# Omega

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 Apr11 comment Linear Algebra Basis Help? $y$ must be free, because its column is null. Apr10 comment What particular solution should I guess for this relation? Ok, doing what you said, for an $RHS = 4+3^n$ it would be $a+b\cdot 3^n$? Apr10 comment What particular solution should I guess for this relation? If I multiply each term by an unknown constant, wouldn't it be $2a+3b\times4^n$ instead? Apr10 accepted About the particular solution given an homogeneous solution in a recurrence relation. Apr10 comment What particular solution should I guess for this relation? In step 1, "try something like RHS", can you elaborate on that? In this example, where $RHS = n2^n$, you decided to use $an2^n$. I do see that your decision is "like the RHS", but what made you specifically choose this and not something else? Or is it just a matter of copy-pasting the RHS? Basically, if I had $RHS = 4+3^n$, would it be $a4+a3^n$? Apr10 comment What particular solution should I guess for this relation? I'm working on one where the homogeneous solution is $(b_0+b_0n)\cdot(-3)^n$ and $RHS = 4 + 3^n$. In this case, I don't need to add a lower order term right? Apr10 comment What particular solution should I guess for this relation? I'm trying to apply this with another exercise at the moment. "If necessary modify this by including "lower order" terms" - what determines whether it is necessary or not to include lower order terms? Apr10 comment What particular solution should I guess for this relation? If, hypothetically, there was yet another shared term between my guess and the $f(n)$, I would have to multiply the guess by $n$ again, right? In that case the guess would be $(an^3+bn^2)2^n$? Apr10 revised What particular solution should I guess for this relation? deleted 1 characters in body Apr10 comment What particular solution should I guess for this relation? @user88595: Oh no, actually, I saw your comment and said "yeah I should invert the sign" so I went and inverted it without realising you already had inverted so it returned to its original form lol. Apr10 revised What particular solution should I guess for this relation? added 3 characters in body Apr10 comment What particular solution should I guess for this relation? How did you rearrange $f(n) = 2f(n-1) + n2^n$ to $f(n)-f(n-1) = n2^n$? I ask because of $2f(n-1)$, how did it become just $f(n-1)$? Apr10 comment What particular solution should I guess for this relation? @ajotatxe: This stuff confuses me. If we chose $b_0$ to replace $n$ there, why aren't we replacing the exponential $n$ too? Apr10 asked What particular solution should I guess for this relation? Apr10 comment Does the order I multiply the characteristic equation's factors in the homogeneous solution matter? @achillehui: But suppose that $b_0 = 1$ and $b_1 = 2$ and $n = 1$. Clearly $$1\cdot2+2\cdot 5 \neq 2\cdot2+ 1 \cdot 5$$ I'm asking about product distribution, not addition. Apr10 asked Does the order I multiply the characteristic equation's factors in the homogeneous solution matter? Apr10 comment About the particular solution given an homogeneous solution in a recurrence relation. Oh, very interesting. This explains $d_0$ on the third one too I guess. But then, what about $d_1n$? Apr10 asked About the particular solution given an homogeneous solution in a recurrence relation. Apr10 comment How to solve non-homogeneous recurrence relations? I'd have $$f(n) = f(n-1) + 2f(n-2) + n + 2^n + K + an + b + cn2^n$$ and still wouldn't have the fainted idea what's going on. Apr10 comment How to solve non-homogeneous recurrence relations? Into what equation? The whole $f(n) = f(n-1) + 2f(n-2) + n + 2^n + K$?