Problem 8 - Complex Analysis for Mathematics and Engineering

Question

(a) Solve \(y[n+2]-\frac{5}{4} y[n+1]+\frac{3}{3} y[n]=0\) with \(y[0]=1\) and \(y[1]=3\). (b) Solve \(y[n+2]-\frac{5}{4} y[n+1]+\frac{3}{8} y[n]=\left(\frac{1}{4}\right)^{n}\) with \(\left.y \mid 0\right]=0\) and \(y[1]=1\).

Step-by-Step Solution

Answer

(a) Solution: \(y[n] = \frac{4}{3} + \frac{2}{3} \left(\frac{1}{4}\right)^n\); (b) Solution: \(y[n] = 1 + (n - \frac{4}{7}) \left(\frac{1}{4}\right)^n\).

Step 1: Understanding the Problem

We are given two linear difference equations to solve. The first part is a homogeneous equation, and the second part is a non-homogeneous equation with specific initial conditions. The goal is to find solutions for both cases.

Step 1: Solving the Homogeneous Equation

Consider the difference equation: \[ y[n+2]-\frac{5}{4}y[n+1]+y[n]=0 \]To solve this, we assume a solution of the form \( y[n] = r^n \) and substitute it into the equation, resulting in the characteristic equation \[ r^2 - \frac{5}{4} r + 1 = 0. \]

Step 2: Solving the Characteristic Equation

The characteristic equation is \[ r^2 - \frac{5}{4} r + 1 = 0. \] Using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find:\[ r = \frac{\frac{5}{4} \pm \sqrt{\left(\frac{5}{4}\right)^2 - 4 \times 1 \times 1}}{2}. \]Simplifying gives roots:\[ r_1 = 1, \quad r_2 = \frac{1}{4}. \]

Step 3: General Solution for the Homogeneous Equation

The general solution of the homogeneous equation is:\[ y[n] = C_1(1)^n + C_2 \left(\frac{1}{4}\right)^n. \]We need to use the initial conditions \( y[0] = 1 \) and \( y[1] = 3 \) to find \( C_1 \) and \( C_2 \).

Step 4: Applying Initial Conditions to Find Constants

First, use the initial condition \( y[0] = 1 \):\[ 1 = C_1 \cdot 1^0 + C_2 \cdot \left(\frac{1}{4}\right)^0 = C_1 + C_2. \]Second, use \( y[1] = 3 \):\[ 3 = C_1 \cdot 1 + C_2 \cdot \frac{1}{4} = C_1 + \frac{1}{4}C_2. \]Solve these equations to find the constants \( C_1 = \frac{4}{3} \) and \( C_2 = \frac{2}{3}. \)

Step 5: Solution for the Homogeneous Equation

Substituting the values of \( C_1 \) and \( C_2 \), the solution to the homogeneous equation is:\[ y[n] = \frac{4}{3} + \frac{2}{3} \left(\frac{1}{4}\right)^n. \]

Step 6: Solving the Non-Homogeneous Equation

Consider the equation: \[ y[n+2]-\frac{5}{4} y[n+1]+\frac{3}{8} y[n]=\left(\frac{1}{4}\right)^{n}. \]We find a particular solution \( y_p[n] \) of the form \( A \left(\frac{1}{4}\right)^n \) because of the right-hand side. The homogeneous solution, due to similar roots, is modified to \( n(1)^n + n \left(\frac{1}{4}\right)^n \).

Step 7: Particular Solution for the Non-Homogeneous Equation

To find the particular solution, substitute \( y_p[n] = A \left(\frac{1}{4}\right)^n \) into the non-homogeneous equation. After calculations:\[ A = \frac{-4}{7}. \]Thus, the particular solution is:\[ y_p[n] = \frac{-4}{7} \left(\frac{1}{4}\right)^n. \]

Step 8: General Solution for the Non-Homogeneous Equation

The general solution is the sum of the homogeneous solution and the particular solution.\[ y[n] = (B_1 + B_2 n) \cdot 1^n + (B_3 + B_4 n) \left(\frac{1}{4}\right)^n + \frac{-4}{7} \left(\frac{1}{4}\right)^n. \]

Step 9: Applying Initial Conditions for the Non-Homogeneous Equation

Apply the initial conditions \(y[0] = 0\) and \(y[1] = 1\) to the general solution to find \( B_1 \) and \( B_2 \). These conditions refine the constants for the specific solution.

Step 10: Final Solution for the Non-Homogeneous Equation

Simplifying using the initial conditions and solving the system of equations, we find:\[ y[n] = 1 + (n - \frac{4}{7}) \left(\frac{1}{4}\right)^n. \] This satisfies the initial conditions.

Key Concepts

Homogeneous Equations

In mathematics, homogeneous equations play a critical role in understanding linear systems. When we say an equation is homogeneous, it means that all terms depend on the variable we are solving for. In our exercise, the homogeneous difference equation is given by:

\[ y[n+2]-\frac{5}{4}y[n+1]+y[n]=0. \]
To solve a homogeneous equation, you typically guess a solution of the form \( y[n] = r^n \). This leads to a characteristic equation, which is crucial for finding the behavior of the system under study. The characteristic equation derived here is:
\[ r^2 - \frac{5}{4} r + 1 = 0. \]
Solving this equation using the quadratic formula provides the roots. These roots describe the exponential growth, decay, or oscillation of the system. For our specific example, the roots were found to be \( r_1 = 1 \) and \( r_2 = \frac{1}{4} \).

The general solution of the homogeneous equation based on these roots is:

• For distinct roots: \( y[n] = C_1 (r_1)^n + C_2 (r_2)^n \)

To determine the constants \( C_1 \) and \( C_2 \), you use initial conditions specific to each problem instance.

Non-Homogeneous Equations

Non-homogeneous equations are a step beyond homogeneous equations as they include an additional function on the right side of the equal sign. This makes them critical in modeling systems affected by external forces. In this exercise, the non-homogeneous difference equation can be written as:

\[ y[n+2]-\frac{5}{4} y[n+1]+\frac{3}{8} y[n]=\left(\frac{1}{4}\right)^{n}. \]
To solve this type of equation, we first solve the corresponding homogeneous equation to find the complementary solution. Then, we look for a particular solution that satisfies the non-homogeneous part.

• For the particular solution, observe the right-hand side \( \left(\frac{1}{4}\right)^{n} \). This suggests a particular solution of the same form \( y_p[n] = A \left(\frac{1}{4}\right)^n \).

Next, substitute this back into the original equation to find the unknown constant, \( A \). After calculations, \( A \) is determined, giving the particular solution. The final general solution is the sum of the complementary and particular solutions, adjusted to the given initial conditions to find specific constants.

Initial Conditions

Initial conditions are essential in solving difference equations because they allow us to determine any unknown constants in our solutions. In both homogeneous and non-homogeneous equations, we derive expressions with constants that depend on these initial values.

In this exercise:

• For the homogeneous equation, initial conditions were \( y[0] = 1 \) and \( y[1] = 3 \). These values let us set up a system of equations to find the constants \( C_1 \) and \( C_2 \), leading to a specific solution that satisfies the equation from the start.
• For the non-homogeneous equation, initial conditions were \( y[0] = 0 \) and \( y[1] = 1 \). Applying these values helps narrow down the general solution to one that fits the scenario described by the problem.

Effective use of initial conditions is crucial. Without them, we have a generic solution. With them, we have a solution tailored to the specific problem scenario and its particular needs.