CONVERSION OF ODE TO RECURRENCE RELATION

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MINATI
MINATI on 2 Jan 2020
Edited: MINATI on 7 Jan 2020
syms x k r f(x) g(x) a b beta b1 M L
syms F(k) G(k)
F(0)=0;F(1)=1;F(2)=a/2;G(0)=0;G(1)=1/2;G(2)=b/2;b1=1/beta;
%%%%dnf=diff(f,x,n)
f=F(k);g=G(k);d1f=(k+1)*F(k+1);d2f=(k+1)*(k+2)*F(k+2);d3f=(k+1)*(k+2)*(k+3)*F(k+3);d1g=(k+1)*G(k+1);
d2g=(k+1)*(k+2)*G(k+2);d3g=(k+1)*(k+2)*(k+3)*G(k+3);
f*d2f=sum((k-r+1)*(k-r+2)*F(r)*F(k-r+2),r,0,k);g*d2g=sum((k-r+1)*(k-r+2)*G(r)*G(k-r+2),r,0,k);
f*d2g=sum((k-r+1)*(k-r+2)*F(r)*G(k-r+2),r,0,k);g*d2f=sum((k-r+1)*(k-r+2)*G(r)*F(k-r+2),r,0,k);
(d1f)^2=sum((k-r+1)*(r+1)*F(r+1)*F(k-r+1),r,0,k);(d1g)^2=sum((k-r+1)*(r+1)*G(r+1)*G(k-r+1),r,0,k);
eqns=simplify((1+b1)*d3f-(d1f)^2+f*d2f+g*d2f-(M+L)*d1f==0,(1+b1)*d3g-(d1g)^2+f*d2g+g*d2g-(M+L)*d1g==0)
Rsolve(eqns,{F(k+3),G(k+2)});
%% I want a recurrence relation in terms of F(k+3) and G(k+2) (k = 0 -> Inf) but unable to code it properly
  1 Comment
MINATI
MINATI on 3 Jan 2020
Edited: MINATI on 7 Jan 2020
is there any way
Any idea

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