Solving the Linear Homogeneous Difference Equation

The given equation is a linear, homogeneous difference equation with constant coefficients:

\begin{equation*} x(t + 2) + 3x (t + 1) + 2x (t) =0 \end{equation*}

with initial conditions \(x(0) = 0\) and \(x(1) = 1\). We assume that the variable \(t\) refers to the discrete time index, often denoted by \(k\). So we can rewrite the equation as:

\begin{equation*} x(k + 2) + 3x (k + 1) + 2x (k) =0 \end{equation*}

To solve this type of equation, we use the characteristic equation method.

Finding the Characteristic Equation and Roots

We assume a solution of the form \(x(k) = r^k\). Substituting this into the difference equation:

\begin{equation*} r^{k+2} + 3r^{k+1} + 2r^k = 0 \end{equation*}

Since \(r^k\) cannot be zero for a non-trivial solution, we can divide the entire equation by \(r^k\):

\begin{equation*} r^2 + 3r + 2 = 0 \end{equation*}

This is the characteristic equation. We can solve this quadratic equation by factoring:

\begin{equation*} (r + 1)(r + 2) = 0 \end{equation*}

The roots of the characteristic equation are \(r_1 = -1\) and \(r_2 = -2\).

Forming the General Solution

Since the roots are distinct and real, the general solution of the difference equation is given by:

\begin{equation*} x(k) = C_1 r_1^k + C_2 r_2^k \end{equation*}

Substituting the roots \(r_1 = -1\) and \(r_2 = -2\):

\begin{equation*} x(k) = C_1 (-1)^k + C_2 (-2)^k \end{equation*}

Here, \(C_1\) and \(C_2\) are constants determined by the initial conditions.

Applying Initial Conditions to Find Constants

We are given the initial conditions \(x(0) = 0\) and \(x(1) = 1\). We substitute these values into the general solution:

For \(k=0\), \(x(0) = 0\):

\begin{equation*} x(0) = C_1 (-1)^0 + C_2 (-2)^0 = 0 \end{equation*}

\begin{equation*} C_1 (1) + C_2 (1) = 0 \end{equation*}

\begin{equation*} C_1 + C_2 = 0 \quad (*)\end{equation*}

For \(k=1\), \(x(1) = 1\):

\begin{equation*} x(1) = C_1 (-1)^1 + C_2 (-2)^1 = 1 \end{equation*}

\begin{equation*} C_1 (-1) + C_2 (-2) = 1 \end{equation*}

\begin{equation*} -C_1 - 2C_2 = 1 \quad (**)\end{equation*}

Now we have a system of two linear equations with two variables \(C_1\) and \(C_2\):

• \(C_1 + C_2 = 0\)
• \(-C_1 - 2C_2 = 1\)

From the first equation, we can express \(C_1\) as \(C_1 = -C_2\).

Substitute this into the second equation:

\begin{equation*} -(-C_2) - 2C_2 = 1 \end{equation*}

\begin{equation*} C_2 - 2C_2 = 1 \end{equation*}

\begin{equation*} -C_2 = 1 \end{equation*}

\begin{equation*} C_2 = -1 \end{equation*}

Now substitute the value of \(C_2\) back into \(C_1 = -C_2\):

\begin{equation*} C_1 = -(-1) = 1 \end{equation*}

The Specific Solution

Substitute the values of \(C_1 = 1\) and \(C_2 = -1\) back into the general solution \(x(k) = C_1 (-1)^k + C_2 (-2)^k\):

\begin{equation*} x(k) = 1 \cdot (-1)^k + (-1) \cdot (-2)^k \end{equation*}

\begin{equation*} x(k) = (-1)^k - (-2)^k \end{equation*}

This is the specific solution that satisfies the given difference equation and initial conditions. The options use the notation \(x(KT)\) or \(x(k)\), which corresponds to our \(x(k)\).

Comparing with Given Options

Let's compare our derived solution \(x(k) = (-1)^k - (-2)^k\) with the given options:

• Option 1: \(x(KT) = (-1)^k + (2)^k\) - Does not match.
• Option 2: \(x(KT) = (-1)^k - (2)^k\) - Does not match.
• Option 3: \(x(KT) = (-1)^k + (-2)^k\) - Does not match.
• Option 4: \(x(KT) = (-1)^k - (-2)^k\) - Matches our solution.

Revision Table: Solving Difference Equations

Additional Information: Types of Difference Equations

Difference equations can be classified based on various properties:

• Linear vs. Nonlinear: A difference equation is linear if it does not involve products or nonlinear functions of the dependent variable \(x(k)\) or its delayed terms. The given equation is linear.
• Homogeneous vs. Non-homogeneous: A linear difference equation is homogeneous if the right-hand side is zero. If the right-hand side is a non-zero function of \(k\), it is non-homogeneous. The given equation is homogeneous.
• Constant Coefficients vs. Variable Coefficients: The coefficients multiplying the \(x\) terms are constants (1, 3, 2) in the given equation, making it a constant coefficient equation. If they were functions of \(k\), it would be variable coefficient.
• Order: The order of a difference equation is the difference between the highest and lowest indices of the dependent variable. In \(x(k+2) + 3x(k+1) + 2x(k) = 0\), the indices are \(k+2\), \(k+1\), and \(k\). The highest is \(k+2\) and the lowest is \(k\). The order is \((k+2) - k = 2\). This is a second-order difference equation.

Solving linear homogeneous difference equations with constant coefficients using the characteristic equation is a standard technique, analogous to solving linear homogeneous differential equations with constant coefficients.