Math Help - affine difference/solution equations

1. affine difference/solution equations

Hey! I am having a problem in my math class understanding affine/difference equations...I can give you an example right out of my text book of one of the problems I'm struggling with:

For each affine difference equation and initial condition, find the solution equation.
a) p(n) = 2.5p(n-1)+900; p(0)= 400 - p(0) is the initial condition
b) p(n) = 0.75p(n-1) - 45; p(0) = 510
c) p(n) = 1.03p(n-1) - 75; p(0) = 75

Maybe it's just me, but i have no idea where to start. I have looked at examples in the first few pages of the chapter, only to be led to the wrong answer! Can anyone help??

2. Originally Posted by emilyerin88
Hey! I am having a problem in my math class understanding affine/difference equations...I can give you an example right out of my text book of one of the problems I'm struggling with:

For each affine difference equation and initial condition, find the solution equation.
a) p(n) = 2.5p(n-1)+900; p(0)= 400 - p(0) is the initial condition
b) p(n) = 0.75p(n-1) - 45; p(0) = 510
c) p(n) = 1.03p(n-1) - 75; p(0) = 75

Maybe it's just me, but i have no idea where to start. I have looked at examples in the first few pages of the chapter, only to be led to the wrong answer! Can anyone help??
For difference equations of the form

$p_n = k \,p_{n-1}$

one typically tries a solution of the form $p_n = c \,\rho^n$ and find the $\rho$ that works. For equations of the form

$p_n = k \,p_{n-1} + A$

where $A$ is a constant, first try $p_q = q_n + \alpha$ and try and find $\alpha$ such that the new equation becomes $q_n = k q_{n-1}$.

Try that and see how you make out.