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Chapter 22: Problem 3

Solve the difference equation $$ y[k+2]+3 y[k+1]+2 y[k]=0 $$ subject to the conditions \(y[0]=0, y[1]=1\).

Short Answer

Expert verified
The solution is \( y[k] = (-1)^k - (-2)^k \).

Step by step solution

01

Write the Characteristic Equation

To solve the homogeneous difference equation \( y[k+2] + 3y[k+1] + 2y[k] = 0 \), start by finding the characteristic equation. Assume a solution of the form \( y[k] = r^k \). Substitute it into the equation to get: \( r^{k+2} + 3r^{k+1} + 2r^k = 0 \). This simplifies to: \( r^k (r^2 + 3r + 2) = 0 \). The characteristic equation is \( r^2 + 3r + 2 = 0 \).
02

Solve the Characteristic Equation

Solve the quadratic equation \( r^2 + 3r + 2 = 0 \) using the factorization method. The equation can be factored as \( (r + 1)(r + 2) = 0 \). Solving for \( r \), we find the roots to be \( r = -1 \) and \( r = -2 \).
03

Write the General Solution

The general solution for the difference equation is based on its characteristic roots. Since both roots \( r_1 = -1 \) and \( r_2 = -2 \) are distinct, the general solution is given by: \( y[k] = A(-1)^k + B(-2)^k \), where \( A \) and \( B \) are constants determined by the initial conditions.
04

Apply Initial Conditions

Use the initial conditions to solve for \( A \) and \( B \). Substituting \( y[0] = 0 \) into the general solution gives: \( A(-1)^0 + B(-2)^0 = A + B = 0 \). Substituting \( y[1] = 1 \), we get: \( A(-1)^1 + B(-2)^1 = -A - 2B = 1 \).
05

Solve for Constants \( A \) and \( B \)

Solve the system of equations from Step 4: \( A + B = 0 \) and \( -A - 2B = 1 \). First, express \( A = -B \) from the first equation. Substitute \( A = -B \) into the second equation: \( -(-B) - 2B = 1 \), simplifying to \( B - 2B = 1 \), resulting in \( -B = 1 \). Thus, \( B = -1 \) and \( A = 1 \).
06

Write the Particular Solution

Substitute \( A = 1 \) and \( B = -1 \) back into the general solution. Thus, the particular solution satisfying the initial conditions is: \( y[k] = (-1)^k - (-2)^k \).

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Key Concepts

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

Characteristic Equation
When solving a homogeneous difference equation like \( y[k+2] + 3y[k+1] + 2y[k] = 0 \), the first goal is to derive the characteristic equation. This equation helps us determine possible patterns in the solution. To do this, assume a solution of the form \( y[k] = r^k \), where \( r \) is an unknown constant. Substituting this form back into the difference equation leads to \( r^{k+2} + 3r^{k+1} + 2r^k = 0 \). Simplifying, we factor out \( r^k \) to get:
  • \( r^k (r^2 + 3r + 2) = 0 \)
The characteristic equation is \( r^2 + 3r + 2 = 0 \). Solving this characteristic equation will allow us to find the values of \( r \), which will guide us toward crafting the general solution of the difference equation.
Initial Conditions
Initial conditions are critical to determining the specific solution that fits a real-world scenario. In this exercise, we're given \( y[0] = 0 \) and \( y[1] = 1 \). Initial conditions help us find the exact constants in our general solution, transforming it from a type of ideal or general solution to a practical, particular solution.After solving the characteristic equation and obtaining the general solution, these values are substituted in to determine specific constants (such as \( A \) and \( B \) in our equation). Applying these conditions ensures that the solution is tailored to the specific sequence at hand and satisfies the scenario given by the problem.
Quadratic Equation
A quadratic equation is vital in the process of solving the characteristic equation. In our case, the characteristic equation \( r^2 + 3r + 2 = 0 \) is indeed quadratic as it is of the form \( ax^2 + bx + c = 0 \). Solving this involves finding the roots, which are the values of \( r \) that satisfy the equation. There are several methods to solve a quadratic equation, such as:
  • Factorization
  • Quadratic formula
  • Completing the square
For this exercise, the factorization method is used:
  • \((r + 1)(r + 2) = 0\)
The roots are \( r = -1 \) and \( r = -2 \). These roots are essential as they directly influence the form of the general solution.
General Solution
The general solution of a linear difference equation is a structured expression using the roots from the characteristic equation. It provides a template for all possible solutions of the difference equation.For our equation, because the roots are distinct \( r_1 = -1 \) and \( r_2 = -2 \), the general solution takes the form:
  • \( y[k] = A(-1)^k + B(-2)^k \)
Here, \( A \) and \( B \) are coefficients, which are determined using the initial conditions. This solution captures the way the values of the sequence will behave over time, showing how they might grow, decay, or oscillate, depending on the specific problem at hand.

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Most popular questions from this chapter

If $$ z[n] z[n-1]=n^{2} \quad z[1]=7 $$ find \(z[2], z[3]\) and \(z[4]\).

Solve the difference equation $$ x[k+2]-3 x[k+1]+2 x[k]=\delta[k] $$ subject to the conditions \(x[0]=x[1]=0\).

Write each equation so that the highest argument of the dependent variable is as specified: (a) \(p[k]-3 p[k+1]=p[k-2]\), highest argument of the dependent variable is to be \(k+2\). (b) \(R[n-1]-R[n-2]-R[n-3]=n, R[0]=1\) \(R[1]=-2\), highest argument of the dependent variable is to be \(n\). (c) \(q[t]+t q[t-1]=3 q[t+1], q[1]=0\), \(q[2]=-2\), highest argument of the dependent variable is to be \(t-1\). (d) \(T[m]+(m-1) T[m-2]=m^{2}\) \(T[0]=T[1]=1\), highest argument of the dependent variable is to be \(m+2\). (e) \(y[k+1]-y[k+3]=(s[k+2]-s[k+4]) / 2\) where \(y\) is the dependent variable and the highest argument of the dependent variable is to be \(k\).

Find the inverse \(z\) transforms of (a) \(\frac{2 z}{(z-2)(z-3)}\) (b) \(\frac{e z}{(e z-1)^{2}}\) (c) \(1-\frac{2}{z}+\frac{z}{(z-3)(z-4)}\) (d) \(\frac{z^{2}}{\left(z^{2}-\frac{1}{9}\right)}\) (e) \(\frac{2 z^{2}}{(z-1)(z-0.905)}\)

Calculate the first five terms of the following difference equations with the given initial conditions: (a) \(x[n+1]-x[n]=2, x[0]=3\) $$ \text { (b) } \begin{aligned} x[n+2]+x[n+1]-x[n]=4 \\ & x[0]=5, x[1]=7 \end{aligned} $$ $$ \text { (c) } x[p+2]-x[p+1]+2 x[p]=2 \text {, } $$ $$ x[0]=1, x[1]=1 $$
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