Removing hypergeometric solutions in RSolve [closed]

I am using RSolve to find solutions, but I am only interested in rational solutions. For example here:

And here is the input code:

input[x_,y_] := (2x^3+13x^2+22x+8)y[x+3]-(2x^3+11x^2+18x+9)y[x+2]+(2x^3+x^2-6x)y[x+1]-(2x^3-x^2-2x+1)y[x]
solutions = RSolve[input[x, y] == 0, y[x], x]


I am only interested in the first term, I would prefer that the result was only $$\frac{(3-2x)c_1}{3(-1+x^2)}$$. I looked around and I could not find a solution, since RationalQ does not exist.

To further clarify, I don't know how to strip the hypergeometric part. Setting any $$c$$ to 0 would not work, since then in one other example I could have:

input[x_, y_] := y[x + 2] - 6 y[x + 1] + 9 y[x];
RSolve[input[x, y] == 0, y[x], x]


Which outputs:

$$\left\{\left\{y(x)\to c_1 3^x+c_2 3^x x\right\}\right\}.$$

Setting $$c_2$$ to zero would remove one solution, which is not what I want.

• Just set $c_2$ and $c_3$ to zero in your solution. – yarchik Mar 28 at 12:35
• That's a great idea, but I am interested in general sense, since this would not work if there were for example 2 polynomial solutions – klemen kobau Mar 28 at 12:49
• Why would not that work? Actually, you do not provide any information and yet expect a full answer. Why don't you copy and paste your actual code you are currently having a problem? This is actually a general requirement for posts here--to provide the code. – yarchik Mar 28 at 14:28
• Please post the code used to generate this result. Copy and paste the code in Raw InputForm rather than a picture of the code. – Bob Hanlon Mar 28 at 14:29
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I have method to only select rational solutions, but it is a bit clumsy and probably can be improved. Here is my code:

ClearAll[onlyrationals, splitC, polynomialQ, rationalQ];

onlyrationals[solns_, vars_] := With[{k = Keys[#][[1]]},
{k -> splitC[Values@#[[1]], vars]}] & /@ solns;

splitC[expr_, vars_] := Module[{cv, cx, c0, cvars = Sort@
Reap[Map[If[Head[#] === C, Sow[#], #] &, expr, Infinity]][[2, 1]]},
c0 = expr - Sum[Coefficient[expr, cv] cv, {cv, cvars}] // Simplify;
c0 + Sum[If[rationalQ[cx = Coefficient[expr, cv], vars], cv cx, 0],
{cv, cvars}]];

polynomialQ[expr_, vars_] := Module[{x, n}, Which[
MatchQ[expr, Power[x_ /; polynomialQ[x, vars],
n_ /; IntegerQ[n] && Positive[n]]], True,
Head@expr === Times, And @@ Table[polynomialQ[x, vars], {x, List @@ expr}],
Head@expr === Plus,  And @@ Table[polynomialQ[x, vars], {x, List @@ expr}],
NumberQ@expr, True,   True, False]];

rationalQ[expr_, vars_] :=  polynomialQ[Numerator@expr, vars] &&
polynomialQ[Denominator@expr, vars];


You would use the code like this:

solutions = RSolve[b[x + 2] - 3 b[x + 1] + 2 b[x] + 4 == 0, b[x], x]
(* {{b[x] -> 4*(1 + x) + C[1] + 2^x*C[2]}} *)
onlyrationals[solutions, {x}]
(* {{b[x] -> 4*(1 + x) + C[1]}} *)

• This does not give me the expected solution, but it answers my question. I will accept this since it is a good step in the right direction – klemen kobau Mar 29 at 19:43
• @klemenkobau Thanks for that helpful comment! I have modified my code for better solution, but it still does not completely do the job. :-( – Somos Mar 29 at 19:52
• @klemenkobau After much work I think my code does what you want. – Somos Mar 30 at 0:10
• uff I will have to go through this, thank you for your work, I will also try to solve it myself using your example – klemen kobau Mar 30 at 20:54