Runge Kutta solving differential equations

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KayLynn
KayLynn 2014-2-11
I am trying to solve differential equations using runge kutta. I have two codes that I am working from . The first code I had an equation and dveloped runge kiutta from that equation. The second code I have four differential equations. I am unsure how to correctly incoroporate the k values from the first code into the second code to solve for k2, k3 and k4 (to which I have labled in the second code as e2,s2,p2,c2.....etc).
First code: %Define f(t,y)
f = @(t,y) (2 - exp(-4*t) - 2*y);
% Define Step Size and final time point (t_final)
h = 0.001;
t_final = 5;
t = 0:h:t_final;
y = zeros(1,numel(t));
y(1) = 1; % y0
% You know the value a t = 0, thats why you'll start: t = h i.e. i = 2
for i = 2:numel(t)
k1 = h*f(t(i-1),y(i-1));
k2 = h*f(t(i-1)+h/2, y(i-1)+k1/2);
k3 = h*f(t(i-1)+h/2, y(i-1)+k2/2);
k4 = h*f(t(i-1)+h, y(i-1)+k3);
y(i) = y(i-1) + (k1+2*k2+2*k3+k4)/6;
disp([t(i) y(i)]);
end
plot(t,y);
Second code: %define your f(t,y)
h = 0.1; % Define Step Size
t = 0:h:200; %define your range of time points (range of t values)
S(1)=100;
E(1)=10;
C(1)=0;
P(1)=0;
k1=0.005;
k2=k1;
k3=0.1;
% You know the value a t = 0, thats why you'll state with t = h i.e. i = 2 %Write for loop for i = 1:numel(t)-1
e1=(k2*C(i)+k3*C(i)-k1*E(i)*S(i));
s1=(k2*C(i)-k1*E(i)*S(i));
c1= k1*S(i) * E(i) -k2*C(i) -k3*C(i) -k2*C(i);
p1=(k3*C(i));
e2=
s2=
c2=
p2=
e3=
s3=
c3=
p3=
e4=
s4=
c4=
p4=
E(i+1) = E(i) + h*e1;
S(i+1) = S(i) + h*s1;
C(i+1) = C(i) + h*c1;
P(i+1)= P(i)+ h*p1;
end % plot each graph plot(t,E)
hold on
plot(t,S)
plot(t,C)
plot(t,P)

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