How to solve the particular solution to a recurrence relation?

Question

Here is my solution to a problem:https://i.stack.imgur.com/9eH7O.jpg, though the answer should be:n+2/3 for the particular solution.

Where am I going wrong?

P.S. I am focusing on the particular solution right now. As that is where I am stuck.

How did you go from $$-10A+21B=0$$ to $$B = \dfrac{120}{441}$$ ???? — AgentS, Aug 03 '19 at 08:23
I first solve A, which is 12/21, then plug it in: -10A+21B=0 and arrive at B=120/441 — , Aug 03 '19 at 08:24
Please take the time to enter important parts of your question—most of the question, in this case—as text instead of forcing people who might wish to help you to chase a link to an image. — amd, Aug 04 '19 at 00:10

score 2 · Accepted Answer · answered Aug 03 '19 at 08:27

2

I don't understand why are you taking it into differential equation.

Directly put $a_n=An+B$ into the recurrence relation.

$$A(n+2)+B-10(An+A+B)+21(An+B)=12n$$ $$\Longrightarrow 12An-8A+12B=12n$$

Comparing coefficients on both sides, we get,

$$A=1, B=\frac{2}{3}$$

as required.

Hope it helps:)

answered Aug 03 '19 at 08:27

Martund

Thanks, went back to the content I was watching and was confusing differential and recurrence. Using the wrong method and the rest is history... – Aug 03 '19 at 08:36

1 Answers1