clcclfclear all% RK4 method% Solve dy/dt = f(t,y). In general it
can be a function of both % variables t and y. If your function is
only a function of t then% you will need to add a 0*y to your
function.% define f(t,y) fcnstr='y-t^2+1' ;
f=inline(fcnstr,'t','y') ;% t0, initial time t0=0 ;% y0,
corresponding value of y at t0 y0=0.5;% tf, upper boundary of time
t (end time). tf=2;% n, number of steps to take
n=5;%**********************************************************************%
Displays title informationdisp(sprintf('\n\nThe 4th Order
Runge-Kutta Method of Solving Ordinary Differential
Equations'))h=(tf-t0)/n ;disp(sprintf(' h = ( tf - t0 ) / n
'))disp(sprintf(' = ( %g - %g ) / %g ',tf,t0,n))disp(sprintf(' =
%g',h))ta(1)=t0 ;ya(1)=y0 ;for i=1:n disp(sprintf('\nStep %g',i))
disp(sprintf('-----------------------------------------------------------------'))%
Adding Step Size ta(i+1)=ta(i)+h ;% Calculating k1, k2, k3, and k4
k1 = f(ta(i),ya(i)) ; k2 = f(ta(i)+0.5*h,ya(i)+0.5*k1*h) ; k3 =
f(ta(i)+0.5*h,ya(i)+0.5*k2*h) ; k4 = f(ta(i)+h,ya(i)+k3*h) ; %
Using 4th Order Runge-Kutta formula
ya(i+1)=ya(i)+1/6*(k1+2*k2+2*k3+k4)*h ; disp('1) Find k1 and k2
using the previous step information.') disp(sprintf(' k1 = f( x%g ,
y%g )',i-1,i-1)) disp(sprintf(' = f( %g , %g )',ta(i),ya(i)))
disp(sprintf(' = %g\n',k1)) disp(sprintf(' k2 = f( x%g + 0.5 * h ,
y%g + 0.5 * k1 * h )',i-1,i-1)) disp(sprintf(' = f( %g + 0.5 * %g ,
%g + 0.5 * %g * %g)',ta(i),h,ya(i),k1,h)) disp(sprintf(' = f( %g ,
%g )',ta(i)+0.5*h,ya(i)+0.5*k1*h)) disp(sprintf(' = %g\n',k2))
disp(sprintf(' k3 = f( x%g + 0.5 * h , y%g + 0.5 * k2 * h
)',i-1,i-1)) disp(sprintf(' = f( %g + 0.5 * %g , %g + 0.5 * %g *
%g)',ta(i),h,ya(i),k2,h)) disp(sprintf(' = f( %g , %g
)',ta(i)+0.5*h,ya(i)+0.5*k2*h)) disp(sprintf(' = %g\n',k3))
disp(sprintf(' k4 = f( x%g + h, y%g + k3*h)',i-1,i-1))
disp(sprintf(' = f( %g , %g )',ta(i)+h,ya(i)+k3*h)) disp(sprintf('
= %g\n',k1)) disp(sprintf('2) Apply the Runge-Kutta 4th Order
method to estimate y%g',i)) disp(sprintf(' y%g = y%g + 1/6*( k1 +
2*k2 + 2*k3 +k4) * h',i,i-1)) disp(sprintf(' = %g + %g * %g *
%g',ya(i),1/6,(k1+2*k2+2*k3+k4),h)) disp(sprintf(' =
%g\n',ya(i+1))) disp(sprintf(' at t%g =
%g',i,ta(i+1)))enddisp(sprintf('\n\n********************************Results**********************************'))%
The following finds what is called the 'Exact' solutiontspan = [t0
tf];[t,y]=ode45(f,tspan,y0);[yfi dummy]=size(y);yf=y(yfi);%
Plotting the Exact and Approximate solution of the ODE.hold
onxlabel('x');ylabel('y');title('Exact and Approximate Solution of
the ODE by the 4th Order Runge-Kutta
Method');plot(t,y,'--','LineWidth',2,'Color',[0 0 1]);
plot(ta,ya,'-','LineWidth',2,'Color',[0 1
0]);legend('Exact','Approximation');disp('The figure window that
now appears shows the approximate solution as ') disp('piecewise
continuous straight lines. The blue line represents the
exact')disp('solution. In this case ''exact'' refers to the
solution obtained by the')disp(sprintf('Matlab function
ode45.\n'))disp('Note the approximate value obtained as well as the
true value and ')disp('relative true error at our desired point x =
xf.')disp(sprintf('\n Approximate = %g',ya(n+1))) disp(sprintf('
Exact = %g',yf))disp(sprintf('\n True Error = Exact -
Approximate')) disp(sprintf(' = %g - %g',yf,ya(n+1)))disp(sprintf('
= %g',yf-ya(n+1)))disp(sprintf('\n Absolute Relative True Error
Percentage'))disp(sprintf(' = | ( Exact - Approximate ) / Exact | *
100'))disp(sprintf(' = | %g / %g | *
100',yf-ya(n+1),yf))disp(sprintf(' = %g',abs( (yf-ya(n+1))/yf
)*100))disp(sprintf('\nThe approximation can be more accurate if we
made our step'))disp(sprintf('size smaller (that is, increasing the
number of steps).\n\n'))
