Why does my graph look funny?

Whenever I try to graph this, something tells me that there's something wrong. Shouldn't Euler's method, Heun's method, etc graph look similar to one another? Here's what I have:

f=inline('t.^2*cos(y)','t','y');   %Define the function
t=linspace(0,10,100);   %Mesh points
y0=0.1;   %Initial condition
ye=EulerODE(f,t, y0)  %Solve using Euler's method
yem=EulerModODE(f, t, y0);   %Solve using Euler's method
yh=HeunODE(f, t, y0);  %Solve using Heun's method
yrk4=RK4(f, t, y0);   %Solve using Runge-Kutta Method
yam=AdamsM(f, t, y0);   %Solve using Adams-Moulton Method
plot(t, ye, 'Linewidth',2)
hold on
plot(t, yem, 'co', 'Linewidth', 2)
hold on
plot(t, yh, 'bx', 'Linewidth', 2)
hold on
plot(t, yrk4, 'k+', 'Linewidth', 2)
hold on
plot(t, yam, 'm*', 'Linewidth', 2)
hold off

Function EulerODE, EulerModODE, HeunODE etc. were taken from a MatLab CD provided from my text book. Below are some of the functions to give you an idea:

function x = EulerODE(f, t, x0)
%
% EULERODE  Uses Euler's method to solve a first-order ODE given in the form
% xdot = f(t, x) subject to initial condition x0.
%
%   x = EulerODE(f, t, x0) where
%
%       f is an inline function representing xdot = f(t, x).
%       t is a vector representing the mesh points.
%       x0 is a scalar representing the initial value of x.  
%
%       x is the vector of x values at the mesh points in t.

x = 0*t;  % Preallocate x
x(1) = x0;
for n = 1:length(t)-1
   h = t(n+1) - t(n);
   x(n + 1) = x(n) + h*f(t(n), x(n));
end

function x = EulerModODE(f, t, x0)
  %
  % EULERMODODE  Uses the modified Euler's method to solve a first-order ODE
  % given in the form xdot = f(t, x) subject to initial condition x0.
  %
  %   x = EulerModODE(f, t, x0) where
  %
  %       f is an inline function representing xdot = f(t, x).
  %       t is a vector representing the mesh points.
  %       x0 is a scalar representing the initial value of x.  
  %
  %       x is the vector of x values at the mesh points in t.

    x = 0*t; %Preallocate x
    x(1) = x0;
    for n = 1:length(t)-1
       h = t(n+1) - t(n);
       x(n + 1) = x(n) + h*f(t(n) + h/2, x(n)+ h/2*f(t(n), x(n)));
    end

Any help would be appreciated. Thanks!

Answer 1

[Grzegorz Knor]

Compare your results with MATLAB ODE functions, e.g.:

ode45(f,[0 10],0.1)

If you increase the number of mesh points:

t=linspace(0,10,1000);   %Mesh points

results will be comparable.

[Andy Jordan]

You are right, however my professor wants "h" to be 0.1(h=0.1) while 0<=t<=10.

Is there any other way of fixing this graph without changing the mesh points?

Thank you

[Andy Jordan]

"On the other hand, it is also important to remember that complicated differential equations should be written an M-file instead of using inline command or function handle."

Could it be that the given differential equation is too complicated, thus inline command doesn't work as accurately?