MATLAB Program for Taylor's Method of Order 2

% 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

>>