First order difference equation

**molimoli** · April 11th 2009, 10:44 AM

I need to solve this using iteration method

y <t> = 0.8 y <t-1> +1

**Soroban** · April 11th 2009, 02:11 PM

Hello, molimoli!

Is there any more information . . . like the first term?

I need to solve this using iteration method: . $y_{t} \:=\: 0.8y_{t-1} +1$

I'm not familiar with the term "interation method", but I can solve it.
. . I'll change the notation to a more convenient form (for me).

We have: . $f(n) \:=\:\tfrac{4}{5}\!\cdot\!f(n\text{-}1) + 1$

We conjecture that $f(n)$ is an exponential function: . $f(n) \:=\:X^n$

$\begin{array}{cccccc}\text{So we have:} & X^n &=& \tfrac{4}{5}X^{n-1} + 1 & [1] \\ \\[-3mm] \text{Write the next term:} & X^{n+1} &=& \tfrac{4}{5}X^n + 1 & [2] \end{array}$

Subtract [2] - [1]: . $X^{n+1} - X^n \:=\:\tfrac{4}{5}X^n - \tfrac{4}{5}X^{n-1} \quad\Rightarrow\quad X^{n+1} - \tfrac{9}{5}X^n + \tfrac{4}{5}X^{n-1} \:=\:0 $

Multiply by 5, divide by $X^{n-1}\quad 5X^2 - 9X + 4 \:=\:0$

Factor and solve: . $(X - 1)(5X - 4) \:=\:0 \quad\Rightarrow\quad X \:=\:1,\:\tfrac{4}{5}$

Hence, the function is of the form: . $f(n) \;=\;A + B\left(\tfrac{4}{5}\right)^n$

Suppose the initial term is: . $f(0) = a$
. . Then: . $f(1) \:=\:\tfrac{4}{5}a + 1$

Then we have:

$\begin{array}{ccccccccccc}f(0)\:=\:a\!: & &A + B(\frac{4}{5})^0 &=&a & \Longrightarrow & A + B &=& a & [3] \\ \\[-3mm] f(1) \:=\:\frac{4}{5}a+1\!: & &A + B(\frac{4}{5})^1 &=& \frac{4}{5}a+1 & \Longrightarrow & A + \frac{4}{5}B &=& \frac{4}{5}a + 1 & [4] \end{array}$

Subtact [3] - [4]: . $\tfrac{1}{5}B \:=\:\tfrac{1}{5}a - 1 \quad\Rightarrow\quad B \:=\:a-5$

Substitute into [3]: . $A + (a-5) \:=\:a \quad\Rightarrow\quad A \:=\:5$

Therefore: . $\boxed{f(n) \;=\;5 + (a-5)\left(\tfrac{4}{5}\right)^n}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

In the original notation: . $y_t \;=\;5 + (y_0-5)\left(\tfrac{4}{5}\right)^t$

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