Homogeneous difference equations. Hello friends, today it’s about homogeneous difference equations. Have a look!!

**Homogeneous difference equations**

Now the general form of any second-order difference equation is:

Also, are constants.

If , then the equation becomes

Then this is an example of second-order homogeneous difference equations.

Now I’ll show how to solve these types of equations.

In some other post, I’ll show * how to solve a non-homogeneous difference equation*.

**Method**

First of all, I’ll choose a general solution to this difference equation. So, let’s say

Next, I’ll put this value of in the difference equation.

So it will be

Here is the common term.

So I can take it out.

Thus the equation will be

Since cannot be , can be .

Therefore what I get is

Now, this is the characteristic equation of this difference equation.

Next, I have to solve this equation to get the values of .

Since has the power , I’ll get two values of .

So let’s choose and .

Then the general solution of the difference equation will be

Now that’s how it works.

Next, I’ll solve some examples of homogeneous difference equations.

**Solved examples in Homogeneous difference equations**

Disclaimer: None of these examples is mine. I have chosen these from some book (s). At the end of the post, there are also due references.

**Example 1**

According to Stroud and Booth (2013)* “Solve the following difference equation: where and .”

**Solution**

Now here the given difference equation is

First of all, I’ll choose the general solution of this equation as

Now I’ll find out the characteristic equation of this difference equation.

**Step 1**

So, for that I’ll put in the difference equation to get

Now I can take out as a common term. Therefore the new equation will be

Since cannot be , can be .

Therefore what I get is

So the characteristic equation of this difference equation will be

Next, I’ll solve this equation to get the general solution of the difference equation.

**Step 2**

So I can factorise this characteristic equation as

Thus the values of will be .

Hence the general solution of the difference equation is

(1)

Now I’ll get the values of and .

So I’ll substitute in the equation (1) to get

Now I already know that .

So this equation will be

(2)

Next, I’ll substitute in the equation (1) to get

Now I already know that .

So this equation will be

(3)

As I can see from equation (2), .

Next, I’ll put back in equation (3).

Thus it will be

So this gives which means

Hence the value of will be

Now I’ll substitute in the equation (1) to get

So this means

is the general solution of the difference equation.

Hence I can conclude that this is the answer to this example.

Now I’ll give another example of homogeneous difference equations.

**Example 2**

According to Stroud and Booth (2013)* “Solve the following difference equation: where and .”

**Solution**

Now here the given difference equation is

First of all, I’ll choose the general solution of this equation as

Now I’ll find out the characteristic equation of this difference equation.

**Step 1**

So, for that I’ll put in the difference equation to get

Now I can take out as a common term. Therefore the new equation will be

Since cannot be , can be .

Therefore what I get is

So the characteristic equation of this difference equation will be

Next, I’ll solve this equation to get the general solution of the difference equation.

**Step 2**

So I can factorise this characteristic equation as

Thus the values of will be .

Hence the general solution of the difference equation is

In other woords, I can also say,

(4)

Now I’ll get the values of and .

So I’ll substitute in the equation (4) to get

Now I already know that .

So this equation will be

(5)

Next, I’ll substitute in equation (4) to get

Now I already know that .

So this equation will be

Next, I’ll simplify it to get

(6)

Now, from equation (6), I can say that

Next, I’ll substitute this value of in equation (5) to get

So this means

and

Now I’ll substitute in equation (4) to get

So this means

is the general solution of the difference equation.

Hence I can conclude that this is the answer to this example.

Now I’ll give another example of homogeneous difference equations.

**Example 3**

According to Stroud and Booth (2013)* “Solve the following difference equation: where and .”

**Solution**

Now here the given difference equation is

First of all, I’ll choose the general solution of this equation as

Now I’ll find out the characteristic equation of this difference equation.

**Step 1**

So, for that I’ll put in the difference equation to get

Now I can take out as a common term. Therefore the new equation will be

Since cannot be , can be .

Therefore what I get is

So the characteristic equation of this difference equation will be

Next, I’ll solve this equation to get the general solution of the difference equation.

**Step 2**

So I can factorise this characteristic equation as

Thus the values of will be .

Hence the general solution of the difference equation is

(7)

Now I’ll get the values of and .

So I’ll substitute in the equation (7) to get

Now I already know that .

So this equation will be

(8)

Next, I’ll substitute in the equation (7) to get

Now I already know that .

So this equation will be

(9)

As I can see from equation (8), .

Next, I’ll put back in equation (9).

Thus it will be

So this gives which means

Hence the value of will be

Now I’ll substitute in the equation (7) to get

So this means

is the general solution of the difference equation.

Hence I can conclude that this is the answer to this example.

Dear friends, this is the end of today’s post on homogeneous difference equations. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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