Question

This Question: 1 pt Consider the initial value problem below to answer to following. a) Approximate the value of y(T) using Euler's method with the given time step on the interval [o.TI b) Using the exact solution given, find the error in the approximation to y(T) (only at the right endpoint of the interval) c) Repeating parts d) Compare the errors in the approximations to y(T) a and b using half the time step used in those calculations, again find an approximation to y(T) a) The approximation with At 0.4 isyT) Round to five decimal places as needed.) b)The error in the approximation to y(T) is□ (Round to five decimal places as needed.) e) The approximation with At 0.2 is y(T) Round to five decimal places as needed.) The error in the approximation to y(T) is Round to five decimal places as needed.) d) How did halving the time step affect the approximations to y(T)? O A Halving the time steps results in approximately doubling the error O B. Halving the time steps results in reducing the error by approximately one tenth. c. Halving the time steps results in reducing the error by approximately one fourth. O D. Halving the time steps result ror in approximately halving thee

Answer 1

SMatlab code for Euler's forward clear all close all %function for Euler equation solution f-e (t,y) -Y %exact solution y_ext-e(t) exp(-t) %initial values t04, %initial t y0-1; %initial y t end values t_end-4; step size h-0.4; end values tn=to : h : t-end; % Euler steps yresult (1)y0; t-result ( 1 )-t0 ; %loop for Euler method for i-1:length (tn) t result(i+1)-t result(i) +h; y-result ( i+1 ) y-result ( ǐ ) +h *double ( f ( t-result ( í ) , y-result ( İ ) ) ) ; end fprint f ( '\ta) Approximate solution for del ( t )-%0.lf is y(82.2f)-82.6f. n',h, t_result (end-1),y result (end-1)) errl-y_ext (t_end) -y_result (end-1); fprint f( "\tb) The error in approximation to y(T) is %2.6f.\n', err!); step size h-0.2; x end values tn-t0:h:t_end; % Euler steps y result( t_ 1)-yo: result (1)-to; loop for Euler method for i-l:length (tn) t result (i+1)t result(i)+h; y-result ( i+1 ,- y-result ( İ ) +h"double ( f ( t-result ( í ) , y-result ( í ) ) ) ; end fprint f ( "\tc ) Approximate solution for del (t)-80.1f ís y(82.2f)-%2.6f. n',h,t_result (end-1),y_result (end-1)) errl-y_ext (t_end)-y_result (end-1); fprint f ("At The error in approximation to y(T) 1s %2.6f.\n", err1); fprintf(td) halving the time steps results approximately halving the errorn')

a) Approximate solution for del(t)-0.4 is y(4.00,-0.00604. b) The error in approximation to y(T) is 0.012269. c) Approximate solution for del (t)-0.2 IS y(4.00,-0.011529 The error in approximation to y (T) is 0.006786. d) halving the time steps results approximately halving the error. Published with MATLABE R2018a

%%Matlab code for Euler's forward
clear all
close all
%function for Euler equation solution
f=@(t,y) -y;
%exact solution
y_ext=@(t) exp(-t);
%Initial values
t0=0; %initial t
y0=1; %initial y
%t_end values
t_end=4;
    %step size
    h=0.4;
    %x end values

    tn=t0:h:t_end;
    % Euler steps
    y_result(1)=y0;
    t_result(1)=t0;
    %loop for Euler method
        for i=1:length(tn)
            t_result(i+1)= t_result(i)+h;
            y_result(i+1)= y_result(i)+h*double(f(t_result(i),y_result(i)));
          
        end
fprintf('\ta) Approximate solution for del(t)=%0.1f is y(%2.2f)=%2.6f.\n',h,t_result(end-1),y_result(end-1))
err1=y_ext(t_end)-y_result(end-1);
fprintf('\tb) The error in approximation to y(T) is %2.6f.\n',err1);
%step size
    h=0.2;
    %x end values

    tn=t0:h:t_end;
    % Euler steps
    y_result(1)=y0;
    t_result(1)=t0;
    %loop for Euler method
        for i=1:length(tn)
            t_result(i+1)= t_result(i)+h;
            y_result(i+1)= y_result(i)+h*double(f(t_result(i),y_result(i)));
        end
fprintf('\tc) Approximate solution for del(t)=%0.1f is y(%2.2f)=%2.6f.\n',h,t_result(end-1),y_result(end-1))
err1=y_ext(t_end)-y_result(end-1);
fprintf('\t The error in approximation to y(T) is %2.6f.\n',err1);

fprintf('\td) halving the time steps results approximately halving the error.\n ')

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