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% Taylor's method of order 2
% 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));
%fprime = @(t,y) (0*y+exp(t));   % f=e^t, f'=e^t

f = @(t,y) (y-t^2+1);
fprime=@(t,y) (y-t^2+1-2*t);    % y'=f=y-t^2+1, f'=y'-2t+0 =y-t^2+1-2t
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=f(t(i),w(i))+(h/2)*fprime(t(i),w(i));
w(i+1)=w(i)+h*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:

>> TaylorMethodofOrder2
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   1.44000000
0.4000   1.96000000
0.6000   2.56000000
0.8000   3.24000000
1.0000   4.00000000
1.2000   4.84000000
1.4000   5.76000000
1.6000   6.76000000
1.8000   7.84000000
2.0000   9.00000000

>>