t n -2t n - 1 =(n+5)3^ n n >= 1 solve using Inhomogenous Recuurence.

To solve the recurrence relation \( t_n - 2t_{n-1} = (n+5)3^n \) for \( n \geq 1 \) using the method of solving inhomogeneous recurrence relations, we will follow these steps:

### Step 1: Solve the Homogeneous Part
First, solve the corresponding homogeneous recurrence relation:
\[
t_n - 2t_{n-1} = 0
\]

The characteristic equation is:
\[
r - 2 = 0
\]
So, \( r = 2 \).

The general solution to the homogeneous equation is:
\[
t_n^{(h)} = C \cdot 2^n
\]
where \( C \) is a constant to be determined by initial conditions.

### Step 2: Find a Particular Solution
Next, we need to find a particular solution \( t_n^{(p)} \) to the inhomogeneous equation:
\[
t_n^{(p)} - 2t_{n-1}^{(p)} = (n+5)3^n
\]

Since the non-homogeneous term is of the form \( (n+5)3^n \), we try a particular solution of the form:
\[
t_n^{(p)} = (An + B) \cdot 3^n
\]
where \( A \) and \( B \) are constants to be determined.

Substitute \( t_n^{(p)} = (An + B)3^n \) into the original recurrence relation:
\[
(An + B)3^n - 2 \cdot \left[ A(n-1) + B \right]3^{n-1} = (n+5)3^n
\]

Simplifying, we get:
\[
(An + B)3^n - 2A(n-1) \cdot 3^n - 2B \cdot 3^{n-1} = (n+5)3^n
\]

\[
3^n \left( An + B - \frac{2A(n-1)}{3} \right) - \frac{2B}{3} \cdot 3^n = (n+5)3^n
\]

Simplifying further:
\[
3^n \left( An + B - \frac{2An}{3} + \frac{2A}{3} - \frac{2B}{3} \right) = (n+5)3^n
\]

Now, equate coefficients of like terms:
1. Coefficient of \( n \cdot 3^n \):
   \[
   A - \frac{2A}{3} = 1 \implies \frac{A}{3} = 1 \implies A = 3
   \]
2. Constant term:
   \[
   B + \frac{2A}{3} - \frac{2B}{3} = 5 \implies B + 2 - \frac{2B}{3} = 5
   \]
   Simplifying:
   \[
   \frac{3B - 2B}{3} = 3 \implies B = 9
   \]

Thus, the particular solution is:
\[
t_n^{(p)} = (3n + 9)3^n
\]

### Step 3: General Solution
The general solution to the original recurrence relation is the sum of the homogeneous and particular solutions:
\[
t_n = t_n^{(h)} + t_n^{(p)} = C \cdot 2^n + (3n + 9)3^n
\]

### Step 4: Apply Initial Conditions (If Given)
If an initial condition like \( t_0 \) is given, you can substitute \( n = 0 \) and solve for the constant \( C \).

### Summary
The general solution to the recurrence relation \( t_n - 2t_{n-1} = (n+5)3^n \) is:
\[
t_n = C \cdot 2^n + (3n + 9)3^n
\]
where \( C \) is determined by initial conditions.

---

Would you like more details or have any questions? Here are some related questions to consider:

1. How would you solve a recurrence relation with a different inhomogeneous term, like \( 3^n \) instead of \( (n+5)3^n \)?
2. What if the characteristic equation had complex roots? How would the solution differ?
3. How can you determine the initial conditions for this problem?
4. What if the recurrence relation had a different coefficient, like \( 3t_{n-1} \) instead of \( 2t_{n-1} \)?
5. How do you verify that the solution satisfies the original recurrence relation?

**Tip:** Always check your particular solution by substituting it back into the original recurrence to verify that it satisfies the equation.

t_n - 2t_{n-1} = (n+5)3^n, n >= 1. Solve using Inhomogeneous Recurrence.

Discover how to solve the inhomogeneous recurrence relation t_n - 2t_{n-1} = (n+5)3^n for n >= 1 using the method of undetermined coefficients. This step-by-step solution includes finding the homogeneous solution, particular solution, and applying initial conditions. Suitable for undergraduate mathematics students.

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