NON-HOMOGENEOUS RECURRENCE RELATIONS - Discrete Mathematics
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it when someone needs a worked walkthrough of the homogeneous-plus-particular method for linear non-homogeneous recurrences.
Recommended by 2 people across 2 different months, 2016 to 2021.
What it is
A university-level discrete mathematics tutorial from TrevTutor on solving non-homogeneous recurrence relations. It builds on the previous video on homogeneous relations and works through finding the homogeneous and particular solutions with examples.
What people said
Recommends David Velasquez · 2016-02 · Mathematics Stack Exchange · on “Finding a solution for a recurrence relation when you have a lone constant in the relation”
Thanks, but I'm specifically trying to solve this using the homogeneous and particular method. I got the homogeneous part as $c^n$ (where $c$ is a constant) but I'm stuck when trying to get the particular part. I was following this video [this video] up to the 7 minute mark but his example uses $2^n$ rather than a constant, so I'm not sure how to continue.
Recommends PrincessEev · 2019-03 · Mathematics Stack Exchange · on “Recurrence relation $a_{n+1}=a_n+(b-a_n)\cdot x$”
Note that $a_n = -qa_{n-1} + kq$. This is a linear, nonhomogenous recurrence relation, for which there are general methods to solve it. Some are done in this video - [this video] If you would like more specific help, you should edit your OP to include the relevant details - including context, what you know, your attempts at the problem, and what exactly is giving you trouble. (MSE dislikes being used as a site where people just expect us to give solutions to your homework and as posted this question looks exactly like that kind of scenario.)
Cites Damian Jacobs · 2021-11 · Mathematics Stack Exchange · on “Solve the following recurrence relation. $a_{n+1} - 2a_{n} = 2^n$”
$a_{n+1} - 2a_{n} = 2^n$ $;\ n\geq0\ ;\ a_0 = 1 $ I worked through the above example which was done in a YouTube video [this video] and managed to reach the answer of: $a_{n}= 2^n + n2^{n-1}$ Which lines up with the answer achieved in the video. I believe the answer to be wrong because if you calculate the values for n = 1, 2, 3 and so on the values between the solution and the recurrence relation differ. Solution: $a_1 = 3 \ a_2 = 8 \ a_3 = 20$ Recurrence Relation: $a_1 = 4 \ a_2 = 8 \ a_3 = 24$ Have I made a stupid mistake somewhere, misunderstood something or is the solution really incorrect? If it is incorrect what is the correct solution and how does one reach it?