# Solve The Difference Equation

• Apr 29th 2009, 04:31 PM
louboutinlover
Solve The Difference Equation
$f_{n+2} - 2f_{n+1}+ 5f_n = 0$

given that $f_0=1$ and $f_1=3$

Auxiliary equation:

$m^2 - 2m + 5 = 0$

Therefore:

$m = 1 + 2i$ or $m = 1 - 2i$

What would the complementry fuction be and paticular solution to work towards the solution of:

$f_n = [(1-i)/2][1+2i]^n + [(1-i)/2][1-2i]^n$
• Apr 29th 2009, 09:55 PM
CaptainBlack
Quote:

Originally Posted by louboutinlover
$f_{n+2} - 2f_{n+1}+ 5f_n = 0$

given that $f_0=1$ and $f_1=3$

Auxiliary equation:

$m^2 - 2m + 5 = 0$

Which is derived from an assumed solution of the form $f_n=m^n$

Quote:

Therefore:

$m = 1 + 2i$ or $m = 1 - 2i$
OK, so you have a general solution of the form:

$f_n=A (1+2\text{i})^n+B(1-2\text{i})^n$

and the initial conditions will allow you to determine $A$ and $B$, which you appear to have done.

Quote:

What would the complementry fuction be and paticular solution to work towards the solution of:

$f_n = [(1-i)/2][1+2i]^n + [(1-i)/2][1-2i]^n$

CB
• Apr 30th 2009, 12:32 AM
louboutinlover
So using the initial conditions would give:

$A+B=1$

$A(1+2i)+B(1-2i)=3$

But then how do you solve these equations ?
• Apr 30th 2009, 01:52 AM
CaptainBlack
Quote:

Originally Posted by louboutinlover
So using the initial conditions would give:

$A+B=1$

$A(1+2i)+B(1-2i)=3$

But then how do you solve these equations ?

$A=1-B$ from 1st,

$(1-B)(1+2\text{i})+B(1-2\text{i})=3$ by substituting from the above into the second.

So:

$
B[-(1+2\text{i})+(1-2\text{i})]=3-(1+2\text{i})
$

$
B=\frac{-1+1\text{i}}{2\text{i}}=\frac{1+i}{2}
$

Assuming my algebra is right (check it, it appears to differ from what you are give by a sign here or there).

CB