Find the general solution of the following non homogeneous recurrence relation with constant coefficient: $$a_n − 11a_{n−1} + 30a_{n−2}= n^2(5^n)$$ I tried the homogeneous part. $a_n−11a_{n−1}+30a_{n−2}=0$ and i get C=5 and D=6 I need full solution
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First we solve the homogeneous part: $$A_n-11A_{n-1}+30 A_{n-2}=0~~~~(1),$$ we let $A_n=x^n$ and get $X_1,x_2=6.$ So the solution of (1) is $A_n=C_1 5^n+ C_2 6^n~~~~(2)$ Next, for $$A_n-11A_{n-1}+30 A_{n-2}=n^2 5^n~~~~(2),$$ we let $A_n=(an^3+bn^2+cn+d)5^n$ in (2), Comparing the co-efficients of various powers of $n$, we get $a=-5/3, b=-65/2, c=-2165/6.$ Sothe total solution of (2) is: $$A_n=C_1 5^n+ C_2 6^n+[-(5/3)n^3-(65/2)n^2-(2165/6)n+d)]~5^n,$$ the undetermined $d$ can be combined with $C_1$.
Z Ahmed
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Thanks Dr Zadar Ahmed DSc – happy Jan 21 '20 at 10:43
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@happy, welcome so now you may accept the answer. – Z Ahmed Jan 21 '20 at 10:45
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you sure about the answer 100% coz I m going to submit to my teacher – happy Jan 21 '20 at 10:58
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Ye, you can check various steps here. – Z Ahmed Jan 21 '20 at 11:00
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Ok thanks good luck – happy Jan 21 '20 at 11:07
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check this question https://math.stackexchange.com/questions/3517057/discrete-mathematics-graph-theory-problem?noredirect=1#comment7233213_3517057 – happy Jan 21 '20 at 11:38