ODE Function time output

Dear all,

I wrote the code below now, I have initial concentration 6mg/L but i need to calculate when it reaches (0.05*6) mg/L. Could you help me please? I need the t value when concentration is equal to (0.05*6) mg/L

Z0=6;
tspan = [0 24]; 
[tZ,Z] = ode45(@ConcDCE,tspan,Z0);
    function dZdt=ConcDCE(t,Z)
k1=1.26;
k2=0.74;
k3=0.22;
X0=1;
Y0=4;
Z0=6;
dZdt=k2*(Y0*exp(-k2*t)+((X0*k1)/(k2-k1)*(exp(-k1*t)-exp(-k2*t))))-k3*(Z0*exp(-k3*t)+((Y0*k2)/(k3-k2)*(exp(-k2*t)-exp(-k3*t)))+X0*k1*k2*((exp(-k1*t))/((k2-k1)*(k3-k1))-(exp(-k2*t))/((k2-k1)*(k3-k2))+(exp(-k3*t))/((k3-k1)*(k3-k2))));
    end

Answer 1

[Stephan]

Use events:

Z0 = 6;
Zt=0.05*6;
tspan = [0 24];
opts = odeset('Events',@(tZ,Z)EventsFcn(tZ,Z,Zt));
[tZ,Z,tZe,Ze,iZe] = ode45(@ConcDCE,tspan,Z0,opts);

plot(tZ,Z)
hold on
scatter(tZe,Ze,'or')

function dZdt=ConcDCE(t,~)
k1=1.26;
k2=0.74;
k3=0.22;
X0=1;
Y0=4;
Z0=6;
dZdt=k2*(Y0*exp(-k2*t)+((X0*k1)/(k2-k1)*(exp(-k1*t)-exp(-k2*t))))-k3*(Z0*exp(-k3*t)+((Y0*k2)/(k3-k2)*(exp(-k2*t)-exp(-k3*t)))+X0*k1*k2*((exp(-k1*t))/((k2-k1)*(k3-k1))-(exp(-k2*t))/((k2-k1)*(k3-k2))+(exp(-k3*t))/((k3-k1)*(k3-k2))));
end
    
function [Conc,isterminal,direction] = EventsFcn(~,Z,Zt)
Conc = Z - Zt; % The value that we want to be zero
isterminal = 0;  % Halt integration 
direction = 0;   % The zero can be approached from either direction
end

[Carey n'eville]

%I couldn't explain myself clearly sorry for this.
%  I actually search for labeling Z=0.3 on graph (in this problem Z),
%  I need to put a dot to graph

[Stephan]

I think there is a mark at the position of Z=0.3 in the graph - maybe i misunderstood. If you only want the second point where Z=0.3 use

scatter(tZe(2),Ze(2),'or')

The t-value is:

tZe(2)

[Stephan]

I edited my code in my answer. For some reason I missed Z0 as initial value - accidentally I used Zt as initial value. This is corrected now. Sorry for this.

[Carey n'eville]

The last one is working properly thank you so much!!