From: &noreply@axelvogt.de   
      
   On 04.04.2014 11:53, Axel Vogt wrote:   
   > On 04.04.2014 06:17, Thomas D. Dean wrote:   
   >> I have a problem,   
   >>   
   >> ode:=diff(y(x),x,x)+(a*x+b)*y(x)=0;   
   >> raw_soln:=dsolve(ode);   
   >> soln1:=convert(raw_soln,BesselI);   
   >>   
   >> This gives an expression with several terms.   
   >>   
   >> Maxima gives a solution,   
   >>   
   >> [y = bessel_y(1/3,2*(a*x+b)^(3/2)/(3*abs(a)))*%k2*sqrt(a*x+b)   
   >>            +bessel_j(1/3,2*(a*x+b)^(3/2)/(3*abs(a)))*%k1*sqrt(a*x+b)]   
   >> where bessel_y is the second kind and bessel_j is the first kind.   
   >>   
   >> If I assume %k2 is _C2 and %k1 is _C1   
   >>   
   >> maxima := y(x) = BesselY(1/3,2*(a*x+b)^(3/2)/(3*abs(a)))*_C2*sqrt(a*x+b)   
   >>      +BesselJ(1/3,2*(a*x+b)^(3/2)/(3*abs(a)))*_C1*sqrt(a*x+b);   
   >> soln2:=convert(maxima,BesselI);   
   >>   
   >> I am having problems determining if these are the same.   
   >>   
   >> Ideas?   
   >>   
   >> Tom Dean   
   >   
   > Better use different constants for Maxima's solution. And convert   
   > to Airy functions. Then comparing the inputs for the Airy functions   
   > show different expressions. May be you need 0 < a to get the 'same'   
   > as your raw_solution   
   >   
   >    convert(maxima, Airy):   
   >    collect(%, [AiryAi, AiryBi]):   
   >    simplify(%, size);   
   >    maxima2:=%;   
      
   You want to compare   
      
   eq1:=( ( (a*x+b)/a )^(3/2)/a )^(2/3);   
   eq1:= simplify(eq);   
   eq2:=(a*x+b)/a^(2/3);   
      
   But these are not the same in general   
      
   [eq1, eq2]; subs(b=0, a=-1, x=-1, %); evalc(%);   
      
                                  1/2                 1/2   
                    [1/2 + 1/2 I 3   , - 1/2 - 1/2 I 3   ]   
      
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