# Difference Equations Assignment Help

## Difference Equation Introduction

The difference equation is a formula for computing an output sample at time n based on past and present input samples and past output samples in the time domain.

## Classiﬁcation of difference Equations:

• First-order equations
• Second-order equations

### First-order difference equations:

A general first-order difference equation takes the form

xt = f (t, xt−1) for all t.

We can solve such an equation by successive calculation: given x0 we have

x1 = f (1, x0)
x2 = f (2, x1) = f (2, f (1, x0))
and so on.

First-order linear difference equations with constant coefficient

A first-order difference equation with constant coefficient takes the form xt = axt−1 + bt where bt for t = 1, are constants.

### Equilibrium

In general, the starting point x0, the value of xt changes with t.

x* = b/(1 − a)

We call x* the equilibrium value of x. We can write the solution as xt = at(x0 − x*) + x*.

### Qualitative behaviour

The qualitative behaviour of the solution path depends on the value of a.

|a| < 1

xt converges to x*: the solution is stable. There are two sub cases:

0 < a < 1

Monotonic convergence.

−1 < a < 0

Damped oscillations.

|a| > 1

Divergence:

a > 1

Explosion.

a < −1

Explosive oscillations.

### First-order linear difference equations with variable coefficient

The solution of the equation xt = atxt−1 + bt. can be found, as before, by successive calculation. We obtain

xt = (Πs=1tas)x0 + ∑k=1ts=k+1tas)bk

Where a product with no terms (e.g. from t+1 to t) is 1.

### Second-order difference equations

A general second-order difference equation takes the form xt+2 = f (t, xt, xt+1).

As for a first-order equation, a second-order equation has a unique solution: by successive calculation we can see that given x0 and x1 there exists a uniquely determined value of xt for all t ≥ 2.

Example

Consider the equation

xt+2 − 5xt+1 + 6xt = 2t − 3.

The associated homogeneous equation is

xt+2 − 5xt+1 + 6xt = 0.

Two solutions of this equation are ut = 2t and vt = 3t. These are linearly independent:

To find a solution of the original equation we can takes the form u*t = at + b.

In order for u*t to be a solution we need

a(t+2) + b − 5[a(t+1) + b] + 6(at + b) = 2t−3.

Equating coefficients, we have a = 1 and b = 0.

Thus u*t = t is a solution.

We conclude that the general solution of the equation is xt = A2t + B3t + t.

### Second-order linear equations with constant coefficients:

A second-order linear equation with constant coefficients takes the form

xt+2 + axt+1 + bxt = ct.

The strategy for solving such an equation is very similar to the strategy for solving a second-order linear differential equations with constant coefficients. We first consider the associated homogeneous equation

xt+2 + axt+1 + bxt = 0.

The homogeneous equation:

We need to find two solutions of the homogeneous equation

xt+2 + axt+1 + bxt = 0.

We can guess that a solution takes the form ut = mt. In order for ut to be a solution, we need

mt(m2 + am + b) = 0

or, if m ≠ 0,

m2 + am + b = 0.

This is the characteristic equation of the difference equation. Its solutions are

−(1/2)a ± √((1/4)a2 − b).