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I'm reading through examples on this site.
In example 2_2, given the recurrence relation $A_n - 2A_{n-1} = 2n^2$, the guess for the particular solution is $A_n= Bn^2 + Cn + D$. Substituting that into the recurrence relation gets you

I just don't know how to continue this to get the answer above.

Extreme112
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2 Answers2

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You want the equation you got by substituting in to the recurrence relation to be true for all $n$, so $B - 2B - 2$, $C + 4B - 2C$ and $D - 2B + 2 C - 2D$ must all be $0$. Now, what are $B$, $C$ and $D$?

Robert Israel
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  • Will the terms always equal to zero regardless of the coefficient in front of the $n^2$? – Extreme112 Mar 17 '15 at 01:03
  • And why can't I just equate coefficients? – Extreme112 Mar 17 '15 at 01:30
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    @Extreme112 It is equating coefficients. On the one hand you have $$Mn^2 + Nn + P = Qn^2 + Rn + S$$ to get $$\begin{cases} M = Q \ N=R \ P = S \end{cases}$$ and on the other hand you have $$Xn^2 + Yn + Z = 0$$ to get $$\begin{cases} X = 0 \ Y=0 \ Z = 0 \end{cases}$$ ...it's just two different ways of looking at the same thing. – DanielV Mar 17 '15 at 02:17
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As Robert Israel pointed out, the following is obtained. Let $f_{n} = a n^{2} + b n + c$ in the difference equation $f_{n} = 2 f_{n-1} + 2 n^{2}$. \begin{align} a n^{2} + b n + c &= 2 a (n-1)^{2} + 2b (n-1) + 2c + 2 n^{2} \\ &= 2a n^{2} - 4b n + 2a + 2b n - 2b + 2 n^{2} \\ &= 2(a+1) n^{2} + (b-2a) n + (a-b+c). \end{align} This leads to \begin{align} (a + 2) n^{2} + (b-4a) n + (2a - 2b + c) = 0. \end{align} Equating coefficients leads to $a = -2$, $b= 4a = -8$, and $c = 2b -2a = -12$ for which \begin{align} f_{n} = -2( n^{2} + 4n + 6) \end{align}

Leucippus
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