# Some more diffeq help

• Jan 25th 2009, 09:11 PM
Codename46
Some more diffeq help
I'm reading this example in my book and I don't get how the author gets this solution.

X1(n) = 2X1(n-1) ; x-sub-1
X2(n) = 2X2(n-1) ; x-sub-2

The solutions to these DEs, according to the book, are

X1(n) = (2^n) * X1(0)
X2(n) = (3^n) * X2(0)

I can't figure out how he makes this leap.

• Jan 26th 2009, 03:18 PM
Jester
Quote:

Originally Posted by Codename46
I'm reading this example in my book and I don't get how the author gets this solution.

X1(n) = 2X1(n-1) ; x-sub-1
X2(n) = 3X2(n-1) ; x-sub-2 probably a three here

The solutions to these DEs, according to the book, are

X1(n) = (2^n) * X1(0) (**)
X2(n) = (3^n) * X2(0) (**)

I can't figure out how he makes this leap.

These are both differrence equations. Both have the same form

$\displaystyle (1)\;\;\;x_n = r x_{n-1},\;\;\;x_0=a$

Here we seek a solution of the form

$\displaystyle x_n = k \rho ^ n$

Subsituting into (1) gives

$\displaystyle k \rho ^{n+1} = r k \rho ^ n\;\;\; \Rightarrow\;\;\;\rho = r$

Thus, $\displaystyle x_n = k r^n$. Now $\displaystyle x_0 = a$ so

$\displaystyle x_0 = k r^0 = k = a$. Thus,

$\displaystyle x_n = a r^n.$ For your solutions (**) $\displaystyle a = x_1(0),\;\;x_2(0)$
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