To solve the recurrence relation $a_n = 2a_{n-1} + 2^n$ with the initial condition $a_0 = 2$, follow these steps:

Step 1: Homogeneous Solution

The first step is to solve the associated homogeneous recurrence relation: $a_n = 2a_{n-1}$ This is a simple first-order recurrence relation. Assuming a solution of the form $a_n = r^n$, we substitute into the homogeneous equation: $r^n = 2r^{n-1}$ This simplifies to: $r = 2$ Thus, the solution to the homogeneous equation is: $a_n^{(h)} = C \cdot 2^n$ where $C$ is a constant to be determined later.

Step 2: Particular Solution

Next, we find a particular solution to the non-homogeneous recurrence relation: $a_n = 2a_{n-1} + 2^n$ Since the non-homogeneous term is $2^n$, we try a particular solution of the form $a_n^{(p)} = A \cdot n \cdot 2^n$, where $A$ is a constant to be determined. Substituting this into the recurrence relation:

$A \cdot n \cdot 2^n = 2(A \cdot (n-1) \cdot 2^{n-1}) + 2^n$

Simplify the terms on the right-hand side:

$A \cdot n \cdot 2^n = 2A \cdot (n-1) \cdot 2^{n-1} + 2^n$ $A \cdot n \cdot 2^n = A \cdot (n-1) \cdot 2^n + 2^n$ $A \cdot n = A \cdot (n-1) + 1$ $A \cdot n = A \cdot n - A + 1$ This simplifies to: $0 = -A + 1$ Thus, $A = 1$.

So the particular solution is: $a_n^{(p)} = n \cdot 2^n$

Step 3: General Solution

The general solution to the recurrence relation is the sum of the homogeneous and particular solutions: $a_n = C \cdot 2^n + n \cdot 2^n$ $a_n = (C + n) \cdot 2^n$

Step 4: Apply Initial Condition

To find $C$, use the initial condition $a_0 = 2$: $a_0 = (C + 0) \cdot 2^0 = C = 2$

Thus, $C = 2$, and the final solution is: $a_n = (2 + n) \cdot 2^n$

Final Answer:

$a_n = (2 + n) \cdot 2^n$

Would you like further details or have any questions?

Here are 5 related questions for practice:

1. Solve the recurrence relation $a_n = 3a_{n-1} + 2^n$, with $a_0 = 1$.
2. Find the general solution for $a_n = 4a_{n-1} + 3^n$, with $a_0 = 0$.
3. Solve the recurrence relation $a_n = a_{n-1} + n^2$, with $a_0 = 5$.
4. Determine the solution for $a_n = 5a_{n-1} + n$, with $a_0 = 10$.
5. Solve the recurrence relation $a_n = 2a_{n-1} + n$, with $a_0 = 3$.

Tip: For non-homogeneous recurrence relations, always try a particular solution form that mirrors the non-homogeneous term (e.g., polynomials for linear terms, exponentials for exponential terms).