The problem requires finding the general solution of the recurrence relation:

$u_n = 4u_{n-1} + 2, \quad n \geq 2$

Step-by-Step Solution

1. Solve the homogeneous part:

The homogeneous part of the recurrence relation is:

$u_n^{(h)} = 4u_{n-1}^{(h)}.$

Assume a solution of the form $u_n^{(h)} = r^n$. Substituting into the homogeneous part:

$r^n = 4r^{n-1}.$

Simplify:

$r = 4.$

Thus, the homogeneous solution is:

u_n^{(h)} = A \cdot 4^n, \quad \text{where A is a constant}.

2. Find the particular solution:

For the non-homogeneous part $2$, assume a particular solution of the form:

u_n^{(p)} = C, \quad \text{where C is a constant}.

Substitute $u_n^{(p)} = C$ into the original equation:

$C = 4C + 2.$

Simplify:

$C - 4C = 2 \implies -3C = 2 \implies C = -\frac{2}{3}.$

Thus, the particular solution is:

$u_n^{(p)} = -\frac{2}{3}.$

3. General solution:

The general solution is the sum of the homogeneous and particular solutions:

$u_n = u_n^{(h)} + u_n^{(p)}.$

Substitute the results:

$u_n = A \cdot 4^n - \frac{2}{3}.$

4. Final general solution:

The general solution of the recurrence relation is:

$u_n = A \cdot 4^n - \frac{2}{3},$

where $A$ is an arbitrary constant determined by initial conditions.

Let me know if you'd like further clarification or help solving for specific values of $A$ using initial conditions!

Related Questions:

1. How do you determine $A$ if $u_1$ is given?
2. How would the solution change if the recurrence relation was $u_n = 4u_{n-1} - 2$?
3. What is the behavior of the general solution as $n \to \infty$?
4. How do you solve non-homogeneous recurrences using generating functions?
5. What are the key differences between linear and non-linear recurrence relations?

Tip: When solving recurrence relations, always check the homogeneous solution first, and then add a particular solution for non-homogeneous terms.