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Re: Bessel And Airy Functions in Solutions
Posted:
Apr 4, 2014 9:59 AM


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 ]



