Homogeneous Difference Equations

Last updated: September 5, 2021 4:41 pm

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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 , (w^2+aw+b) can be .

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 .

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

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 , (w^2+5w+6) can be .

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 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 and .

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 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.

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Prabhu TL is an author, digital entrepreneur, and creator of high-value educational content across technology, business, and personal development. With years of experience building apps, websites, and digital products used by millions, he focuses on simplifying complex topics into practical, actionable insights. Through his writing, Dilip helps readers make smarter decisions in a fast-changing digital world—without hype or fluff.

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