Impact-Site-Verification: dbe48ff9-4514-40fe-8cc0-70131430799e

Search This Blog

MATLAB Program for Midpoint method


% Midpoint 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/2)*f(t(i),w(i));
  %w(i+1)=w(i)+(h/2)*f( t(i), w(i)+ f(t(i+1),wprime) );
  w(i+1)=w(i)+h*f( t(i)+(h/2), 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:
>> midpoint_method
Enter left end ponit, a:  0
Enter right end point, b:  2
Enter no. of subintervals, n: 10
Enter the initial condition, alpha:  .5
0.0000   0.50000000
0.2000   0.82800000
0.4000   1.21136000
0.6000   1.64465920
0.8000   2.12128422
1.0000   2.63316675
1.2000   3.17046344
1.4000   3.72116540
1.6000   4.27062178
1.8000   4.80095857
2.0000   5.29036946
>> 


No comments

Popular Posts

Followers