1.  dy
   ---- = f(t,y)    Let f(t,y) = t  . The y(t) = t^2 /2 + c1.
    dt 

In MATLAB  for f(t,y)=t we can solve symbolically this equation as follows.

  >> y = dsolve('Dy = t ', 't')    %D is d 
                                   %     ----   for MATLAB
                                   %     dt    

     y = 1/2 t^2 + c1

  >> eq1 = 'Dy=t';
  >> y = dsolve (eq1,'t');

  % If initial condition is y(0)=1
  % then

  >> y = dsolve(eq1,'y(0)=1', 't')

    y = 1/2 t^2 + 1

  >> init1 = 'y(0)=1';

  >> y = dsolve(eq1,init1,'t');


  >> t = linspace(0,1,20);
  >> z = eval(vectorize(y));

  >> plot(t,z);


     f(t,y) is      then dy/dt= f(t,y) has solution  y(t) equal to  (for y(0)=1)

        0                                             1
        t                                             t^2 /2 +1
        y                                             exp(t)
       -y                                             exp(-t)

2.  dy    ya - ya
   ---  = ------- = f(t,y)  = -y    . For this example we use f(t,y)=y  i.e. solution will be y(t)=exp(-t)
    dt    ta - tb


    Numerical approximation

    >> t = linspace(0,1,20);
    >> y = exp(-t);
    >> plot(t,y); hold on;
    >> dyoverdt = diff(y) ./ diff(t)  % for   y=[y(1)      ....  y(n-1)     y(n)]
                                      % diff(y)=[y(2)-y(1) ... y(n)-y(n-1)]

    >> plot(t,-y);
    >> plot(t(1:end-1),dyoverdt,'r--');


  In other words  dy    yn+1 - yn        '
                 ---- = ------------- = y  = f(t,y) = f(tn,yn)
                  dt    tn+1 - tn

  from which we get that
  yn+1 = yn + (tn+1 - tn) f(tn,yn) = yn + h f(tn,yn) if we denote by h = tn+1 - tn


3. Solver  [ t y ]  = solver ( @function , span, initialy0 ) 

 solver can be  ode45  ode23  oder113 etc

   >> f = @(t, y)  (t)                  % f(t,y) =t i.e. y(t) = t^2 /2 +1

   >> g = @ (t)  ( 0.5 .* t .^ 2  +1 )  % g(t) is thus y(t) exact solution!

   >> ta = [ 0: 0.1 : 1];

   >> ya = g(ta)                        % compute exact solution !


   >> [ t y ] = ode45 ( f, [ 0 1], 1)   % approximate solution

   >> plot(ta,ya,'r--');  hold on ;  % plot exact solution

   >> plot(t,y, 'b--');   % plot approximate solution


Note
If you change
   >> ta = [ 0: 0.1 : 1];
into say
   >> ta = [ 0: 0.1 : 5];

then change
   >> [ t y ] = ode45 ( f, [ 0 1], 1)   % approximate solution
into
   >> [ t y ] = ode45 ( f, [ 0 5], 1)   % approximate solution


If initial condition is not y(0)=1,

then g(t) needs to be recomputed! (use section 1 )

and of course    this -------V-and-V will need to be adjusted appropriately
                             V     V
   >> [ t y ] = ode45 ( f, [ 0 1], 1)   


4. -------------------------------------

   Last fall semester , some students found this code useful

% math2.m
y=0.5;
for t=0.01:0.01:25
  n = round(t/0.01);
  y = y + 0.01*(3.0-5*sqrt(y));
  fprintf('t= %5.2f n= %3d y= %8.4f\n',t,n,y);
end


% math0.m
T=25;
H=0.25;
STEP=H;
Y0=0.5;
y=Y0;
for t=H:STEP:T
  n = floor(t/0.01);
  y = y + 0.01*(3.0-5*sqrt(y));
  if (mod(n,100) == 0)
   fprintf('t= %5.2f n= %6d y= %8.4f\n',t,n,y);
  end
end



%% math1.m
T=25;
H=0.25;
STEP=H;
y=0.5;
for t=H:STEP:T
  n = floor(t/0.01);
  y = y + 0.01*(3.0-5*sqrt(y));
  if (mod(n,100) == 0)
   fprintf('t= %5.2f n= %6d y= %8.4f\n',t,n,y);
  end
end


%%