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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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    Stranger things have happened. All software has bugs. Mathematica is software.2012-10-08

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You can try a solution on the form $y(t)=At^2+Bt+C$. Substituting that in the recurrence relation, you get $2At+A+B-a=b\sqrt{cAt^2+cBt+cC+d}$. Square this, collect similar powers of $t$, and compare coefficients to get three equations to determine $A$, $B$, $C$.

But then you need a fourth equation to satisfy the initial condition, which indicates that you cannot find a solution on this form in general. For special choices of the data it will be possible, however. Trying higher degree polynomials seems to lead nowhere.

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    Tha$n$ks a lot for your help! kind regards.2012-10-16