Second-Order Linear Difference Equations

This is an interactive guide to solving a second-order linear difference equation. Click on the steps below to reveal the solution process.

Problem: Solve \(y_{t+2} = 4y_{t+1} - 8y_t\) with initial conditions \(y_0 = 1\) and \(y_1 = 4\).

Step 1: Characteristic Eq. Explaining Step 1 Step 2: Find Roots Step 3: General Solution Step 4: Solve Constants Explaining Step 4 Why Add Equations? Step 5: Final Solution

Step 1: Find the Characteristic Equation

We convert the difference equation into a polynomial known as the characteristic equation, which helps us find the general pattern of the solution.

The characteristic equation for our problem is: \(\lambda^2 - 4\lambda + 8 = 0\)

Explaining Step 1: What is a Characteristic Equation?

Think of \(y_{t+2} = 4y_{t+1} - 8y_t\) as a rule for a sequence. To find its underlying pattern, assume \(y_t = \lambda^t\) and plug it in.

First rearrange: \(y_{t+2} - 4y_{t+1} + 8y_t = 0\).

Then substitute \(y_t = \lambda^t\):

\(\lambda^{t+2} - 4\lambda^{t+1} + 8\lambda^t = 0\)

Factor out \(\lambda^t\):

\(\lambda^t(\lambda^2 - 4\lambda + 8) = 0\)

The term in brackets is the characteristic equation: \(\lambda^2 - 4\lambda + 8 = 0\).

Step 2: Finding the Roots (\(\lambda\))

We solve \(\lambda^2 - 4\lambda + 8 = 0\) using the quadratic formula.

\(\lambda = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(8)}}{2} = \frac{4 \pm \sqrt{-16}}{2} = 2 \pm 2i\)

The roots are: \(\lambda_1 = 2 + 2i\) and \(\lambda_2 = 2 - 2i\).

Step 3: General Solution

The general solution of a second-order linear difference equation is:

\(y_t = c_1(\lambda_1)^t + c_2(\lambda_2)^t\)

Substitute the roots:

\(y_t = c_1(2+2i)^t + c_2(2-2i)^t\)

Step 4: Find Constants (\(c_1, c_2\))

Using \(y_0 = 1\) and \(y_1 = 4\), we form two equations:

\(c_1 + c_2 = 1\)
\(c_1 - c_2 = -i\)

Explaining Step 4: Building the Equations

For \(t = 0\), \(y_0 = c_1 + c_2 = 1\).
For \(t = 1\), \(y_1 = 2(c_1 + c_2) + 2i(c_1 - c_2) = 4\).

From this we find \(c_1 - c_2 = -i\).

Why Add Equations?

Adding equations helps eliminate one variable, making it easier to solve. It’s like combining two clues to find one missing piece.

Add (I) and (II): \(2c_1 = 1 - i \Rightarrow c_1 = \frac{1 - i}{2}\).

Step 5: Final Solution

We find:

\(c_1 = \frac{1 - i}{2}, \quad c_2 = \frac{1 + i}{2}\)

So the final solution is:

\(y_t = \left(\frac{1-i}{2}\right)(2+2i)^t + \left(\frac{1+i}{2}\right)(2-2i)^t\)