# Math Help - Difference Equations

1. ## Difference Equations

The question states -

For each of rhe following difference equations find:
(i) the general solutions
(ii) the solution corresponding to the given initial conditions

(c) $x_{n+2} - 6x_{n+1} + 8x_{n} = 3.5^n, x_{0} = 3, x_{1} = 6$

Thanks for the help.. im not too sure what to do.. so any help wouldbe appreciated

2. first we solve the homogeneous difference equation:

$
x_{n + 2} - 6x_{n + 1} + 8x_n = 0

$

we assume the solution is of the following form:

$
r^n
$

we now plug in this expression into the DE:

$\begin{gathered}
r^{n + 2} - 6r^{n + 1} + 8r^n = 0 \hfill \\
\hfill \\
\Leftrightarrow r^n \left( {r^2 - 6r + 8} \right) = 0 \hfill \\
\end{gathered}$

and solve for r:

r = 4, 2 thus:

$
x_h = C_1 2^n + C_2 4^n
$

now to find the particular solution we use the method of undetermined coefficients namely we assume that the solution has the form of the inhomogeneous term:

$
x_p = D3.5^n

$

we proceed to substituting the particular solution in the DE:

$
\begin{gathered}
D3.5^{n + 2} - 6D3.5^{n + 1} + 8D3.5^n = 3.5^n \hfill \\
\hfill \\
\Leftrightarrow D3.5^n \left( {3.5^2 - 6 \cdot 3.5 + 8} \right) = 3.5^n \hfill \\
\hfill \\
\Leftrightarrow D = \frac{1}
{{3.5^2 - 6 \cdot 3.5 + 8}} \hfill \\
\end{gathered}

$

so the general solution is:

$
x = x_h + x_p = C_1 2^n + C_2 4^n + D3.5^n
$

now all you have to do is apply the initial conditions to find the constants...

3. i was able to do that bit by simply factorising the orginial equation, whilst treating it as a quadratic, thanks for showing me a new method though. How do i then find the constants?

4. well I've already told you: apply the initial conditions:

you're told that: $
x_0 = 3
$

which means that:
$
x[n = 0] = 3

$

thus: $
x_0 = C_1 2^0 + C_2 4^0 + D3.5^0 = C_1 + C_2 + D = 3

$

after applying the second initial condition you'll get another equation ....

5. I wrote down the wrong initial conditions, but i got the answer eventually!.. Thanks for the help