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Magazine Chapter 8: Problem 29
a) Find all solutions of the recurrence relation \(a_{n}=2 a_{n-1}+3^{n} .\) b) Find the solution of the recurrence relation in part (a) with initial condition \(a_{1}=5\)
Short Answer
Expert verified
The solution is \(a_n = -2 \times 2^n + 3^{n+1}\).
Step by step solution
01
- Understand the Recurrence Relation
The recurrence relation given is \(a_{n}=2a_{n-1}+3^n\). This means the value of \(a_n\) depends on the previous term \(a_{n-1}\) and the term \(3^n\).
02
- Find the Homogeneous Solution
Find the solution to the homogeneous part of the recurrence relation \(a_{n}^{(h)}=2a_{n-1}^{(h)}\). The characteristic equation is \(r=2\), so the homogeneous solution is \(a_{n}^{(h)}=A \times 2^n\), where \(A\) is a constant.
03
- Find the Particular Solution
For the non-homogeneous part \(a_{n}=2a_{n-1}+3^n\), assume a particular solution of the form \(a_{n}^{(p)}=B \times 3^n\). Substitute this into the original recurrence relation:\(B \times 3^n = 2B \times 3^{n-1} + 3^n\).Divide through by \(3^{n-1}\) to find \(B\).
04
- Solve for B
Simplify the previous equation from Step 3. \(3B = 2B +3\). This simplifies to \[3B - 2B = 3\], which results in \[B = 3\]. So, the particular solution is \(a_{n}^{(p)} = 3 \times 3^n = 3^{n+1}\).
05
- Combine Solutions
The general solution is the sum of the homogeneous and particular solutions:\(a_n = a_{n}^{(h)} + a_{n}^{(p)} = A \times 2^n + 3^{n+1}\).
06
- Apply Initial Condition
Use the initial condition \(a_1=5\) to find \(A\).Substitute \(n=1\) into the general solution: \(a_1 = A \times 2^1 + 3^2 = 5\).So, \(2A + 9 = 5\), \[2A = -4\], \[A = -2\].
07
- Write Final Solution
Substitute \(A\) back into the general solution: \(a_n = -2 \times 2^n + 3^{n+1}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Solution
The first step in solving a recurrence relation is to handle the homogeneous part. A homogeneous recurrence relation doesn't include the independent term like \(3^n\) in our given example (where \(a_n = 2a_{n-1} + 3^n\)). Instead, it looks something like \(a_n^{(h)} = 2a_{n-1}^{(h)}\).
The characteristic equation helps us here. For our homogeneous relation \(a_n^{(h)} = 2a_{n-1}^{(h)}\), setting \(a_n = r^n\) and substituting it into the relation gives: \[r^n = 2r^{n-1}\rightarrow r = 2.\]
So, the homogeneous solution is: \[a_n^{(h)} = A \times 2^n.\]
Here, \(A\) is a constant that will be determined later using the initial conditions.
The characteristic equation helps us here. For our homogeneous relation \(a_n^{(h)} = 2a_{n-1}^{(h)}\), setting \(a_n = r^n\) and substituting it into the relation gives: \[r^n = 2r^{n-1}\rightarrow r = 2.\]
So, the homogeneous solution is: \[a_n^{(h)} = A \times 2^n.\]
Here, \(A\) is a constant that will be determined later using the initial conditions.
Particular Solution
The next step is to find the particular solution. This part deals with the non-homogeneous element of the recurrence relation鈥擻(3^n\) in this case.
We assume a particular solution of the form \(a_n^{(p)} = B \times 3^n\).
We substitute this into the non-homogeneous recurrence relation: \[B \times 3^n = 2B \times 3^{n-1} + 3^n.\]
By dividing both sides by \(3^{n-1}\), we get: \[3B = 2B + 3.\]
Simplifying this, we find: \[B = 3.\]
So, the particular solution is: \[a_n^{(p)} = 3 \times 3^n = 3^{n+1}.\]
We assume a particular solution of the form \(a_n^{(p)} = B \times 3^n\).
We substitute this into the non-homogeneous recurrence relation: \[B \times 3^n = 2B \times 3^{n-1} + 3^n.\]
By dividing both sides by \(3^{n-1}\), we get: \[3B = 2B + 3.\]
Simplifying this, we find: \[B = 3.\]
So, the particular solution is: \[a_n^{(p)} = 3 \times 3^n = 3^{n+1}.\]
Characteristic Equation
The characteristic equation is integral to finding the homogeneous solution. In general, for a recurrence relation of the form \(a_n = C \times a_{n-1}\), we propose a solution \(a_n = r^n\). Substituting this into the recurrence relation gives us our characteristic equation: \[r^n = C \times r^{n-1} \rightarrow r = C.\]
For our example, where \(a_n = 2a_{n-1}\), the characteristic equation yields: \[r = 2.\]
The roots of this equation give us the basis for our homogeneous solution. Each root \(r\) corresponds to a solution of the form \(A \times r^n\), where \(A\) is a constant determined by initial conditions.
For our example, where \(a_n = 2a_{n-1}\), the characteristic equation yields: \[r = 2.\]
The roots of this equation give us the basis for our homogeneous solution. Each root \(r\) corresponds to a solution of the form \(A \times r^n\), where \(A\) is a constant determined by initial conditions.
Initial Condition
Finally, we apply the initial condition to find the constant in our general solution. For our problem, the initial condition given is \(a_1 = 5\).
Our general solution combines both the homogeneous and the particular solutions: \[a_n = A \times 2^n + 3^{n+1}.\]
Substituting \(n = 1\) and using the initial condition: \[5 = A \times 2^1 + 3^{2}.\]
We get: \[5 = 2A + 9.\]
Simplifying further: \[2A = -4 \rightarrow A = -2.\]
Thus, the final solution is: \[a_n = -2 \times 2^n + 3^{n+1}.\]
Our general solution combines both the homogeneous and the particular solutions: \[a_n = A \times 2^n + 3^{n+1}.\]
Substituting \(n = 1\) and using the initial condition: \[5 = A \times 2^1 + 3^{2}.\]
We get: \[5 = 2A + 9.\]
Simplifying further: \[2A = -4 \rightarrow A = -2.\]
Thus, the final solution is: \[a_n = -2 \times 2^n + 3^{n+1}.\]
