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Chapter 6: Problem 17

Find the particular solution. \(x_{n+1}=2 x_{n}+8 x_{n-1} ; x_{0}=1, x_{1}=-2\)

Short Answer

Expert verified
The particular solution starts from the initial conditions: \(x_2 = 4\), \(x_3 = -8\).

Step by step solution

01

Understand the Recursive Relation

The given recursive relation is \(x_{n+1} = 2x_{n} + 8x_{n-1}\), with initial conditions \(x_{0} = 1\) and \(x_{1} = -2\). The goal is to find the particular solution for \(n \geq 0\).
02

Apply Initial Conditions

Start by using the initial conditions to find the next term, \(x_2\). Using the relation \(x_{n+1} = 2x_{n} + 8x_{n-1}\), substituting \(n = 1\) gives us \(x_2 = 2x_1 + 8x_0\). Substitute \(x_0 = 1\) and \(x_1 = -2\) into the formula: \(x_2 = 2(-2) + 8(1) = -4 + 8 = 4\).
03

Calculate Further Terms if Necessary

Continue to apply the recursive relation to calculate additional terms if required. For example, to find \(x_3\), use \(n = 2\): \(x_3 = 2x_2 + 8x_1\). Substitute \(x_2 = 4\) and \(x_1 = -2\) into the formula: \(x_3 = 2(4) + 8(-2) = 8 - 16 = -8\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

initial conditions
Initial conditions in a recursive sequence help us start the chain of calculations. For the given problem, the initial conditions are provided as follows:
  • The value of the first term, denoted as \(x_0\), is 1.
  • The value of the second term, denoted as \(x_1\), is -2.
These initial conditions are essential because they allow us to compute the subsequent terms using the recursive relation.
Consider them as the starting point or the seeds that help the sequence grow. Without initial conditions, we would not have the necessary information to progress with the calculations.
recursive relation
A recursive relation defines how each term in the sequence is related to the previous terms. For the given exercise, the recursive relation is:
\(x_{n+1} = 2x_{n} + 8x_{n-1}\).
This means any term (\(x_{n+1}\)) can be calculated using the two terms before it (\(x_{n}\) and \(x_{n-1}\)).
Let鈥檚 go through an example:
  • If we know the values of \(x_0\) and \(x_1\), we can use the relation to find \(x_2\).
  • Substitute \(x_0 = 1\) and \(x_1 = -2\) into the relation, which gives us \(x_2 = 2(-2) + 8(1) = -4 + 8 = 4\).
The recursive relation simplifies complex sequences into manageable calculations, making it easier to generate further terms.
particular solution
A particular solution refers to the specific sequence derived from applying the initial conditions and the recursive relation. This solution provides us with the exact terms in the sequence that fit the given criteria.
In this exercise, we鈥檝e computed the first few terms based on initial conditions and the recursive relation:
  • \(x_0 = 1\)
  • \(x_1 = -2\)
  • \(x_2 = 4\)
  • \(x_3 = -8\)
Because these terms follow both the initial conditions and the recursive relation, they are part of the particular solution. This explicit sequence allows us to understand the behavior of the entire sequence over time, ensuring that all terms comply with the defined rules.

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Most popular questions from this chapter

Solve using Gaussian elimination. $$ \begin{array}{rr} x+y-2 z= & 4 \\ 4 x+7 y+3 z= & 3 \\ 14 x+23 y+5 z= & 10 \end{array} $$

Home Air Quality: Chemicals enter a house's basement air. Let \(\mathrm{F}(t)\) and \(B(t)\) be the amount of the chemical (in \(\mathrm{mg} / \mathrm{m}^{3}\) ) in the first-floor air and the basement air after \(t\) minutes, respectively. The rate of change of \(F\) and \(B\) are given by the equations \(^{8}\) \(B^{\prime}=-0.01 B+0.1\) and \(F^{\prime}=0.01 B-0.02 F\) Find the equilibrium values for \(B\) and \(\mathrm{F}\)

Let \(A=\left[\begin{array}{ll}4 & -1 \\ 7 & -9\end{array}\right], B=\left[\begin{array}{rr}3 & 9 \\ 2 & -2\end{array}\right], C=\left[\begin{array}{rr}8 & 3 \\ 0 & -3\end{array}\right]\) \(D=\left[\begin{array}{rrr}5 & -6 & 1 \\ 10 & 3 & -1\end{array}\right], E=\left[\begin{array}{rr}-7 & 4 \\ -3 & 2 \\ 2 & -1\end{array}\right]\) \(F=\left[\begin{array}{rrr}-4 & 2 & 3 \\ 0 & -1 & 2 \\ -7 & -2 & 5\end{array}\right]\), and \(v=\left[\begin{array}{r}2 \\\ -3\end{array}\right]\). Compute \(\mathrm{B}-\mathrm{C}\) and \(\mathrm{B}+(-1)\) C. Compare your answers.

If the characteristic equation for a second-order linear difference equation has a double root \(r\), then the general solution is of the form $$ x_{n}=c_{1} r^{n}+c_{2} n r^{n} . $$ Find the particular solution of \(x_{n+1}=6 x_{n}-9 x_{n-1}, x_{0}=7, x_{1}=21\)

The matrix \(A\) has eigenvalues \(r_{1}\) and \(r_{2}\) with corresponding eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), respectively. Compute \(\mathbf{A}^{n} \mathbf{w}\). $$ \begin{array}{l} r_{1}=2, r_{2}=1, v_{1}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right], v_{2}=\left[\begin{array}{l} 0 \\ 1 \end{array}\right], n=10, \\ w=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \end{array} $$
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