# Non-homogeneous recurrence relations

• Oct 24th 2009, 06:39 AM
Rmathk
Non-homogeneous recurrence relations
Hello everyone (Hi)

I really need help on this work:

Solve 4an+2 - 4an+1 + an = (1/3)^n

Thanks in anticipation :)
• Oct 24th 2009, 09:19 AM
HallsofIvy
Quote:

Originally Posted by Rmathk
Hello everyone (Hi)

I really need help on this work:

Solve 4an+2 - 4an+1 + an = (1/3)^n

Thanks in anticipation :)

If $a_n$ is the general solution to the associated homogeneous equation, $4a_{n+2}- 4a_{n+1}+ a_n= 0$, and $b_n$ is any solution to the entire equation, $4a_{n+2}- 4a_{n+1}+ a_n= (\frac{1}{3})^n$, then $a_n+ b_n$ is the general solution to the entire equation.

Do you know how to find the general solution to the associated homogeneous equation?

To find a solution to the entire equation, try a particular solution $b_n= A\left(\frac{1}{3}\right)^n$.
• Oct 24th 2009, 07:55 PM
Rmathk
Quote:

Originally Posted by HallsofIvy
If $a_n$ is the general solution to the associated homogeneous equation, $4a_{n+2}- 4a_{n+1}+ a_n= 0$, and $b_n$ is any solution to the entire equation, $4a_{n+2}- 4a_{n+1}+ a_n= (\frac{1}{3})^n$, then $a_n+ b_n$ is the general solution to the entire equation.

Do you know how to find the general solution to the associated homogeneous equation?

To find a solution to the entire equation, try [tex]a_n= A\left(\frac{1}{3}\right)^n[/itex].

I obtained the characteristic equation of the homogeneous equation as
4(x^2) – 4x + 1 = 0
x = (1/2)

I suppose the general solution is then:
an = (a+bn)[(1/2)^(n)]

How do i find a solution to the entire equation? Am stuck :(
• Oct 24th 2009, 11:52 PM
CaptainBlack
Quote:

Originally Posted by Rmathk

I obtained the characteristic equation of the homogeneous equation as
4(x^2) – 4x + 1 = 0
x = (1/2)

I suppose the general solution is then:
an = (a+bn)[(1/2)^(n)]

How do i find a solution to the entire equation? Am stuck :(

go back and read HallsofIvy's post again

CB