1. ## More Difference Equations

I thought id post this in a new thread, i got the answer to the last one, but am kinda unsure of what to do in this -

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

(A) $
x_{n} - 3x_{n-1} = 2.3^n + n, x_{0} = 1/4
$

I am able to get Yn, but im confused with the rest of the question

Also, with this question i am unsure of what to do with the '4', should i represent it as C?

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

(B) $
x_{n+2} - 2x_{n+1} + x_{n} = 4, x_{0} = 3, x_{1} = 6
$

Thanks everybody!

2. $
x_n - 3x_{n - 1} = 2.3^n + n

$

solve the homogeneous equation, using the method I've already shown you.

Then assume a particular solution of the following form:

$
x_p = C_1 2.3^n + C_2 n
$

and apply the initial conditions...

(B) $x_{n + 2} - 2x_{n + 1} + x_n = 4$

we know that the difference equation should hold for any positive n, in particular for n=n+1:

$x_{n + 3} - 2x_{n + 2} + x_{n + 1} = 4
$

subtract the two equations:

$
x_{n + 3} - 3x_{n + 2} + 3x_{n + 1} - x_n = 0

$

and this is a homogeneous equation you've already know how to solve...

3. For the second question, i resulted in $(x-1)^3$ is that right?
And then i used $A(1)^n + Bn(1)^n + Cn^2(1)^n$.. does that seem right?

4. Right

6. This is how i did the following question, but i seem to get stuck -

$x_{n} = 3x_{n-1} + 2.3^n + n, x_{0} = 1/4$

$y_{n} = A3^n$

$z_{n} = Bn3^n + Cn$

$Bn3^n + Cn - 3(B(n-1)3^{n-1} + C(n-1)) = 2.3^n + n$

$B3^n - 2Cn + 3C = 2.3^n + n$

$B=2$

How do i obtain C, is that right so far?