I'm trying to solve a second order differential equation and I got a recurrence. Can someone help to solve $$n(n-1+q)a_{n}-a_{n-3}+e\cdot a_{n-2}=0$$ where $q$, $e$, and $a_{0}$ are some real numbers with $a_{1}=0$ and $a_{2}=-e\cdot a_{0}/(2(1+q))$.
recurrence relation with non constant coefficients
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recurrence-relations
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3Perhaps you should edit your question so that it looks a little cleaner. Some $a_0$'s just floating in there – 2012-02-11
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1Recurrences like this don't always have neat solutions. One approach is to calculate the next few terms to see whether there is a pattern. – 2012-02-12
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0I actually calculate the first six terms, the dependence on n and q clears up but it remains unclear in "e" – 2012-02-12
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0Maybe if you edit into your question what you have found out about $n$ and $q$, someone will be able to help you with $e$. – 2012-02-13
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1I'm very hapy becouse I found a solution in "Combinatorics Function technics" hfa1.physics.msstate.edu/064.pdf – 2012-02-15
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0This recurrence relation comes from what second order differential equation? – 2012-05-31
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0I know that this recurrence relation comes from solving the ODE $xy''+qy'+x(e-x)y=0$ by power series method. Am I guessing right? – 2013-04-22
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0Usually, what you get out of such an exercise is the differential equation you started with... – 2013-05-08