My Math Forum Question regarding Particular solution in recurrence relation
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 May 23rd, 2017, 01:54 AM #1 Newbie   Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0 Question regarding Particular solution in recurrence relation Hey all, I need help with solving a recurrence relation using homogeneous+particular function. * I have no issue finding the homogeneous coefficients, I only need help with the particular (non-homogeneous addition) $\displaystyle f(n) = 5f(n-1)-6(n-2)+n \cdot 2^{n-1} \\ Homogenous\, roots: (x-2)(x-3)$ The non-homogeneous addition: $\displaystyle n \cdot 2^{n-1} \rightarrow \frac{1}{2} \cdot n \cdot 2^n$ Now my question: I need to find a generic polynomial that I can use in my recurrence relation. From what I know, it's like this (correct me if I am wrong): $\displaystyle \frac{1}{2}\, =\, C\, (constant) \\ n\, =\, n^1\, =\, (ax+b)\, a\, polynomial\, with\, degree\, 1 \\ 2^n,\, 2\, is\, a\, root\, with\, degree\, 1\, (x-2) = 2^n$ So my guess is that the generic polynomial is: $\displaystyle (an+b) \cdot C \cdot 2^n$ But this is my guess, I don't know if it's true... can someone help me find the correct way? Trying to continue with putting the polynomial back into the recurrence would give me: $\displaystyle (an+b) \cdot C \cdot 2^n = 5(a(n-1)+b) \cdot c \cdot 2^{n-1} - 6(a(n-1)+b) \cdot c \cdot 2^{n-2} + n \cdot 2^{n-2}$ From here I would divide by $\displaystyle 2^{n-2}$, do all the algebra and at the end I would get $\displaystyle 0=2ac+2n$ what now? My guess is that the entire process is wrong from the start! So, can someone help me understand how to get the generic polynomial? when I have an extra unknown n? or different combinations of the non-homogenous particular addition? Any formula to follow? Thanks!
 May 24th, 2017, 12:53 AM #2 Newbie   Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0 Nobody ?
 May 24th, 2017, 09:05 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,064 Thanks: 1621 Did you intend the recurrence relation to be $f(n) = 5f(n-1) - 6f(n-2) + n \cdot 2^{n-1}$?
 May 25th, 2017, 07:00 AM #4 Member   Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 The form of your particular solution is wrong. If you've had elementary differential equations, the proper technique is entirely similar to finding a particular solution to a constant coefficient linear differential equation with non-zero forcing function. The following shows the proper form for a particular solution and then finds said solution: Thanks from shanytc
May 30th, 2017, 01:34 PM   #5
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Quote:
 Originally Posted by johng40 The form of your particular solution is wrong. If you've had elementary differential equations, the proper technique is entirely similar to finding a particular solution to a constant coefficient linear differential equation with non-zero forcing function. The following shows the proper form for a particular solution and then finds said solution:
Thanks guys,
I understand it now!

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