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The given recurrence relation is:

\[
y_{n+3} + 3y_{n+2} + 3y_{n+1} + y_n = 0
\]

with the initial conditions:

\[
y_0 = 1, \quad y_1 = -2, \quad y_2 = -1.
\]

### Step 1: Find the characteristic equation

We solve the recurrence relation by first finding its characteristic equation. The general form of the recurrence relation suggests using:

\[
y_n = r^n
\]

Substitute \(y_n = r^n\) into the recurrence relation:

\[
r^{n+3} + 3r^{n+2} + 3r^{n+1} + r^n = 0
\]

Factor out \(r^n\):

\[
r^n (r^3 + 3r^2 + 3r + 1) = 0
\]

The characteristic equation is:

\[
r^3 + 3r^2 + 3r + 1 = 0
\]

### Step 2: Solve the characteristic equation

Factor the characteristic equation:

\[
r^3 + 3r^2 + 3r + 1 = (r + 1)^3 = 0
\]

This gives a triple root:

\[
r = -1
\]

### Step 3: General solution

Since the root \(r = -1\) has multiplicity 3, the general solution to the recurrence relation is:

\[
y_n = (A + Bn + Cn^2)(-1)^n
\]

### Step 4: Use initial conditions to solve for \(A\), \(B\), and \(C\)

We use the initial conditions \(y_0 = 1\), \(y_1 = -2\), and \(y_2 = -1\) to find the constants \(A\), \(B\), and \(C\).

- For \(n = 0\):
  \[
  y_0 = A = 1
  \]

- For \(n = 1\):
  \[
  y_1 = (A + B + C)(-1)^1 = -(1 + B + C) = -2
  \]
  Solving for \(B + C\):
  \[
  1 + B + C = 2 \quad \Rightarrow \quad B + C = 1
  \]

- For \(n = 2\):
  \[
  y_2 = (A + 2B + 4C)(-1)^2 = 1 + 2B + 4C = -1
  \]
  Solving for \(2B + 4C\):
  \[
  2B + 4C = -2 \quad \Rightarrow \quad B + 2C = -1
  \]

Now solve the system of equations:
1. \(B + C = 1\)
2. \(B + 2C = -1\)

Subtract the first equation from the second:
\[
(B + 2C) - (B + C) = -1 - 1 \quad \Rightarrow \quad C = -2
\]

Substitute \(C = -2\) into \(B + C = 1\):
\[
B - 2 = 1 \quad \Rightarrow \quad B = 3
\]

### Step 5: Final solution

Substitute \(A = 1\), \(B = 3\), and \(C = -2\) into the general solution:

\[
y_n = (1 + 3n - 2n^2)(-1)^n
\]

This is the solution to the recurrence relation.

Would you like any further details or explanations?

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Here are five related questions you might consider next:
1. How does the multiplicity of roots affect the form of the solution?
2. Can this solution be verified by plugging values of \(n\) into the recurrence relation?
3. What happens if the characteristic equation has distinct roots instead of repeated ones?
4. How do initial conditions influence the final form of the solution?
5. How can this method be extended to non-homogeneous recurrence relations?

**Tip**: When dealing with repeated roots in recurrence relations, you modify the solution by multiplying terms by powers of \(n\) to account for the multiplicity.

Solve recurrence relation: y_{n+3} + 3y_{n+2} + 3y_{n+1} + y_n = 0 with initial conditions y_0 = 1, y_1 = -2, y_2 = -1.

Learn how to solve the recurrence relation y_{n+3} + 3y_{n+2} + 3y_{n+1} + y_n = 0 with initial conditions y_0 = 1, y_1 = -2, y_2 = -1. This solution includes the characteristic equation and accounts for repeated roots.

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