Recurrence relation for linear difference equation

Question

Given $u_{n+1}=-3u_n+32n^2, u_0=2$

$u_{n+1} = -3u_n \implies u_{homo} = a(-3)^n$

$u_{part} = An^2 + Bn + C \implies A(n+1)^2 + B(n+1) + C = -3(An^2 + Bn + C) + 32n^2 \implies A = 8, B=-4, C = 1$

So we get $u_n = a(-3)^n+8n^2-4n+1$ and solving when $n=0 \implies a = 1$however $u_n = (-3)^n+8n^2-4n+1$ does not work for any other n.

Where am I going wrong here?

Using the original equation we get $u_1=26$ , but using the non recursive we get $u_1 = 2$ — master_jasper, Feb 04 '24 at 07:41

score 0 · Answer 1 · answered Feb 04 '24 at 08:59

0

Your value for $\ C\ $ has the wrong sign. With the correct values of $\ C\ $ and $\ a\ $ (i.e. $\ C=-1,a=3\ $) you should find that your formula for $\ u_n\ $ and stepping the recursion both give you the same value, $\ u_1=-6\ $.

answered Feb 04 '24 at 08:59

lonza leggiera

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Recurrence relation for linear difference equation

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