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Using generating function determine $u_n$ $$u_{n+2}+8u_{n+1}-9u_n=8 \cdot 3 \cdot 3^n$$ $$u_0 =2 $$ $$u_1 = -6$$ And my attempt: $$u(x) = \sum^\infty_{n=0} u_nx^n$$ $$u_{n+2}+8u_{n+1}-9u_n=8 \cdot 3 \cdot 3^n$$ $$\sum^\infty_{n=0} u_{n+2}x^{n} + 8 \sum^\infty_{n=0} u_nx^{n-1} - 9u(x) = 24 \frac{1}{1-3x}$$ After transformations: $$u(x) = \frac{\frac{24x^2}{1-3x} +10x +2}{1+8x-9x^2}$$ And I don't know it is good solution and how to finish it. If it is bad, please help me.

user180834
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1 Answers1

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If you wrote the first term on the LHS as $\frac{u(x) - u_0 - u_1 x }{x^2}$ and the second as $ \frac{8(u(x) -u_0)}{x}$ and then did the algebra, that should be fine. Don't forget expand the fraction on the RHS.

Alex
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  • I don't understand what do you mean. – user180834 Jan 02 '15 at 16:41
  • you have a typo in the second term on the LHS: it should be $\sum_{k=1}^{\infty} u_{k+1}x^k$ – Alex Jan 02 '15 at 16:42
  • What exactly is you complication? What do you get? – Alex Jan 02 '15 at 17:17
  • $\frac{u(x)−u_0−u_1x}{x^2}+ 8\frac{(u(x)−u_0)}{x} - 9u(x) = 24 \frac{1}{1-3x}$ – user180834 Jan 02 '15 at 17:20
  • Note $1+8z -9 z^2 = (1-z)(1+9z) = \phi_1$, so on the RHS you'll have $\frac{24}{(1-3z) \phi_1}$- split it in three, the other ones $\frac{a_0}{\phi_1} +\frac{a_1 z}{\phi_1} + \frac{8 a_0 z}{\phi_1}$, which are elementary – Alex Jan 02 '15 at 17:43