I'm getting a wrong table result from my function.

I made a table of results that includes h and the error: Eh = max|y(t) − y ̃(t)|, where y ̃ is my approximated solution. I'm using twostepmethod function with startingtimet =0, endingtimet =20, and stepsizes: h= 1/2^n for n=1, . . . , 10.

And I'm getting a totally wrong table which is

 Table with h & Eh
    h          Eh
1.0e+225 *

    0.0000    0.0000
    0.0000    0.0000
    0.0000    0.0000
    0.0000    4.2965
    0.0000       Inf
    0.0000       Inf
    0.0000       Inf
    0.0000       Inf
    0.0000       Inf
    0.0000       Inf

Can you please fix my code to get a proper table?

this is my code

f=@(t,y) -y-3*t;
F=@(t) 3-3*t-2*exp(-t);
t0=0; y0=1;
te=20;
nn=1:10;
h = 1./2.^nn;
for ii=1:length(nn)
    [ys, ts] = MyTwoStepMethod(f, t0, te, y0, h(ii));
    Eh(ii) = max(abs(F(ts)-ys));
end

disp('Table with h & Eh'), disp('    h          Eh')
disp([h',Eh'])

and this is my function

function [ys, ts] = MyTwoStepMethod(f, t0, te, y0, h)
ts = t0:h:te;
ys(1,1) = y0;
ys(1) = y0;
ys(2) = y0 + h*f(ts(1),ys(1));
for i = 2:length(ts)-1
    ys(i+1) = -4*ys(i) + 5*ys(i-1) + h*(4*f(ts(i),ys(i))+2*f(ts(i-1),ys(i-1)));
end
end		

[Torsten]

Does your strange "TwoStepMethod" have a name ?

[Seungryul Lee]

Will you plz give me any thoughts for that function:(

[Torsten]

Where did you get your TwoStepMethod from ? It looks wrong to me.

[Seungryul Lee]

Write a function: function [y, ts] = MyTwoStepMethod(f, t0, te, y0, h) that implements a two-step method with the following form: yi+1 =−4yi +5yi−1 +h(4f(ti,yi)+2f(ti−1,yi−1)) using forward Euler for the first step. This function should accept as arguments the function f(t,y), the starting time t0, the ending time te, the initial condition y0, and the step size h. It should return the approximate solution y at all time points as well as the time points ts.

this was the exact problem I had, so I've made some changes from my original Euler method function, and know I have no idea how to code it . .

[Torsten]

If this is the task, your code is correct.

But your method is unstable because the characteristic polynomial has the form x^2+4x-5 with roots -5 and 1.

[Seungryul Lee]

That's good. What about my code for displaying the table? Is that also correct?

Because I'm keep getting a silly table which I posted. It's giving me back 0s for h, which should not be 0s.

[Torsten]

Didn't you see that the values are to be multiplied by 1e225 ?

Give the 0's more digits by using

format long