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Did anyone ever come across the global solution for a non-linear difference equation that looks like this:

$y(t+1)=y(t)+a+b \sqrt{c y(t)+d}$.

The initial condition is $y(0)=y_0$, and a,b,c and d are real numbers.

Any help is more than welcome!

Thanks a lot in advance.

PS: Mathematica is able to find a solution this that looks like $ \left\{\left\{y[t]\to a t^2+\text{y0}-\frac{2 t \sqrt{a (d+c \text{y0})}}{\sqrt{c}}\right\},\left\{y[t]\to a t^2+\text{y0}+\frac{2 t \sqrt{a (d+c \text{y0})}}{\sqrt{c}}\right\}\right\}. $ I could use this of course but I'd like to know whether it is possible to solve this analytically by hand. Or, basically, how to get Mathematica's solution. There should be some way of transforming the problem by substitution. But - how?

PPS: Harald Hanche-Olsen just noticed that Mathematica's solution only holds under a special case for $t=1$.

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    I rather suspect that there is no closed form solution. If I am right, you will need to work with the difference equation itself to find out whatever you need to know, such as asymptotic behaviour and other properties of the sequence.2012-10-08
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    Well, I can obviously linearize it, characterize the limit etc. But I am really searching for a way to get the global solution that Mathematica spits out. Which basically means that there exists this global solution and thus there should potentially exist a way to find it...2012-10-08
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    Is it just me, or does the Mathematica solution only look right for $t=1$ when $2\sqrt{a/c}=b$?2012-10-08
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    That's odd but you are correct. The only assumptions I added in Mathematica were non-negativity ones to allow Mathematica to simplify the results. Hm - I am not really used to Mathematica giving `wrong' results.2012-10-08
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    Stranger things have happened. All software has bugs. Mathematica is software.2012-10-08

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