# how to extract F(k+3) and G(k+2) from symsum code and applying Initial Condition to find series solution

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MINATI on 10 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=sym('F(k)');g=sym('G(k)');L=sym('L');M=sym('M');b1=sym('b1');k=sym('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);
fd2f=symsum(((k-r+1)*(k-r+2)*F(r)*F(k-r+2)),r,0,k);%%% f*d2f;gd2g=symsum((k-r+1)*(k-r+2)*G(r)*G(k-r+2),r,0,k);fd2g=symsum((k-r+1)*(k-r+2)*F(r)*G(k-r+2),r,0,k);
gd2f=symsum((k-r+1)*(k-r+2)*G(r)*F(k-r+2),r,0,k); d1fd1f=symsum((k-r+1)*(r+1)*F(r+1)*F(k-r+1),r,0,k);d1gd1g=symsum((k-r+1)*(r+1)*G(r+1)*G(k-r+1),r,0,k);
%%%%%%%
eqn1=simplify((1+b1)*d3f-d1fd1f+fd2f+gd2f-(M+L)*d1f==0);
eqn2=simplify((1+b1)*d3g-d1gd1g+fd2g+gd2g-(M+L)*d1g);
eqns=[eqn1 eqn2];
solve([eqns,{F(k+3),G(k+2)}])
f=sum(x^k*F(k),k,0,inf);g=sum(x^k*G(k),k,0,inf); %%%%% k=10