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To solve the given recurrence relation \( d_n = -d_{n-1} + 6d_{n-2} \), follow these steps:

### Part (a): Find the Characteristic Equation
For a recurrence relation of the form \( d_n = a d_{n-1} + b d_{n-2} \), the characteristic equation is given by:
\[ r^2 - a r - b = 0 \]

For the given relation \( d_n = -d_{n-1} + 6d_{n-2} \), the coefficients are \( a = -1 \) and \( b = 6 \). Plug these into the characteristic equation:
\[ r^2 + r - 6 = 0 \]

### Part (b): Find the Roots of the Characteristic Equation
Solve the quadratic equation \( r^2 + r - 6 = 0 \) to find the roots:
\[ r^2 + r - 6 = (r + 3)(r - 2) = 0 \]
So, the roots are:
\[ r_1 = -3 \]
\[ r_2 = 2 \]

### Part (c): Find the Closed-Form Expression for \( d_n \)
The general solution to the recurrence relation is of the form:
\[ d_n = A r_1^n + B r_2^n \]
Substituting the roots:
\[ d_n = A (-3)^n + B (2)^n \]

To determine the constants \( A \) and \( B \), use the initial conditions \( d_0 = 1 \) and \( d_1 = 4 \):

For \( n = 0 \):
\[ d_0 = A (-3)^0 + B (2)^0 = A + B = 1 \]

For \( n = 1 \):
\[ d_1 = A (-3)^1 + B (2)^1 = -3A + 2B = 4 \]

Solve these equations simultaneously:
1. \( A + B = 1 \)
2. \( -3A + 2B = 4 \)

From equation (1):
\[ B = 1 - A \]

Substitute \( B = 1 - A \) into equation (2):
\[ -3A + 2(1 - A) = 4 \]
\[ -3A + 2 - 2A = 4 \]
\[ -5A + 2 = 4 \]
\[ -5A = 2 \]
\[ A = -\frac{2}{5} \]

Substitute \( A = -\frac{2}{5} \) back into equation (1):
\[ -\frac{2}{5} + B = 1 \]
\[ B = 1 + \frac{2}{5} = \frac{5}{5} + \frac{2}{5} = \frac{7}{5} \]

Thus, the closed-form expression for \( d_n \) is:
\[ d_n = -\frac{2}{5} (-3)^n + \frac{7}{5} (2)^n \]

### Part (d): Find \( d_{10} \) and \( d_{14} \)
To find \( d_{10} \):
\[ d_{10} = -\frac{2}{5} (-3)^{10} + \frac{7}{5} (2)^{10} \]

Calculate each term separately:
\[ (-3)^{10} = 59049 \]
\[ 2^{10} = 1024 \]

Then:
\[ d_{10} = -\frac{2}{5} \cdot 59049 + \frac{7}{5} \cdot 1024 \]
\[ d_{10} = -\frac{2 \cdot 59049}{5} + \frac{7 \cdot 1024}{5} \]
\[ d_{10} = -\frac{118098}{5} + \frac{7168}{5} \]
\[ d_{10} = -23619.6 + 1433.6 \]
\[ d_{10} = -22186 \]

To find \( d_{14} \):
\[ d_{14} = -\frac{2}{5} (-3)^{14} + \frac{7}{5} (2)^{14} \]

Calculate each term separately:
\[ (-3)^{14} = 4782969 \]
\[ 2^{14} = 16384 \]

Then:
\[ d_{14} = -\frac{2}{5} \cdot 4782969 + \frac{7}{5} \cdot 16384 \]
\[ d_{14} = -\frac{2 \cdot 4782969}{5} + \frac{7 \cdot 16384}{5} \]
\[ d_{14} = -\frac{9565938}{5} + \frac{114688}{5} \]
\[ d_{14} = -1913187.6 + 22937.6 \]
\[ d_{14} = -1890250 \]

So, the answers are:
\[ d_{10} = -22186 \]
\[ d_{14} = -1890250 \]

Do you want further details or have any questions about this solution? Here are some related questions:

1. How would the characteristic equation change if the recurrence relation had a different form?
2. What are the general steps to solve a linear recurrence relation with constant coefficients?
3. How can we verify the solution to a recurrence relation?
4. What if the characteristic equation had repeated roots? How would the solution change?
5. Can you explain how the initial conditions affect the solution to a recurrence relation?
6. How would you solve a non-homogeneous recurrence relation?
7. What methods are available for solving higher-order recurrence relations?
8. Can you describe the significance of the characteristic roots in the solution of recurrence relations?

**Tip:** Always double-check your calculations, especially when working with large powers, to avoid arithmetic mistakes.

Learn how to solve the recurrence relation d_n = -d_{n-1} + 6d_{n-2} step-by-step with a detailed solution. This problem covers key concepts in recurrence relations and characteristic equations. Suitable for undergraduate students and those interested in mathematical problem-solving.

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