Question

Solve the following initial value problem over the interval from t= 0 to 1 where y(O) = 1 using the following methods with a step size of 0.25 4) dt

Display all methods listed below in ONE GRAPH:

1. Analytical method

2. Euler's method

3. Heun's method without iteration

4. Ralston's method

5. Fourth-order RK method

Metlab preferred

Answer 1

SMatlab code for Euler, Mid Point, RK2 and RK4 method clear all close all %Function for which solution have to do f-L(t,y) (1+4.*t).*sqrt(y)i h=0.25; % amount of intervals fprint f( "\nSolution for Euler method using step size fprintf( Initial condition y (0)-1 at t-o.n %Euler method 용음음응용号음응용号음응용 %2.2f.\n',h) 8 initial t initial y y= 1 ; t_eval-1; nm(t-eval-t)/h; t2(1)=t; % at what point we have to evaluate % Number of steps for i=1:n %Eular Steps m-double(f (t,y)); fprint f ( '\t t-%2.2f value of y(32.2f,-%f\n',t2(i+1),t2(i at end %Huen method 88各움움各各움움各各 fprint f( ,\nSolution for Heun method using step size fprintf Initial condition y (0)-1 at t-o.n) t-0; y=1; t_eval-1 n«(t-eval-t)/h; % Number of steps t3 (1)-t y3 ( 1 )=y; for i-1: n %Huen steps %2.2f.\n', h) initial t % initial y at what point we have to evaluate ml-double(f (t,y)); m2-double (E ((tth), (y+h*ml))); y-y+double (h ((ml+m2)/2)); t-t+h; t3(1+1)=t ; fprintf ('\t t-%2.2f value of y(32.2f,-%f\n',t3(i+1),t3(i at end

RK4 method tprint f ( "\nS olution for RK4 method using step size %2.2f.\n',h) fprint f ( "Initial condition y(0)=1 at t=0.\n') t-0; Y-1; t_eval-1; n=(t-eval-t)/h; % Number of steps t4 (1)t; % initial t % initial y at what point we have to evaluate for i-1:n RK4 Steps kl-h*double (f (t,y)) k2 h*double (f ( (tth/2), (y+k1/2))); k3-h double (f ( (t+h/2), (y+k2/2); k4-h double (f ((t+h), (ytk3))); dx= (1/6)*(k1+2+k2+2+k3+k4); t=t+h ; y-y+dx; fprintf ('\t t=82.2f value of y(82.2f)=%f\n',t4(1+1),t4(i at end %Ralston method fprintf( "\nSolution for Ralston method using step size %2.2f. n',h) fprint f ( "Initial condition % initial t % initial y % at what point we have to evaluate % Number of y(0)=1 t-o.\n') at y-l; t_eval-1 n-(t-eval-t)/h ; t5 ( 1 )=t; steps for i=1:n RK2 Steps k1=h-double ( f (t,y)); k2-h*double (f ( (t+h), (y+kl))); dx-(1/2) t=t+1 ; y-y+dx; (kl+k2); fprintf ('\t at t 82.2f value of y(82.2f)-%f\n.,tS(1+1),t5 ( i end %%Exact solution syms y(t)

eqn -diff (y,t)(1+4*t)sqrt(y)i cond y (0)1; ySol(t) = dsolve (eqn, cond); fprintf( 'Exact solution for given ode is y(t)- disp ySol) yy ext (t)-(t (2*t 1) + 2)*2/4; fprintf Initial condition y (0)-1 at t-o.\n') for ii-l:length (t4) y6 (ii)-double (yyext(t4 (ii))); fprintf( "\t at t-82.2f value of y(82.2f,-%f in',t4 (ii),A(ii),y6 ( і i) ) end &Plotting solution using Euler method figure(1) hold on plot (t2,y2, Linewidth',2) plot (t3,y3, 'Linewidth',2) plot (t4,y4, Linewidth',2) plot (t5,y5, 'linewidth',2) plot (t4,y6, 'linewidth',2) xlabel( 't' ylabel'y (t)') title( Solution plot y(t) vs. t') legend( 'Euler Method', 'Heun Method', 'RK4 Method', 'Ralston Method', Exact solution', 'Location", "northwest') grid on ,,음各各움음各各움음各各음음各各 End of Code 韽움음음各움움움움움음음움움 Solution for Euler method using step size 0.25. Initial condition y(0)-1 at t-0. at t 0.25 value of y(0.25)-1.250000 at t-0.50 value of y (0.50)1.809017 at t-0.75 value of y(0.75)-2.817765 at t-1.00 value of y (1.00) 4.496385 Solution for Heun method using step size 0.25 Initial condition y(0)-1 at t o. at t-0.25 value of y(0.25)=1.404508 at t 0.50 value of y(0.50)-2.230730 at t 0.75 value of y (0.75) 3.706089 at t-l.00 value of y(1.00)-6.151785 Solution for RK4 method using step size 0.25. Initial condition y(0)-1 at t 0. at t-0.25 value of y (0.25)-1.410089 at t-0.50 value of y (0.50)-2.249786 at t-0.75 value of y(0.75)-3.753411 at t-l.00 value of y (1.00) 6.249054

Solution for Ralston method using step size 0.25 initial condition y(0)=1 at t=0. at t 0.25 value of y(0.25)-1.404508 at t-0.50 value of y (0.50) 2.230730 at t-0.75 value of y(0.75)-3.706089 at t-1.00 value of y (1.00) 6.151785 Exact solution for given ode is y(t)- (t*(2*t 1) + 2)2/4 symbolic function inputs: t Initial condition y(0)-1 at t-0. at t 0.00 value of y (0.00) 1.000000 at t 0.25 value of y(0.25)=1.410156 at t 0.50 value of y (0.50) 2.250000 at t-0.75 value of y(0.75)-3.753906 at t-l.00 value of y (1.00)-6.250000 Solution plot yft) vs. t Euler Method Heun Method RK4 Method Ralston Method Exact solution 6 06 0 0.2 0.3 0.4 0.5 0.7 0.8 0.9 Published with MATLAB R2018a

%%Matlab code for Euler, Mid Point, RK2 and RK4 method
clear all
close all
%Function for which solution have to do
f=@(t,y) (1+4.*t).*sqrt(y);

    h=0.25; % amount of intervals
    fprintf('\nSolution for Euler method using step size %2.2f.\n',h)
    fprintf('Initial condition y(0)=1 at t=0.\n')
    %Euler method
    %%%%%%%%%%%%%%%        
    t=0;             % initial t
    y=1;             % initial y
    t_eval=1;        % at what point we have to evaluate
    n=(t_eval-t)/h; % Number of steps
    t2(1)=t;
    y2(1)=y;
    for i=1:n
        %Eular Steps
        m=double(f(t,y));
        t=t+h;
        y=y+h*m;
        t2(i+1)=t;
        y2(i+1)=y;
        fprintf('\t at t=%2.2f value of y(%2.2f)=%f\n',t2(i+1),t2(i+1),y2(i+1))
    end

  
    %Huen method
    %%%%%%%%%%%%%%%
    fprintf('\nSolution for Heun method using step size %2.2f.\n',h)
    fprintf('Initial condition y(0)=1 at t=0.\n')
    t=0;             % initial t
    y=1;             % initial y
    t_eval=1;        % at what point we have to evaluate
    n=(t_eval-t)/h; % Number of steps
    t3(1)=t;
    y3(1)=y;
    for i=1:n
    %Huen steps
       m1=double(f(t,y));
       m2=double(f((t+h),(y+h*m1)));
       y=y+double(h*((m1+m2)/2));
       t=t+h;
       y3(i+1)=y;
       t3(i+1)=t;
       fprintf('\t at t=%2.2f value of y(%2.2f)=%f\n',t3(i+1),t3(i+1),y3(i+1))
    end

  
    %RK4 method
    %%%%%%%%%%%%%%%
    fprintf('\nSolution for RK4 method using step size %2.2f.\n',h)
    fprintf('Initial condition y(0)=1 at t=0.\n')
    t=0;             % initial t
    y=1;             % initial y
    t_eval=1;        % at what point we have to evaluate
    n=(t_eval-t)/h; % Number of steps
    t4(1)=t;
    y4(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)));
       dx=(1/6)*(k1+2*k2+2*k3+k4);
       t=t+h;
       y=y+dx;
       t4(i+1)=t;
       y4(i+1)=y;
       fprintf('\t at t=%2.2f value of y(%2.2f)=%f\n',t4(i+1),t4(i+1),y4(i+1))
    end
  
    %Ralston method
    %%%%%%%%%%%%%%%
    fprintf('\nSolution for Ralston method using step size %2.2f.\n',h)
    fprintf('Initial condition y(0)=1 at t=0.\n')
    t=0;             % initial t
    y=1;             % initial y
    t_eval=1;        % at what point we have to evaluate
    n=(t_eval-t)/h; % Number of steps
    t5(1)=t;
    y5(1)=y;
    for i=1:n
    %RK2 Steps
       k1=h*double(f(t,y));
       k2=h*double(f((t+h),(y+k1)));
     
       dx=(1/2)*(k1+k2);
       t=t+h;
       y=y+dx;
       t5(i+1)=t;
       y5(i+1)=y;
       fprintf('\t at t=%2.2f value of y(%2.2f)=%f\n',t5(i+1),t5(i+1),y5(i+1))
    end

  
     %%Exact solution
     syms y(t)
        eqn = diff(y,t) == (1+4*t)*sqrt(y);
        cond = y(0) == 1;
        ySol(t) = dsolve(eqn,cond);
        fprintf('Exact solution for given ode is y(t)=')
        disp(ySol)
        yy_ext(t)=(t*(2*t + 1) + 2)^2/4;
        fprintf('Initial condition y(0)=1 at t=0.\n')
        for ii=1:length(t4)
            y6(ii)=double(yy_ext(t4(ii)));
            fprintf('\t at t=%2.2f value of y(%2.2f)=%f\n',t4(ii),t4(ii),y6(ii))
        end

%%Plotting solution using Euler method
figure(1)
hold on
plot(t2,y2,'Linewidth',2)
plot(t3,y3,'Linewidth',2)
plot(t4,y4,'Linewidth',2)
plot(t5,y5,'linewidth',2)
plot(t4,y6,'linewidth',2)

xlabel('t')
ylabel('y(t)')
title('Solution plot y(t) vs. t')
legend('Euler Method','Heun Method','RK4 Method','Ralston Method','Exact solution','Location','northwest')
grid on

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