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 $a_{2}=-e\cdot a_{0}/(2(1+q))$.
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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 – Patrick Da Silva Feb 11 '12 at 14:51
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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. – Gerry Myerson Feb 12 '12 at 00:59
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I actually calculate the first six terms, the dependence on n and q clears up but it remains unclear in "e" – mohamed benbitour Feb 12 '12 at 10:13
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Maybe if you edit into your question what you have found out about $n$ and $q$, someone will be able to help you with $e$. – Gerry Myerson Feb 13 '12 at 03:55
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1I'm very hapy becouse I found a solution in "Combinatorics Function technics" hfa1.physics.msstate.edu/064.pdf – mohamed benbitour Feb 15 '12 at 13:49
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This recurrence relation comes from what second order differential equation? – doraemonpaul May 31 '12 at 04:22
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I 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? – doraemonpaul Apr 22 '13 at 04:17
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Usually, what you get out of such an exercise is the differential equation you started with... – vonbrand May 08 '13 at 12:37