h=0.5; % step size
x = 0:h:100; % Calculates upto y(3)
Y = zeros(1,length(x)); %y(1) = [0.2;0.3;0.2];
y(1) = 0.2; % redo with other choices here.
% initial condition
F_xy = @(t,r) 3.*exp(-t)-0.4*r; % change the function as you desire
for i=1:(length(x)-1) % calculation loop
k_1 = F_xy(x(i),y(i)); k_2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k_1); k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2)); k_4 = F_xy((x(i)+h),(y(i)+k_3*h)); y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h; % main equation
end% validate using a decent ODE integrator
tspan = [0,100]; y0 = -0.5;[tx, yx] = ode45(F_xy, tspan, y0)plot(x,y,'o-', tx, yx, '--')
how can i run this code for three initial value and plot it in same graph?
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