Question

PROBLEMS 22.1 Solve the following initial value problem over the interval from 0to2 where yo) 1.Display all your results on the same graph. dy=vr2-1.ly dt (a) Analytically. (b) Using Euler's method with h 0.5 and 0.25. (c) Using the midpoint method with h 0.5 (d) Using the fourth-order RK method with h 0.5.

Answer 1

SMatlab code for Euler, Mid Point, RK2 and RK4 method clear all close all %Function for which solution have to do h-0.5; % amount of interva ïs fprintf( "\nSolution for Euler method using step size fprintf( Initial condition y (0)-1 at t-o.n %Euler method 용음음응용号음응용号음응용 t=0; %2.2f.\n',h) 8 initial t initial y Y-1; t_eval-2; 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)); y2(i+1)=y; fprint f ( "\t t=%2.2f value of y(32.2f)=%f\n',t2(i+1),t2(i at end %Euler method 88各움움各各움움各各 h=0.25; fprintf( "\nSolution for Euler method using step size fprintf( Initial condition y (0)-1 at t-o.In t-0 y-l; t eval-2; nm(t-eval-t ) /h ; % Number of steps t3 (1)-t %2.2f.\n',h) % initial t % initial y % at what point we have to evaluate for i=1:n %Euler steps m-double(f (t, y); y-y+h*m,; 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 ,,88움움88움움움을움움움 h=0.5; % amount of intervals fprintf( "\nSolution for RK 4 method using step size %2.2f.\n',h) fprintf( Initial condition y (0)-1 at t-o.In initial t % initial y % at what point we have to evaluate % Number of t eval-2 nm(t-eval-t)/h; t4 (1)-t steps for i-1: n RK4 Steps k1-h double(f (t,y)); k2-h double (f ( (t+h/2), (y+kl/2); k3-h* double ( f ( ( tth/ 2 ) , ( y+k2 /2 ) ) ) ; k4-h*double (f ( (t+h), (y+k3))); dx-(1/6) (k1+2 k2+2wk3+k4) t=t+h ; Y y+dx; fprintf("\t t=%2.2f value of y(32.2f)=%f\n',t4(i+1), t4(i at %Midpoint method h-0.5; 各各各各各各各各各各各各各 fprint f( "\nSolution for Midpoint 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-2 n-(t-eval-t ) /h ; t5 ( 1 )=t; steps for i=1:n Midpoint Steps kl-h*double(f (t,y))i k2-h*double (f ( (t+h), (y+kl))); dx-(1/2) (kl+k2); t-t+h; y-y+dx; y5 (i+l)-yi fprintf ('\t at t 82.2f value of y(82.2f) %f\n',tS(1+1),t5 ( i end

%%Exact solution syms y(t) cond y(0)1; ySolt)dsolve (eqn, cond) fprintf ( Exact solution for given ode is y(t) disp(ysol) yy_ext(t)-ysol; 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 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', Euler Method, RK4 Method', 'Midpoint Method', Exact solution', Location", "northwest') grid on Solution for Euler method using step size 0.50 initial condition y(0)=1 at t=0. at t-0.50 value of y(0.50) -0.450000 at t-1.00 value of y (1.00) 0.258750 at t-1.50 value of y(1.50)-0.245812 at t-2.00 value of y (2.00) 0.387155 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) 0.725000 at t 0.50 value of y(0.50)-0.536953 at t 0.75 value of y (0. 75) 0.422851 at t=1.00 value of y(1.00)-0.366030 at t-l.25 value of y (1.25)-0.356879 at t=1.50 value of y(1.50)-0.398143 at t-l.75 value of y (1.75)-0.512610 at t-2.00 value of y (2.00) 0.764109

Solution for RK4 method using step size 0.50. Initial condition y(0)-1 at t o. at t-0.50 value of y(0.50)=0 . 601570 at t-1.00 value of y(1.00)-0.464524 at t=1.50 value of y(1.50)-0.591380 at t-2.00 value of y(2.00)-1.584452 Solution for Midpoint method using step size 0.50. Initial condition y(0)-1 at t-0. at t 0.50 value of y(0.50)-0.629375 at t-1.00 value of y (1.00) 0.486586 at t-1.50 value of y(1.50)-0.607320 at t-2.00 value of y (2.00) 1.475407 Exact solution for given ode is y(t)-exp((t (10*t 2 - 33))/30) 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-0.50 value of y (0.50)-0.601497 at t-1.00 value of y (1.00) 0.464559 at t-l.50 value of y (1.50) 0.591555 at t-2.00 value of y (2.00) 1.594670 Solution plot yt) vs. t 1.6 Euler Method Euar Method RK4 Method 14FMidpoint Method Exact solution 1.2 0.8 0.6 0.4 0.2 04 02 0.6 08 14 1.6 Published with MATLABR 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) y.*t.^2-1.1.*y;

    h=0.5; % 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=2;        % 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

  
    %Euler method
    %%%%%%%%%%%%%%%
    h=0.25;
    fprintf('\nSolution for Euler 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=2;        % 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
    %Euler steps
       m=double(f(t,y));
       y=y+h*m;
       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
    %%%%%%%%%%%%%%%
    h=0.5; % amount of intervals
    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=2;        % 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
  
    %Midpoint method
    h=0.5;
    %%%%%%%%%%%%%%%
    fprintf('\nSolution for Midpoint 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=2;        % 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
    %Midpoint 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) == y*t^2-1.1*y;
        cond = y(0) == 1;
        ySol(t) = dsolve(eqn,cond);
        fprintf('Exact solution for given ode is y(t)=')
        disp(ySol)
        yy_ext(t)=ySol;
        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','Euler Method','RK4 Method','Midpoint Method','Exact solution','Location','northwest')
grid on

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