using ode45

I have used successfully used ode45 on several occasions. Now I am trying to "improve" my utilization. Previously I had something like ---

      for i = 1 : dx 
          [t,y] =  ode45( 'ffssl' , [0,xinit(i)], yinit ) ;
          fprintf ( fp , ' %f\t%f\n' , xinit(i) , y(end) ) ;
      end ;

I would like to use something such as the following ---

      for i = 1 : dx
          [t,y] = ode45 ( 'ffssl' , [xinit(i),xinit(i+1)],  <yinit> ) ;
          fprintf  ( fp , ' %f\t%f\n' , xinit(i+1) ,  <y(end)> ) ;
      end ;

Is this even possible? If so, I do not know how do provide the proper y initial value yinit or the y calculated value y(end). Any suggestions would be appreciated. Thanks very much in advance.

[Matt Tearle]

Can you explain what you're trying to achieve? It looks like you're writing out the solution at some given times (x values). But you wouldn't need a loop for that.

Answer 1

[Walter Roberson]

At first glance it appears you could use

      xi = [0, xinit];
      for i = 1 : length(xi)-1
          [t,y] = ode45 ( 'ffssl', [xi(i),xi(i+1)],  yinit(i) ) ;
          fprintf( fp, ' %f\t%f\n', xi(i+1), y(end) );
      end

but I have no idea what your proper yinit values should be if they are not to be constant.