Finding a specific ode solution

Kara Scully

2021 3 10

0 个回答

4 次查看（30 天）

0 个投票

Hello! I am trying to solve an ode given starting data and some of the final data, how would I get it to display the time and the y matrix given that y(1) final should be 1e6?

clc; clear;
t = linspace(0,1e7);
C0 = [1e-9,2e6]; %initial concentrations, Z close to zero
%y(1) = Z y(2) = H 
%want the time and other concentrations when H=1e6
[t,y] = ode45(@odefcn,t,C0);
function dydt = odefcn(t,y) %y=Cj
kj = [5e-7,4.7e-7];
 r = [kj(1)*y(1)*y(2);kj(2)*y(1)*y(2)];
 stoictransposed = [1 -1 ;
     -1 0 ;
     0 1 ];
 Rj = stoictransposed*r;
 dydt=[Rj];
end

3 个评论
显示 1更早的评论隐藏 1更早的评论

Kara Scully 2021-3-10

i forgot another initial parameter, it is just very close to zero too like the first element in C0

I know how to plot this but do not know how to output the solution i need, i can get close by selecting a point on the graph but cannot get very accurate that way as it is not the exact number i need. Is there a command that would allow me to plug in one parameter into this solution that would then give the other parameters?

Alan Stevens 2021-3-10

If you make the first term of stoictransposed 1.44 instead of 1, then y(1) ends up at 10^6. However, this is quite arbitrary. Can you explain more about the underlying system that is being modelled?

请先登录，再进行评论。

请先登录，再回答此问题。

Follow Question