yes i know it is a puplication of that,but i still have the same problem,my main question is :

when i decrease initial conditons to 1e-45,i can evaluate till the time 192 seconds,but now i face another problem, and this is that my answers are too small to be used in a reall system,and i think it is beacuase of my initial conditions,for example for the time=24sec the answers are : 

what should i do to make the answers more logical?




i changed the intial condition to 1e-15 , now my answers are more logical,but i have a problem of max fun in this case , and it is not exceeds more than .11 sec 
could u please help me?thnx alot.


my equations are very very long,maybe more than 30 pages to write them here,if it is need then i can reduced in terms of approximation so that i can write down them here.thnx alot,

@Markiyan Hirnyk my equations are vey long to type them here,maybe more than 10 pages, but when i decrease initial conditons to 1e-45,i can evaluate till the time 192 seconds,but now i face another problem, and this is that my answers are too small to be used in a reall system,and i think it is beacuase of my initial conditions,for example for the time=24sec the answers are : 

what should i do to make the answers more logical?




i changed the intial condition to 1e-15 , now my answers are more logical,but i have a problem of max fun in this case , and it is not exceeds more than .11 sec 
could u please help me?thnx alot.
 

@Markiyan Hirnyk my equations are vey long to type them here,maybe more than 10 pages, but when i decrease initial conditons to 1e-45,i can evaluate till the time 192 seconds,but now i face another problem, and this is that my answers are too small to be used in a reall system,and i think it is beacuase of my initial conditions,for example for the time=24sec the answers are : 

what should i do to make the answers more logical?




i changed the intial condition to 1e-15 , now my answers are more logical,but i have a problem of max fun in this case , and it is not exceeds more than .11 sec 
could u please help me?thnx alot.
 

@Markiyan Hirnyk thnx for your help and advice,i changed the maxfun to 0,now i face this problem,


MMM := dsolve(AA,numeric,method = rkf45, maxfun = 0);

Error, (in MMM) cannot evaluate the solution further right of .14055561, probably a singularity

is there any problem with my equations?or it is involved with my initial conditions? why singularity happens here?

@Markiyan Hirnyk thnx for your help and advice,i changed the maxfun to 0,now i face this problem,


MMM := dsolve(AA,numeric,method = rkf45, maxfun = 0);

Error, (in MMM) cannot evaluate the solution further right of .14055561, probably a singularity

is there any problem with my equations?or it is involved with my initial conditions? why singularity happens here?

hi,there was a problem with my equations,in my code,and i solve it,then i used dslove(numeric) and solvee my equations,but now i have a question,my answers do not go more than the time 0.14 ,what can i do?i do need my answers at least for the time 5 seconds ,

i have write my code in this way :

MMM := dsolve(AA,numeric,method = rkf45, maxfun = 500000);


but when i want to evaluate MMM(1) i face this error : 

Error, (in MMM) cannot evaluate the solution further right of .14055561, maxfun limit exceeded (see ?dsolve,maxfun for details)

 can i do anything to increase the amount of maxfun ?

i am really thankful for your help and attention,i will do as u helped me,really thank u.

i am really thankful for your help and attention,i will do as u helped me,really thank u.

@Preben Alsholm 

these are 6 ODEs with 6 unkown functions :

first)  -(125000/3)*a[2, 1](t)-(250000/3)*a[1, 1](t)+104.1666667*a[2, 2](t)+208.3333333*a[1, 2](t)+(25000/3)*(a[2, 3](t))^2+(50/3)*(diff(a[2, 1](t), t, t))+(100/3)*(diff(a[1, 1](t), t, t))+(100/3)*(diff(a[2, 2](t), t, t))+(200/3)*(diff(a[1, 2](t), t, t)) = 0;

second)  -10.41666667*a[2, 1](t)+91.69270833*a[2, 2](t)+166.7187500*a[1, 2](t)-4.464285714*a[2, 3](t)^2-10.41666667*a[1, 3](t)^2+(25/3)*a[2, 3](t)+(50/3)*(diff(a[2, 1](t), t, t))+(100/3)*(diff(a[1, 1](t), t, t))+(100/3)*(diff(a[2, 2](t), t, t))+(200/3)*(diff(a[1, 2](t), t, t)) = 0;

third) (125000/3)*a[1, 1](t)^2*(-2*a[1, 3](t)+2*a[2, 3](t))-(125000/3)*a[1, 1](t)*a[1, 3](t)+133.9285714*a[2, 2](t)*a[2, 3](t)-(250000*(a[1, 1](t)+.5*a[1, 3](t)^2))*a[1, 3](t)+(31.2500*(-2*a[1, 2](t)+2*a[2, 2](t)))*(-2*a[1, 3](t)+2*a[2, 3](t))-(100/3)*(diff(a[1, 3](t), t, t))+37500*a[2, 1](t)*a[1, 3](t)+25000*a[2, 1](t)*(-2*a[1, 3](t)+2*a[2, 3](t))+37500*a[1, 1](t)*a[2, 3](t)-(62500/3)*a[1, 1](t)*(-2*a[1, 3](t)+2*a[2, 3](t))-(1250/3)*a[2, 3](t)-(25/3)*a[2, 2](t)+(52.08333333*(-2*a[1, 2](t)+2*a[2, 2](t)))*a[1, 3](t)+(250000*(-a[1, 1](t)-a[2, 1](t)+.5*(-a[1, 3](t)-a[2, 3](t))^2))*(-a[1, 3](t)-a[2, 3](t))+(325000/7)*a[2, 1](t)^2*(-2*a[1, 3](t)+2*a[2, 3](t))-175000*a[1, 1](t)^2*a[2, 3](t)-93.7500*a[2, 2](t)*a[1, 3](t)-(2500/3)*a[1, 3](t)-(375000/7)*a[2, 1](t)*a[2, 3](t)+104.1666667*a[1, 2](t)*a[1, 3](t)-(50/3)*(diff(a[2, 3](t), t, t))+(50000*(4*a[1, 1](t)^2+4*a[2, 1](t)^2))*a[2, 3](t)-(12500*(4*a[1, 1](t)^2+4*a[2, 1](t)^2))*(-2*a[1, 3](t)+2*a[2, 3](t))-(62500/3*(-2*a[1, 1](t)+2*a[2, 1](t)))*a[1, 3](t)-(62.5000*(-2*a[1, 2](t)+2*a[2, 2](t)))*a[2, 3](t)-(12500*(-2*a[1, 1](t)+2*a[2, 1](t)))*(-2*a[1, 3](t)+2*a[2, 3](t))-62.5000*a[2, 2](t)*(-2*a[1, 3](t)+2*a[2, 3](t))-93.7500*a[1, 2](t)*a[2, 3](t)+(25000*(-2*a[1, 1](t)+2*a[2, 1](t)))*a[2, 3](t)+52.08333333*a[1, 2](t)*(-2*a[1, 3](t)+2*a[2, 3](t))-187500*a[2, 1](t)^2*a[2, 3](t) = 0;

fourth) -(100000/3)*a[2, 1](t)-(125000/3)*a[1, 1](t)+83.33333333*a[2, 2](t)+104.1666667*a[1, 2](t)+(50000/7)*a[2, 3](t)^2+(25000/3)*a[1, 3](t)^2+(200/21)*(diff(a[2, 1](t), t, t))+(50/3)*(diff(a[1, 1](t), t, t))+(400/21)*(diff(a[2, 2](t), t, t))+(100/3)*(diff(a[1, 2](t), t, t)) = 0;

fifth) 10.41666667*a[1, 1](t)+66.68154762*a[2, 2](t)+75.02604166*a[1, 2](t)-2.604166667*a[2, 3](t)^2-5.208333333*a[1, 3](t)^2-(25/3)*a[1, 3](t)+(200/21)*(diff(a[2, 1](t), t, t))+(50/3)*(diff(a[1, 1](t), t, t))+(400/21)*(diff(a[2, 2](t), t, t))+(100/3)*(diff(a[1, 2](t), t, t)) = 0;

sixth) (87500/3)*a[1, 1](t)^2*(-2*a[1, 3](t)+2*a[2, 3](t))-(62500/3)*a[1, 1](t)*a[1, 3](t)+100.4464286*a[2, 2](t)*a[2, 3](t)-(250000*(a[1, 1](t)+.5*a[1, 3](t)^2))*a[1, 3](t)-(50/3)*(diff(a[1, 3](t), t, t))+25000*a[2, 1](t)*a[1, 3](t)+(125000/7)*a[2, 1](t)*(-2*a[1, 3](t)+2*a[2, 3](t))+25000*a[1, 1](t)*a[2, 3](t)-12500*a[1, 1](t)*(-2*a[1, 3](t)+2*a[2, 3](t))-(1300/3)*a[2, 3](t)+(25/3)*a[1, 2](t)+(250000*(-a[1, 1](t)-a[2, 1](t)+.5*(-a[1, 3](t)-a[2, 3](t))^2))*(-a[1, 3](t)-a[2, 3](t))+(250000/7*(4*a[1, 1](t)^2+4*a[2, 1](t)^2))*a[2, 3](t)-(25000/3*(4*a[1, 1](t)^2+4*a[2, 1](t)^2))*(-2*a[1, 3](t)+2*a[2, 3](t))+31250*a[2, 1](t)^2*(-2*a[1, 3](t)+2*a[2, 3](t))-125000*a[1, 1](t)^2*a[2, 3](t)+(125000/7*(-2*a[1, 1](t)+2*a[2, 1](t)))*a[2, 3](t)-(25000/3*(-2*a[1, 1](t)+2*a[2, 1](t)))*(-2*a[1, 3](t)+2*a[2, 3](t))-(44.64285714*(-2*a[1, 2](t)+2*a[2, 2](t)))*a[2, 3](t)-(12500*(-2*a[1, 1](t)+2*a[2, 1](t)))*a[1, 3](t)+(20.83333333*(-2*a[1, 2](t)+2*a[2, 2](t)))*(-2*a[1, 3](t)+2*a[2, 3](t))+(31.2500*(-2*a[1, 2](t)+2*a[2, 2](t)))*a[1, 3](t)-62.5000*a[2, 2](t)*a[1, 3](t)-(2750/3)*a[1, 3](t)-(281250/7)*a[2, 1](t)*a[2, 3](t)+52.08333333*a[1, 2](t)*a[1, 3](t)-(200/21)*(diff(a[2, 3](t), t, t))-44.64285714*a[2, 2](t)*(-2*a[1, 3](t)+2*a[2, 3](t))-62.5000*a[1, 2](t)*a[2, 3](t)+31.2500*a[1, 2](t)*(-2*a[1, 3](t)+2*a[2, 3](t))-(937500/7)*a[2, 1](t)^2*a[2, 3](t) = 0;

thnx for your help,

i reduced some terms in my approximation,so these equations reduced to six,and i need to solve them so that i can find six "a".note that these equations are individually equall to zero.i am really thankfull for your attention.