Second-order difference equations. Hi guys, today it’s all about the second-order difference equations. Have a look!!

**Second-order difference equations**

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

Also, are constants. And is a function of .

If I want to solve this equation, first I have to solve its homogeneous part.

Now the homogeneous part of the equation is

In one of my earlier posts, I have shown how to solve a homogeneous difference equation.

Then I’ll get the particular solution of the difference equation.

So if is the solution of the difference equation, then

where is the general solution of the homogeneous part of the equation and is the particular integral of the equation.

Now I’ll briefly describe how to get the value of .

**Method**

- If the non-homogeneous part is a constant, say , the particular integral is
- For the non-homogeneous part to be, say , the particular integral is
- If the non-homogeneous part is, say, , the particular integral is

Next, I’ll give an example of second-order difference equations.

**Solved example on second-order difference equations**

Note: None of these examples is mine. I have chosen these from some books. I have also given the due reference at the end of the post.

So here is the example.

**Example 1 **

According to Stroud & Booth (2011)* “Solve the following difference equation in the form of the homogeneous solution plus the particular solution: , where and .”

**Solution**

Now here the given difference equation is

First of all, I’ll solve the homogeneous part of the equation.

Now the homogeneous part of the equation is

So I’ll choose the general solution of this equation as

Next, 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 is

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

So I can factorise this characteristic equation as

Thus the values of will be .

Hence the general solution of the homogeneous part is

So I can rewrite the general solution of the homogeneous part of the difference equation as

(1)

Next, I’ll get the particular solution of the difference equation.

**Step 2**

Since in this example the difference equation is , I’ll choose the particular solution as

Next, I’ll put back the value of in the equation to get

Now I’ll simplify it. So it will be

Then I’ll compare the coefficient of the variable and the constant part on both sides.

At first, I’ll compare the coefficient of the variable to get

(2)

Next, I’ll compare the constant part on both sides of the equation to get

So this means

Now, from equation (2), I already know that .

Thus the value of will be

(3)

Hence the particular integral is

(4)

Therefore the general solution of the difference equation is

So I can substitute equations (1) and (4) in to get

(5)

Now I’ll get the values of and .

**Step 3**

So I’ll substitute in the equation (5) to get

Now I already know that .

So this equation will be

which means

(6)

Next, I’ll substitute in the equation (5) to get

Now I already know that .

So this equation will be

Now this gives

(7)

Next, I’ll subtract equation (6) from equation (7) to get

So this gives

Then I’ll substitute in the equation (6) to get

Therefore I’ll put back in the equation (5) to get

So this means

is the solution of the difference equation.

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

Dear friends, this is the end of my today’s post on the second-order 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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