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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: 20,935 Thanks: 2209 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: 93 Thanks: 48 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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 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! Tags question, recurrence, relation, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post uniquegel Algebra 4 September 8th, 2014 04:18 PM Tiome_nguyen Calculus 2 May 23rd, 2012 09:45 PM Dragonkiller Linear Algebra 2 May 15th, 2012 10:49 AM ChessTal Linear Algebra 5 July 4th, 2011 07:03 AM kec11494 Applied Math 6 December 17th, 2010 11:57 PM

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