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May 23rd, 2017, 02:54 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 (nonhomogeneous addition) $\displaystyle f(n) = 5f(n1)6(n2)+n \cdot 2^{n1} \\ Homogenous\, roots: (x2)(x3) $ The nonhomogeneous addition: $\displaystyle n \cdot 2^{n1} \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\, (x2) = 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(n1)+b) \cdot c \cdot 2^{n1}  6(a(n1)+b) \cdot c \cdot 2^{n2} + n \cdot 2^{n2} $ From here I would divide by $\displaystyle 2^{n2}$, 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 nonhomogenous particular addition? Any formula to follow? Thanks! 
May 24th, 2017, 01:53 AM  #2 
Newbie Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0 
Nobody ? 
May 24th, 2017, 10:05 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1981 
Did you intend the recurrence relation to be $f(n) = 5f(n1)  6f(n2) + n \cdot 2^{n1}$?

May 25th, 2017, 08:00 AM  #4 
Member Joined: Jan 2016 From: Athens, OH Posts: 92 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 nonzero forcing function. The following shows the proper form for a particular solution and then finds said solution: 
May 30th, 2017, 02:34 PM  #5  
Newbie Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0  Quote:
I understand it now!  

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