Subject: finding out if expr has the form F(y/x^a)
From: Stavros Macrakis
Date: Wed, 12 Dec 2012 18:25:05 -0500
Nice, but it fails for some trivial cases (as does rich.hennessy's version):
solution(0) => error but h(x):=0, m=<anything> is a solution
solution(y/x) => error but h(x):=1, m=<anything> is a solution
-s
On Wed, Dec 12, 2012 at 3:52 PM, Daniel Rupistraliz Avez <
danielrupistralizavez at yahoo.es> wrote:
> This is the correct one
>
> set_display('none)$
> load(pdiff)$
>
> solution(y/x*h(y/x^m));
>
>
> solution(g):=block([g2,g3,g6,a],g2:g*x/y,g3:at(g2,y=1/x),g6:at(diff(at(g,y=x),x)/diff(g3,x),x=1),
> a:-ratsimp((g6+1)/(g6-1)),f:at(g*x/y,y=x^(a+1)),['a=a,'f(x)=f]);
>
> Example:
>
> solution(y/x*h(y/x^m));
>
> Example:
>
> u(x):=(8*x^4+5*x^3+6*x^2+3)/(5*x^3+2*x^2+x);
>
> solution(ratsimp(y/x*u(y/x^4)));
>
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