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% Heun's method
% Example 1:  Approximate the solution to the initial-value problem
% dy/dt=e^t ; 0<=t<=2 ; y(0)=1;

% Example 2:  Approximate the solution to the initial-value problem
% dy/dt=y-t^2+1 ; 0<=t<=2 ; y(0)=0.5;

%f = @(t,y) (0*y+exp(t));   %Example 1
f = @(t,y) (y-t^2+1);      %Example 2
a = input('Enter left end ponit, a:  ');
b = input('Enter right end point, b:  ');
n = input('Enter no. of subintervals, n: ');
alpha = input('Enter the initial condition, alpha:  ');

h = (b-a)/n;
t=[a zeros(1,n)];
w=[alpha zeros(1,n)];

for i = 1:n+1
t(i+1)=t(i)+h;
wprime=w(i)+(h/3)*f(t(i),w(i));
w(i+1)=w(i)+(h/4)*(f(t(i),w(i))+3*f(t(i)+(2*h/3), w(i)+(2*h/3)*f(t(i)+h/3, wprime)));
fprintf('%5.4f  %11.8f\n', t(i), w(i));
plot(t(i),w(i),'r*'); grid on;
xlabel('t values'); ylabel('w values');
hold on;
end

OUTPUT:
>> heun_method
Enter left end ponit, a:  0
Enter right end point, b:  2
Enter no. of subintervals, n: 10
Enter the initial condition, alpha:  1
0.0000   1.00000000
0.2000   1.43991111
0.4000   1.95980255
0.6000   2.55966996
0.8000   3.23950802
1.0000   3.99931024
1.2000   4.83906868
1.4000   5.75877366
1.6000   6.75841334
1.8000   7.83797327
2.0000   8.99743580

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