i\sqrt{10}}{3} \).

### Step 3: Write the general solution
Since the roots are complex, the general solution to the differential equation is of the form:

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

where \( \alpha = -1 \) and \( \beta = \frac{\sqrt{10}}{3} \). Therefore, the solution is:

\[
y(t) = e^{-t}\left(C_1 \cos\left(\frac{\sqrt{10}}{3} t\right) + C_2 \sin\left(\frac{\sqrt{10}}{3} t\right)\right)
\]

This is the general solution to the initial value problem.

Would you like to know more details, or perhaps explore further steps in solving differential equations?

Here are 5 related questions:
1. How do initial conditions modify the solution to a differential equation?
2. What is the role of the characteristic equation in solving second-order differential equations?
3. Can you explain how complex roots affect the form of the solution?
4. What methods are available for solving non-homogeneous differential equations?
5. How do damping and oscillation appear in the solutions of differential equations?

**Tip**: Always check if roots of the characteristic equation are real or complex, as this affects the structure of the general solution.

Solve the initial value problem: 9y'' + 18y' + 19y = 0

Learn how to solve the second-order linear homogeneous differential equation 9y'' + 18y' + 19y = 0. This solution involves finding complex roots using the characteristic equation and deriving the general solution. Suitable for college-level students studying differential equations.

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