Solve the following difference equation to the specific $x$ values  $x_0 = 2$ $x_1 = 5$ $x_n-4x_{n-1}+3x_{n-2}=0$   I need some help and guidance to get this problem started and to understand the proper way to solve these equations.
Q1. Find the general solution to the difference equation  $$ a_{n} - 4a_{n-1} + 3a_{n-2} = 6 $$    Q2. Solve the difference equation   $$ a_{n} - a_{n-1} - 2a_{n-2} = 0, a_0 = 2, a_1 = 4 $$    I am completely lost in solving recurrence relation questions. Can anyone guide me in steps to solve the following 2 questions?
ATI Radeon HD 7770 no GPU rendering? I've removed the poll style answers in favour of open discussion.  I realise now this isn't the best way to have a proper discussion about this issue - sorry to anyone who commented/voted.    This is mostly a followup on this question:    As I've answered, the reason this is happening is that the root database (in the production Stack Exchange database server) is still called StackExchange.Programmers - a database name that propagates to SEDE and through it to the data dump.  The question here is what to do about it - the current situation is somewhat broken for users of the data dump.  These are the options we have:   Do nothing Update the root database name (requires downtime - so very undesirable) Update the update scripts so the SEDE database name is the new name, change will flow to the data dump export (breaks any queries using the old name) Update the export to the data dump so only it gets the new name Something else we haven't thought about   I have added all of these options as answers below - please vote on the one you most prefer, we will try to accommodate the top voted answer. Native Youtube app suddenly stopped working on iPad 1 It's been some weeks now, whenever I try to used my Youtube app, it tells me it's impossible to connect to Youtube. Since it's an iPad 1, I have had trouble finding information online.  Did Google or Apple stop supporting the app? Is it finally obsolete? Is there another app I could use that works on iOS 5?  Just to be clear, I did not tinker with my device. I did not change its parameters, I did not jailbreak it, it connects to the Internet just fine and I've tried the app on different Internet connections. Also, this is not a login problem, the app just refuses to let me browse videos, whether I'm logged in or not.  Thanks for your help. Why does my iPhone power off when the battery is down to 30%? "Pick up something" or "pick something up"? How can I use Photoshop to imitate a photo taken with a low-quality camera? How can I create an image in Photoshop that imitates a photo taken with a low-quality phone camera? Examples of qualities I want to simulate:   blurring in some areas caused by a dirty lens low quality exposure handling in the darkroom noise   Some examples: Does "a couple" always mean two? Get steam to realize I have games already installed I uninstalled and re-installed steam. I didn't uninstall my games, and they still exist on my hard drive (I checked).  I have an SSD and a HDD, and to save space I install programs (including steam) to C:\Program Files (x86)\Steam drive (SSD) and games to D:\Games\SteamApps (HDD). How to I let steam know the programs are installed there? Duplicate user in reputation league So I understand that this has happened in the past: , but all of the &quot;duplicate&quot; posts are marked as  so here is a new bug report: On the , I am listed twice:  While that's fun and all, it isn't exactly right. How to find all positive integers $n$ such that $(n-1)!+1$ can be written as $n^k$ for some positive integer $k$? the least possible value for :$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $ If we know that for every $a,b,c&gt;0$ ,how we can find  the least possible value for :  $$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $$ Grammatical number agreement in a complex phrase using singular "they" another question, I used the following phrase:     Your reader is [...], but they are a busy person.   I have two difficulties here:   In the first phrase, a reader, being singular noun, certainly implies is; In the second phrase, they, being a plural pronoun (regardless that it refers a single person), still requires are, but are implies the plural for busy persons which sounds awkward (a reader &lrarr; persons);   Obviously, if I used "...but he is a busy person", everything works fine, but I'm trying to use a gender-agnostic form.  What is the proper number agreement here? Put the book back on the table.   I'm having trouble. I think it is a preposition. I have already read        and many others   But I still have the following problem:   On one hand I've been told that if pure rotation takes place then the object goes on forever. This must imply no frictional force in any form acting on the body. On the second hand I've been told that during pure rotation(no external forces like wind) on a plane surface friction acts opposite to velocity of center of mass (accompanied by change of normal) and slows it down.     Both seem logical enough: The point of contact doesn't slip therefore no friction is required; So no retarding force and it must go on. Things don't go on forever; Friction must be present to slow it down.    So, which hand is correct? Or is there a third hand altogether? Is  && statement() the same as if() statement()? Can I cast a spell while another spell or ability is resolving? My opponent has a  with scads of loyalty. I've got a  and a  on the battlefield.  He activates Sorin's "ultimate" ability, targeting Gideon.     -6: Destroy up to three target creatures and/or other planeswalkers. Return each card put into a graveyard this way to the battlefield under your control.   Is there any way I can use the Spellbomb to prevent my opponent from stealing Gideon after he is sent to my graveyard? I'm working in Lee's book on topological manifolds and have gotten stumped on the first question in chapter 5, the chapter on cell complexes.  The problem is:   Let $D$ and $D'$ be two closed cells not necessarily of the same dimension.     Show that any continuous map $f:\partial D \to \partial D'$ can be extended to a continuous map $F:D \to D'$ such that $F(Int \; D) \subseteq Int \; D'$. Given points $p \in Int \; D$ and $p' \in Int \; D'$, show that $F$ can be chosen so $F(p) = p'$. Show that if $f$ is a homeomorphism then $F$ can be chosen to be a homeomorphism.   I proved the first part roughly as follows: First suppose $D$ and $D'$ are convex and each contain $0$ in their interior (in their respective ambient spaces).  Then every element other than $0$ in $D$ can be expressed uniquely in the form $\lambda q$ where $q \in \partial D$ and $\lambda\in (0,1]$ (an equivalence relation on the chords connecting $0$ to boundary points partitions $D- \{0\}$).  I then define the map $F(\lambda q) = \lambda f(q)$ which is continuous since if $\lambda_n q_n \to \lambda q$ in $D$ then $F(\lambda_n q_n) \to F(\lambda q)$.  Lastly $\lambda q \in Int \; D$ implies $\lambda &lt;1$ and so $\lambda f(q)$ is an interior point of $D'$ since $f(q)$ is a boundary point and $D'$ is convex.    Now if we suppose $D$ and $D'$ are arbitrary closed cells there are homeomorphisms $g_1: \overline{\mathbb{B}^n} \to D$ and $g_2: \overline{\mathbb{B}^m} \to D'$ (where possibly $m=n$) then we have that $g_2^{-1}\circ f \circ g_1$ is a continuous map between the boundaries of two closed balls and so by the first part can be extended to a continuous map $F:\overline{\mathbb{B}^n} \to \overline{\mathbb{B}^m}$.  The mapping $g_2 \circ F \circ g_1^{-1}$ is a continuous map that satisfies the desired claim.    My issue is that parts 2 and 3 don't seem (to me at least) to follow that easily from the proof I constructed.  Maybe my proof is wrong and I'm not seeing it, but more likely I think that I'm missing the spirit in which Lee is intending for us to approach this problem. When I press F3 to see the debug screen in I use Minecraft 1.8.1, some information is missing.   Most notably, there are no X, Y, Z coordinates.     What happened to my debug overlay? Turn off text message (SMS) sync to Exchange email account I recently started using a Samsung Galaxy S4 and one of the features in a synchronization between text messages and email accounts.  When a text message comes in it throws a copy into my Exchange email accounts that are one the phone. I want to turn this feature off but cannot find a setting for it anywhere.  Any ideas? How can I see flags I cast from all sites?