Given the recurrence: $$x_{n+1} = ax_n+\frac{b}{x_n}$$ How do I determine the general solution? Is it possible to solve by generating functions? If I could have a small hint that would be great. Thanks.

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    $\begingroup$ There is not going to be a general solution. Just the asymptotics, maybe. $\endgroup$ – Ivan Neretin Oct 20 '17 at 10:16
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    $\begingroup$ Nah, there is going to be a solution. Or I will make one. $\endgroup$ – mtheorylord Oct 20 '17 at 10:39
  • $\begingroup$ I like your attitude. As to the solution, let's wait for some third opinion. $\endgroup$ – Ivan Neretin Oct 20 '17 at 10:55
  • $\begingroup$ For $\,a\neq 1\,$ it's enough to discuss the special case $\,\displaystyle z_{n+1} = az_n+\frac{1-a}{z_n}\,$ because with $\,\displaystyle z_n:=x_n/\sqrt{\frac{b}{1-a}}\,$ for all $\,n\,$ one gets the initial recurrence. $\endgroup$ – user90369 Oct 20 '17 at 12:24


Similar to Does this recurrence have a closed form limit $x_{n+1}=x_n-\frac{a}{3^{2n+1}x_n}$?,

Let $x_n=ku_n$ ,

Then $ku_{n+1}=aku_n+\dfrac{b}{ku_n}$



Case $1$: $ab>0$

Take $k=\dfrac{\sqrt b}{\sqrt a}$ , the recurrence becomes


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    $\begingroup$ I'm sorry, I don't see how this is any easier than the original $\endgroup$ – mtheorylord Oct 23 '17 at 19:39

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