CONVERSION OF ODE TO RECURRENCE RELATION
1 view (last 30 days)
Show older comments
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
Answers (0)
See Also
Categories
Find more on Ordinary Differential Equations in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!