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To solve the differential equation:

\[
y''' - 3y'' - 9y' + 27y = 0,
\]

we'll follow these steps:

### Step 1: Assume a Solution of the Form \( y = e^{rt} \)

This assumption allows us to convert the differential equation into an algebraic equation. Plugging \( y = e^{rt} \), \( y' = re^{rt} \), \( y'' = r^2 e^{rt} \), and \( y''' = r^3 e^{rt} \) into the equation yields:

\[
r^3 e^{rt} - 3r^2 e^{rt} - 9r e^{rt} + 27 e^{rt} = 0.
\]

Since \( e^{rt} \neq 0 \), we can divide through by \( e^{rt} \):

\[
r^3 - 3r^2 - 9r + 27 = 0.
\]

### Step 2: Solve the Characteristic Polynomial

The characteristic polynomial is:

\[
r^3 - 3r^2 - 9r + 27 = 0.
\]

To factor this polynomial, let's test possible roots using the Rational Root Theorem. Testing \( r = 3 \):

\[
3^3 - 3 \cdot 3^2 - 9 \cdot 3 + 27 = 27 - 27 - 27 + 27 = 0.
\]

So, \( r = 3 \) is a root. We can factor \( r - 3 \) from the polynomial.

### Step 3: Perform Synthetic Division

Divide \( r^3 - 3r^2 - 9r + 27 \) by \( r - 3 \):

1. **Divide:** \( r^3 - 3r^2 - 9r + 27 \) by \( r - 3 \).
2. The quotient is \( r^2 - 0r - 9 \), so we have:

\[
r^3 - 3r^2 - 9r + 27 = (r - 3)(r^2 - 9).
\]

### Step 4: Factor Further

Notice that \( r^2 - 9 \) can be factored as:

\[
r^2 - 9 = (r - 3)(r + 3).
\]

So, the characteristic polynomial factors completely as:

\[
(r - 3)^2 (r + 3) = 0.
\]

### Step 5: Solve for \( r \)

The roots are:
- \( r = 3 \) (with multiplicity 2),
- \( r = -3 \).

### Step 6: Write the General Solution

Since we have a repeated root at \( r = 3 \), we use the form \( y(t) = (c_1 + c_2 t)e^{3t} \) for this root. For the distinct root \( r = -3 \), the solution is \( c_3 e^{-3t} \). Thus, the general solution is:

\[
y(t) = (c_1 + c_2 t)e^{3t} + c_3 e^{-3t}.
\]

### Final Answer

\[
y(t) = (c_1 + c_2 t)e^{3t} + c_3 e^{-3t}.
\]

Would you like a breakdown of any specific steps, or further explanation on factoring the characteristic polynomial?

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Here are five related questions to explore further:
1. How do we determine the general solution for a differential equation with complex roots?
2. What is the method to solve non-homogeneous differential equations?
3. How does the multiplicity of roots affect the solution form in differential equations?
4. Can this approach be used for higher-order differential equations?
5. What role does the exponential function play in the solutions of linear differential equations?

**Tip:** Always check for root multiplicity carefully; it affects the form of the general solution significantly.

Find the general solution of the differential equation: y''' - 3y'' - 9y' + 27y = 0.

Learn how to find the general solution to the differential equation y''' - 3y'' - 9y' + 27y = 0. This solution involves finding roots of the characteristic polynomial and accounting for root multiplicity, suitable for college-level students studying differential equations.

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