Question

Matlab & Differential Equations Help Needed

I need help with this Matlab project for differential equations. I've got 0 experience with Matlab other than a much easier project I did in another class a few semesters ago. All we've been given is this piece of paper and some sample code. I don't even know how to begin to approach this. I don't know how to use Matlab at all and I barely can do this material.

Here's the handout:

Matlab Project 2 due Monday 5.20.2019 (As with the first project, this project counts as another Quiz) For the initial value problem 1. Solve the IVP, and define a Matlab function to compute the exact values of the solution o(t) 2. Use Euler's method to find an approximation to the solution at t 0.5, 1.0, 1.5, and 2.0, using a. h 0.025 b. h 0.0125 For each step size, use your solution from part 1 to compare to this approximate solution by computing E-p(t)-y(t) at each value of t 3 Repeat #1 using the improved Euler method. 4. Use the Runge-Kutta method as in part 1 and 2, but using step sizes a, h = 0.1 b.. h = 0.05 Use Matlab to generate your results, save them into a diary.m file. For part 4, use a Matlab function to compute the exact solution and print them to the screen. 冫t -2t te.at V, 인 dt dan: s of dt tc งเ

Here's the sample code:

function [t, y] = myeuler(f, tinit, yinit, b, n)

% Euler approximation for initial value problem

% dy/dt = f(t, y), y(tinit) = yinit

% Approximations y(1),..., y(n+1) are calculated at

% the n+1 points t(1),..., t(n+1) in the interval

% [tinit, b]. The right-hand side of the differential

% equation is defined as an anonymous function f.

% Calculation of h from tinit, b, and n.

h = (b - tinit)/n;

% Initialize t and y as length n+1 column vectors.

t = zeros(n+1, 1);

y = zeros(n+1, 1);

% Calculation of points t(i) and the corresponding

% approximate values y(i) from the Euler Method formula.

t(1) = tinit;

y(1) = yinit;

for i = 1:n

t(i+1) = t(i) + h;

y(i+1) = y(i) + h*f(t(i), y(i));

end

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) 0.5-t+2.*y; initial guess yinit-1 ; tinit-o; Answering Question 1 syms y(t) eqndiff (y,t 0.5-t+2.*y cond -y(0) == yínit ; ySolt) dsolve (eqn, cond) %displaying result fprintf( Exact solution phi(t)- disp (ySol) Answering Question 2 tall t values tt-0.5 1.0 1.5 2.0]; All step size hh-[0.025 0.0125]; fprintf(nSolution Using Euler Method\n %loop for all step size and tend for ii-l:length(hh ) h-hh (ii)i for jj-1:length tt) tend-tt (33)i It_euler,y_euler ]-Euler (fun, tinit,yinit,tend, h); y_ext double (ySol tend)); error abs (y_ext-y euler (end)); fprint f ( '\tFor h=82.4f at t=%2.2f value of y=%f.\n ,h, t-euler ( end ) , y-euler ( end) ) fprint f ( '\t Error E- %f"\n\n',error) end end Answering Question 3 fprintf (nSolution Using Improved Euler Methodn') %loop for all step size and tend for ii-1length(hh) hshh ( İİ); for jj-l:length(tt) tend-tt (jj);

It_euler,y euler J-ImpEuler (fun,tinit,yinit,tend, h)i y_ext-double (ySol(tend)); error abs(y_ext-y_euler end)); fprint f ('\tFor h-82.4f at t-%2.2f value of _eule r (end) ) fprintft Error E f.\n\n',error) y-Af.\n h,tend,y , , end end Answering Question 4 fprintf( InSolution Using RK4 MethodIn' %All step size hh-0.1 0.051 %loop for all step size and tend for ii-1length(hh) for jj-l:length(tt) tend-tt (jj); lt_euler,y euler]-RK4 (fun,tinit,yinit,tend,h)i y_ext-double (ySol(tend)); error abs (y_ ext-y_euler (end) fprint f ( '\tFor h=%2.4f at t=%2.2f value of y=%f.\n , ,h, tend, y-euler (end) ) fprintft Error E .nn',error) end end 응음움各各各各各各各各各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음음各各음 %%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;i t eval-tend; n-(t-eval-t ) /h ; t_euler (1)-t; y_euler(1)-y; for i-1: n % initial y % at what point we have to evaluate % Number of steps Eular Steps m-double(f (t,y)) t_euler(it1)-t; y_euler(i+1)-y;

윰움움응各움움各各움움움움움움움움움움응움움움各움움움움움움움움움움움움움各움움움움움움움움움움움움움움움움움움움움各움움 %%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; t_eval-tend; nm(t-eval-t)/h; t imp(1)-t; y imp(l)-y; % initial y % at what point we have to evaluate % Number of steps for i-l:n %improved Euler steps ml-double(f (t,y)) m2-double (E ((tth), (y+h*ml))); y-y+double (h ((ml+m2)/2)); t-t+h; y-imp (i+1,-y; t_imp(i+1)t; end end 8888옹옹옹옹옹옹88%옹88888888888888888888옹%88옹%88%%888888옹옹888888: %%Matlab function for Runge Kutta Method function [t_rk,yrk]=RK4(f,tinit,yin it ,tend , h) %Euler method h amount of intervals t-tinit; % initial t y-yinit; t eval-tend; % initial y % at what point we have to evaluate ns(t-eval-t)/h; t_rk (1)-t y rk(1)-Yi % Number of steps for i= 1 : n RK4 Steps k1=h-double ( f (t,y)); k2-h*double (f ((t+h/2), (ytkl/2)); k 3-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 rk(i+1)-y 88888888888888888888 End of Code 8888888888888888888888

Exact solution phi(t)-t/2 exp(2*t) symbolic function inputs: t Solution Using Euler Method For h-o.0250 at t-0.50 value of y-2.903298 Error E 0.064984. For h=0 .0250 at t=1.00 value of y=7.539989 Error E 0.349067 For h-o.0250 at t1.50 value of y-19.429186. Error E = 1.406351. For h=0 .0250 at t-2.00 value of Error E-5.036709 y-50.56 1441. For h=0 .0125 t=0.50 value of y=2.935064. at ErrorE0.033218. For h 0.0125 at t-1.00 value of y 7.709568. Error E = 0.179488. For h=0 .0125 at t=1.50 value of y=20.108150. Error E0.727387. For h=0 .0125 t=2.00 value of y=52.977868. at Error E = 2.620282. Solution Using Improved Euler Method For h=0 . 0 250 at t=0.50 value of y=2 . 9 6 7 19 1 . Error E = 0.001091. For h=0 .0250 at t-1.00 value of y-7 .883127. Error E 0.005929 For h 0.0250 at t-1.50 value of y 20.811367. Error E 0.024170 For h-o.0250 at t-2.00 value of y-55.510568. Error E 0.087582. For h 0.0125 at t-0.50 value of y 2.968004 Error E- 0.000278. For h 0.0125 at t 1.00 value of y 7.887545. Error E = 0.001511. h=0 .0125 at t-1.50 value of y-20.829378. For Error E 0.006159. For h=0.0125 at t-2.00 value of y-55.575828. Error E 0.022322.

Solution Using RK4 Method For h-0 .1000 at tー0.50 value of y"2.968251. Error E-0.000031 For h-o.1000 at t-.00 value of y-7-888889. Error E-0.000167. For h-o.1000 at t-1.50 value of y-20.834857 Error E = 0.000680. For h 0.1000 at t-2.00 value of y 55.595684. Error E-0.002466 h-0.050 o at t-0.50 value of For y-2.968280. Error E = 0.000002. For h-0.0500 at t-1.00 value of y-7.889045. Error E -0.000011 For h=0 .0500 at t=1.50 value of y=20.835491. Error E0.000046 For h-o.0500 at t-2.00 value of y-55.597983. Error E = 0.000 167. 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) 0.5-t+2.*y;
%initial guess
tinit=0; yinit=1;

%Answering Question 1.
syms y(t)
eqn = diff(y,t) == 0.5-t+2.*y;
cond = y(0) == yinit;
ySol(t) = dsolve(eqn,cond);

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

%Answering Question 2.

%all t values
tt=[0.5 1.0 1.5 2.0];
%All step size
hh=[0.025 0.0125];

fprintf('\nSolution Using Euler Method\n')
%loop for all step size and tend
for ii=1:length(hh)
    h=hh(ii);
    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
end

%Answering Question 3.

fprintf('\nSolution Using Improved Euler Method\n')
%loop for all step size and tend
for ii=1:length(hh)
    h=hh(ii);
    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
end      

%Answering Question 4.

fprintf('\nSolution Using RK4 Method\n')
%All step size
hh=[0.1 0.05];
%loop for all step size and tend
for ii=1:length(hh)
    h=hh(ii);
    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
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 %%%%%%%%%%%%%%%%%%%%%%%