Could someone please help me fix this code?

function dy=f(t,y)
L=2;
V1=10;
V2=5;
dydt= [-L/V1*y(1)  -L/V2*(y(2)-y(1))];
end

%% preparations
a=0; b=10; N=100;             % t0, t final, N values
h=(b-a)/N;                    % step size
t=linspace(a,b,N+1);          
y0=[0.3  0];                  % initial value(s)
m=length(y0);                 % size of ODE system
y=zeros(m,N+1);               % numerical approx. vector
y(:,1)=y0;
L=2;
V1=10;
V2=5;

%% RK2 method algorithm
f=@(t,y) [-L/V1*y(1)  -L/V2*(y(2)-y(1))];
for j=1:N
	k1=f(t(j),y(:,j));
	k2=f(t(j)+0.5*h,y(:,j)+0.5*h*k1);
        y(:,j+1)=y(:,j)+h*k2;    
end

plot(t, y(1,:), t,y(2,:), 'g')
hold on
exact1=0.3*exp(-0.2*t);
exact2=0.6*(exp(-0.2*t)-exp(-0.4*t));
Plot (exact 1, y(1, :), exact2, y(2,:))
hold off
axis square
grid

And this is the message I get: "Unable to perform assignment because the size of the left side is 2-by-1 and the size of the right side is 2-by-2."

I would really appreciate your help.

D.

Answer 1

[Torsten]

function main

%% preparations
a=0; b=10; N=100;             % t0, t final, N values
h=(b-a)/N;                    % step size
t=linspace(a,b,N+1);          
y0=[0.3  0];                  % initial value(s)
m=length(y0);                 % size of ODE system
y=zeros(N+1,m);               % numerical approx. vector
y(1,:)=y0;
L=2;
V1=10;
V2=5;

%% RK2 method algorithm
f=@(t,y) [-L/V1*y(1)  -L/V2*(y(2)-y(1))];
for j=1:N
    k1=f(t(j),y(j,:));
    k2=f(t(j)+0.5*h,y(j,:)+0.5*h*k1);
    y(j+1,:)=y(j,:)+h*k2;    
end

exact1=0.3*exp(-0.2*t);
exact2=0.6*(exp(-0.2*t)-exp(-0.4*t));
plot (t, exact1, t, y(:, 1), t, exact2, t, y(:,2))
hold off
axis square
grid

end

[Diana Danila]

OMG, it works! Thank you :D