Question

Question 1: Given the initial-value problem 12-21 0 <1 <1, y(0) = 1, 12+10 with exact solution v(t) = 2t +1 t2 + 1 a. Use Euler's method with h = 0.1 to approximate the solution of y b. Calculate the error bound and compare the actual error at each step to the error bound. c. Use the answers generated in part (a) and linear interpolation to approximate the following values of y, and compare them to the actual value of y(0.04). d. Compute the value of h necessary for y(ti) - wil <0.1. Question 2 : Use the Modified Euler method to approximate the solutions to the above IVP with h=0.1, and compare the results to the actual values. (use MATLAB). Question 3: Use the Runge-Kutta method of order four to approximate the solutions to the above IVP with h=0.1, and compare the results to the actual values.

Answer 1

% % Matlab code for solving ode using Euler Improved Euler and RK4 clear all close all %function for which Solution have to do fun=@(t,y) (2-2.*t.*y)./(t.^2+1); initial guess tinit=0; yinit=1; %Answering Question 1. syms y(t) eqn = diff(y,t) == (2-2.*t.*y)./(t.^2+1); cond = y(0) == yinit; ySol(t) = dsolve (eqn, cond); %displaying result fprintf('Exact solution phi(t)=') disp(y Sol) Answering Question 1. %all t values tt=[0.4 1.0]; All step size h=0.1; fprintf('\nSolution Using Euler Method\n') % loop for all step size and tend for jj=1:length(tt) tend=tt(jj); [t_euler,y_euler ]=Euler(fun, tinit,yinit, tend, h); y_ext=double (ySol(tend)); error = abs(y_ext-y_euler (end) ); fprintf('\tFor h=%2.4f at t=%2.2f value of y=%f. In ',h, t_euler (end),y_euler(end)) fprintf('\t Error E = %.\n\n',error) end %Answering Question 2. fprintf('\nSolution Using Improved Euler Method\n') %loop for all step size and tend for jj=1:length(tt) tend=tt(jj); [t_euler,y_euler ]=ImpEuler (fun, tinit,yinit, tend, h); Y_ext=double (ySol(tend) ); error = abs(y_ext-y_euler(end));

fprintf('\tFor h=82.4f at t=%2.2f value of y=%f.\n ',h, tend, y_euler(end)) fprintf('\t Error E = %f.\n\n',error) end Answering Question 4. fprintf('\nSolution Using RK4 Method\n') All step size %loop for all step size and tend for jj=1:length(tt) tend=tt(jj); [t_euler,y_euler ]=RK 4 (fun, tinit, yinit, tend, h); y_ext=double (ySol(tend)); error = abs(y_ext-y_euler (end) ); fprintf('\tFor h=%2.4f at t=%2.2f value of y=%f. In ',h, tend, y_euler(end)) fprintf('\t Error E = %f.\n\n',error) end %%Matlab function for Euler Method function [t_euler,y_euler ]=Euler (f, tinit,yinit, tend, h) Euler method % h amount of intervals t=tinit; % initial t y=yinit; % initial y t_eval=tend; % at what point we have to evaluate n=(t_eval-t)/h; % Number of steps t_euler(1)=t; y_euler(1)=y; for i=1:n %Eular Steps m=double(f(t,y)); t=t+h; y=y+h*m; t_euler(i+1)=t; Y_euler(i+1)=y; end end % Matlab function for Improved Euler Method function [t_imp,y_imp]=ImpEuler(f, tinit,yinit, tend, h) Euler method % h amount of intervals trtinit; % initial t y=yinit; % initial y t_eval=tend; % at what point we have to evaluate

% Number of steps n=(t_eval-t)/h; t_imp (1)=t; y_imp (1)=y; for i=1:n %improved Euler steps ml=double(f(t,y)); m2=double(f( (t+h), (y+h*m1))); y=y+double(h* ((m1+m2)/2)); t=t+h; y_imp (i+1)=y; timp (i+1)=t; end end 응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응응 % % Matlab function for Runge kutta Method function [t_rk,y_rk)=RK4 (f, tinit,yinit, tend, h) %Euler method % h amount of intervals trtinit; % initial t y=yinit; % initial y t_eval=tend; % at what point we have to evaluate n= (t eval-t) /h; % Number of steps t_rk (1)=t; y_rk (1)=y; for i=1:n 8RK4 Steps kl=h* double(f(t,y)); k2=h* double(f( (t+h/2), (y+k1/2))); k3=h* double(f( (t+h/2), (y+k2/2))); k4=h* double(f( (t+h), (y+k3))); dy=(1/6) * (k1+2*k2+2*k3+k4 ); t=t+h; y=y+dy; t_rk(i+1)=t; Y_rk(i+1)=y; end end %%%%%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%%%%%%% Exact solution phi(t)=(2*t)/(t^2 + 1) + 1/(t^2 + 1) symbolic function inputs: t Solution Using Euler Method For h=0.1000 at t=0.40 value of y=1.613872. Error E = 0.062148.

For h=0.1000 at t=1.00 value of y=1.579669. Error E = 0.079669. Solution Using Improved Euler Method For h=0.1000 at t=0.40 value of y=1.549061. Error E = 0.002663. For h=0.1000 at t=1.00 value of y=1.498431. Error E = 0.001569. Solution Using RK4 Method For h=0.1000 at t=0.40 value of y=1.551723. Error E = 0.000001. For h=0.1000 at t=1.00 value of y=1.499998. Error E = 0.000002. Published with MATLAB® R2018a

%%Matlab code for solving ode using Euler Improved Euler and RK4
clear all
close all

%function for which Solution have to do
fun=@(t,y) (2-2.*t.*y)./(t.^2+1);
%initial guess
tinit=0; yinit=1;

%Answering Question 1.
syms y(t)
eqn = diff(y,t) == (2-2.*t.*y)./(t.^2+1);
cond = y(0) == yinit;
ySol(t) = dsolve(eqn,cond);

%displaying result
fprintf('Exact solution phi(t)=')
disp(ySol)

%Answering Question 1.

%all t values
tt=[0.4 1.0];
%All step size
h=0.1;

fprintf('\nSolution Using Euler Method\n')
%loop for all step size and tend
    for jj=1:length(tt)
        tend=tt(jj);
      
        [t_euler,y_euler]=Euler(fun,tinit,yinit,tend,h);
        y_ext=double(ySol(tend));
      
        error = abs(y_ext-y_euler(end));
        fprintf('\tFor h=%2.4f at t=%2.2f value of y=%f.\n ',h,t_euler(end),y_euler(end))
        fprintf('\t Error E = %f.\n\n',error)
    end

%Answering Question 2.

fprintf('\nSolution Using Improved Euler Method\n')
%loop for all step size and tend
    for jj=1:length(tt)
        tend=tt(jj);
      
        [t_euler,y_euler]=ImpEuler(fun,tinit,yinit,tend,h);
        y_ext=double(ySol(tend));
      
        error = abs(y_ext-y_euler(end));
        fprintf('\tFor h=%2.4f at t=%2.2f value of y=%f.\n ',h,tend,y_euler(end))
        fprintf('\t Error E = %f.\n\n',error)
    end      

%Answering Question 4.

fprintf('\nSolution Using RK4 Method\n')
%All step size
%loop for all step size and tend
    for jj=1:length(tt)
        tend=tt(jj);
      
        [t_euler,y_euler]=RK4(fun,tinit,yinit,tend,h);
        y_ext=double(ySol(tend));
      
        error = abs(y_ext-y_euler(end));
        fprintf('\tFor h=%2.4f at t=%2.2f value of y=%f.\n ',h,tend,y_euler(end))
        fprintf('\t Error E = %f.\n\n',error)
    end      

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Matlab function for Euler Method

function [t_euler,y_euler]=Euler(f,tinit,yinit,tend,h)
    %Euler method
    % h amount of intervals
    t=tinit;         % initial t
    y=yinit;         % initial y
    t_eval=tend;     % at what point we have to evaluate
    n=(t_eval-t)/h; % Number of steps
    t_euler(1)=t;
    y_euler(1)=y;
    for i=1:n
        %Eular Steps
        m=double(f(t,y));
        t=t+h;
        y=y+h*m;
        t_euler(i+1)=t;
        y_euler(i+1)=y;
    end
  
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Matlab function for Improved Euler Method

function [t_imp,y_imp]=ImpEuler(f,tinit,yinit,tend,h)
    %Euler method
    % h amount of intervals
    t=tinit;         % initial t
    y=yinit;         % initial y
    t_eval=tend;     % at what point we have to evaluate
    n=(t_eval-t)/h; % Number of steps
    t_imp(1)=t;
    y_imp(1)=y;
  
    for i=1:n
    %improved Euler steps
       m1=double(f(t,y));
       m2=double(f((t+h),(y+h*m1)));
       y=y+double(h*((m1+m2)/2));
       t=t+h;
       y_imp(i+1)=y;
       t_imp(i+1)=t;
    end
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Matlab function for Runge Kutta Method

function [t_rk,y_rk]=RK4(f,tinit,yinit,tend,h)
    %Euler method
    % h amount of intervals
    t=tinit;         % initial t
    y=yinit;         % initial y
    t_eval=tend;     % at what point we have to evaluate
    n=(t_eval-t)/h; % Number of steps
    t_rk(1)=t;
    y_rk(1)=y;
  
    for i=1:n
    %RK4 Steps
       k1=h*double(f(t,y));
       k2=h*double(f((t+h/2),(y+k1/2)));
       k3=h*double(f((t+h/2),(y+k2/2)));
       k4=h*double(f((t+h),(y+k3)));
       dy=(1/6)*(k1+2*k2+2*k3+k4);
       t=t+h;
       y=y+dy;
       t_rk(i+1)=t;
       y_rk(i+1)=y;
    end
end

%%%%%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%%%%%%%