need to use this analytical solution to plot the absolute percent error in the concentration A at each time step for Euler's method

%% Euler

nsteps = 12;

t = zeros (nsteps,1);

A = zeros (nsteps,1);

B = zeros(nsteps, 1);

P = zeros(nsteps,1);

A(1) = 1;

B(1) = 3;

C(1) = 0;

K = 5*10^-5;

for k = 2:13

t(k) = t(k-1)+3600;

A(k) = A(k-1)+(-K*A(k-1)*B(k-1))*3600;

B(k) = B(k-1)+(-Yb*(K*A(k-1)*B(k-1)))*3600;

P(k)= P(k-1)+ Yp*(K*A(k-1)*B(k-1))*3600;

end

timestep is 3600 sec

[Jan]

This is some code. What is your question?

[Naveen Krish]

I need to plot the absolute percent error in the concentration of A at each timestep of 3600sec from ti=0 to tf=12 hours using Eulers method

[Sam Chak]

Hi @Naveen Krish

You have been posting some "questions" in the last few days. If you don't want the question get "pruned" again, I suggest you to use the following template:

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Answer 1

[VBBV]

nsteps = 12;
t = zeros (nsteps,1);
A = zeros (nsteps,1);
B = zeros(nsteps, 1);
P = zeros(nsteps,1);
A(1) = 1;
B(1) = 3;
C(1) = 0;
K = 5*10^-5;
Yp = 1.34; % e,g, values
Yb = Yp/2; % e.g 
for k = 2:13
    t(k) = t(k-1)+3600;
    Aold = A(k-1);
    A(k) = A(k-1)+(-K*A(k-1)*B(k-1))*3600;
    B(k) = B(k-1)+(-Yb*(K*A(k-1)*B(k-1)))*3600;
    P(k)= P(k-1)+ Yp*(K*A(k-1)*B(k-1))*3600;
    
    Aps(k) = abs((A(k)-Aold)/A(k))*100;  % absolute percent error
end

plot(Aps);title('Absoulte % error'); ylabel('%'); xlabel('timestep')

Check with values for Yb and Yp . i have used them here for e.g. purpose

[Torsten]

@VBBV

A(k-1) and A(k) are the solutions for A at times t(k-1) and t(k). Thus the difference A(k) - A(k-1) is an indicator on how much the solution changes with time (so something like A'(t(k))*dt), but not an error in the Euler integration.