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Homogeneous Difference Equations

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

\[af(n+2)+b f(n+1) +cf(n) = d.\]

Also, a,b,c, d are constants.

If d = 0, then the equation becomes 

\[f(n+2)+a f(n+1) +bf(n) = 0.\]

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

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

METHOD

First of all, I’ll choose a general solution to this difference equation. So, let’s say f(n) = k w^{n}.

Next, I’ll put this value of f(n) in the difference equation.

So it will be

\[kw^{n+2}+akw^{n+1}+bkw^{n} =0.\]

Here kw^{n} is the common term.

So I can take it out.

Thus the equation will be

\[kw^{n}(w^2+aw+b) = 0.\]

Since kw^{n} cannot be 0, (w^2+aw+b) can be 0.

Therefore what I get is

\[(w^2+aw+b) = 0.\]

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

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

Since w has the power 2, I’ll get two values of w.

So let’s choose w = m_1 and m_2.

Then the general solution of the difference equation will be

\[f(n) = A \times m_1^{n} + B \times m_2^{n}.\]

Now that’s how it works.

EXAMPLE

According to Stroud and Booth (2013)* “Solve the following difference equation: f(n+2)+5f(n+1)+6f(n) = 0 where f(0) = 0 and f(1) = 1.”

SOLUTION

Now here the given difference equation is

\[ \boxed{f(n+2)+5f(n+1)+6f(n) = 0,~~~~~f (0) = 0, f(1) = 1.} \]

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

\[f(n) = kw^n.\]

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

STEP 1

So, for that I’ll put f(n) = kw^n in the difference equation to get

\[kw^{n+2}+5kw^{n+1}+6kw^{n} = 0.\]

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

\[kw^n(w^2+5w+6) = 0.\]

Since kw^{n} cannot be 0, (w^2+5w+6) can be 0.

Therefore what I get is

\[(w^2+5w+6) = 0.\]

So the characteristic equation of this difference equation will be 

\[ \boxed{w^2+5w+6 = 0.} \]

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

\[(w+2)(w+3) = 0.\]

Thus the values of w will be w = -2, -3.

Hence the general solution of the difference equation is

(1) \begin{equation*}f (n) = A \times (-2)^{n}+B\times (-3)^{n}.\end{equation*}

Now I’ll get the values of A and B.

So I’ll substitute n = 0 in the equation (1) to get

\[f(0) = A \times (-2)^{0}+B\times (-3)^{0}.\]

Now I already know that f (0) = 0.

So this equation will be

\[0 = A \times 1 + B \times 1.\]

Hence this means 

(2) \begin{equation*}A + B = 0.\end{equation*}

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

\[f(1) = A \times (-2)^{1}+B\times (-3)^{1}.\]

Now I already know that f (1) = 1.

So this equation will be

\[1 = A \times (-2) + B \times (-3).\]

Hence this means 

(3) \begin{equation*}-2 A -3 B = 1.\end{equation*}

As I can see from equation (2), A = -B.

Next, I’ll put back A = -B in equation (3).

Thus it will be

\[-2 (-B) -3 B = 1.\]

So this gives -B = 1 which means

\[B = -1.\]

Hence the value of A will be

\[A = 1.\]

Now I’ll substitute A = 1, B = -1 in the equation (1) to get

\[f (n) = 1 \times (-2)^{n}+(-1)\times (-3)^{n}.\]

So this means 

\[f (n) = (-2)^{n}- (-3)^{n}\]

is the general solution of the difference equation.

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