# Linear Difference Equation

• Apr 28th 2009, 03:03 AM
LooNiE
Linear Difference Equation
Ok I have a linear difference equation, which is as follows:
f_t - f_(t+2) = 2sin(t*(pi/2))
I am not given any conditions. All I am asked to do is solve it.
• Apr 28th 2009, 04:10 AM
simplependulum
f(t) = cos[(t-1)*pi/2] + k sin[pi*(t+w)] + c , c,k,w are constants
• Apr 28th 2009, 05:38 AM
LooNiE
Quote:

Originally Posted by simplependulum
f(t) = cos[(t-1)*pi/2] + k sin[pi*(t+w)] + c , c,k,w are constants

I'm sorry, but that has completely confused me. I have solved the A.E. and the C.F, which turns out to be A(1)^t + B(-1)^t, but afterwards I get lost. I don't understand how you got that answer and would appreciate some more detail on how you arrived at it. Cheers
• Apr 28th 2009, 05:48 AM
Jester
Quote:

Originally Posted by LooNiE
Ok I have a linear difference equation, which is as follows:
f_t - f_(t+2) = 2sin(t*(pi/2))
I am not given any conditions. All I am asked to do is solve it.

Is $\displaystyle t$ an integer?
• Apr 28th 2009, 05:53 AM
LooNiE
I believe t is an integer yes.
• Apr 28th 2009, 07:38 AM
HallsofIvy
First solve $\displaystyle f(t)- f(t+2)= 0$ which is equivalent to [tex]f(t+2)= f(t). Try a solution of the form $\displaystyle a^t$ so that $\displaystyle f(t+2)= a^{t+2}= a^2a^t$ and the equation becomes $\displaystyle a^2a^t= a^t$ so that $\displaystyle a^2= 1$ and a= 1 or a= -1. The general solution to the "associated homogenous equation" is $\displaystyle C+ D(-1)^t$ exactly as you have.

To find a specific solution to [tex]f(t)- f(t+2)= 2sin((\pi/2)t), try a solution of the form $\displaystyle f(t)= A sin((\pi/2)t)$ so that $\displaystyle f(2t)= A sin((\pi/2)(t+2))= A sin((\pi/2)t+ \pi)= -A sin((\pi/2)t)$ so the equation becomes f(t)- f(t+2)= 2A sin(\pi/2)t)= 2 sin((\pi/2)t) and is satisfied if A= 1.

The solution is $\displaystyle f(t)= C+ D(-1)^t+ sin((\pi/2)t)$.