# 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.

## Classication of difference Equations:

- First-order equations
- Second-order equations

### First-order difference equations:

A general first-order difference equation takes the form

x_{t} = f (t, x_{t}1) for all t.

We can solve such an equation by successive calculation: given x_{0} 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 x_{t} = ax_{t}1 + b_{t} where b_{t} for t = 1, are constants.

### Equilibrium

In general, the starting point x_{0}, the value of x_{t} changes with t.

x* = b/(1 a)

We call x* the equilibrium value of x. We can write the solution as x_{t} = a^{t}(x0 x*) + x*.

### Qualitative behaviour

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

|a| < 1

x_{t} 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 x_{t} = a_{t}x_{t}1 + bt. can be found, as before, by successive calculation. We obtain

x_{t} = (_{s}=1_{t}a_{s})x0 + k=1^{t}(_{s}=k+1^{t}a_{s})b_{k}

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 x_{t}+2 = f (t, x_{t}, x_{t}+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 x_{t} 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) = 2t3.

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

x_{t}+2 + ax_{t}+1 + bx_{t} = c_{t}.

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 u_{t} = m^{t}. In order for ut to be a solution, we need

m^{t}(m^{2} + am + b) = 0

or, if m 0,

m^{2} + am + b = 0.

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

(1/2)a ((1/4)a^{2} b).