Question

1. For the initial-value problem, ?'(?) = ?? sin(5?) , ?(0) = 0.5,
2. Use the Euler Method with step size of 0.1 to calculate the approximate solution for the

differential equation for the initial condition y(0) = 0.5 and plot the result on the direction

field graph you produced in paragraph 1. Then do the same for the results of using the Euler Method with step sizes of 0.01 and then 0.005.

3. For the initial-value problem in paragraph 2 above, ?'(?) = ?? sin(5?) , ?(0) = 0.5,

find the approximate the value of y when x = 4 using the Euler Method for each of the three different step sizes you used in paragraph 2.

Type or write:
“For the Euler Method with a step size of 0.1, y(4) =”, followed by the answer accurate

to five decimal places.
“For the Euler Method with a step size of 0.01, y(4) =”, followed by the answer accurate

to five decimal places.
“For the Euler Method with a step size of 0.005, y(4) =”, followed by the answer accurate

to five decimal places.

4. Use the fourth-order Runge-Kutta Method with step sizes of 0.1, 0.01, and 0.005 to calculate approximate solutions for the same initial-value problem as paragraphs 2 and 3 and plot these approximate solutions on the direction field graph.
   • Attach a direction field graph showing the fourth-order Runge-Kutta Method results using a step size of 0.1.
   • Attach a second direction field graph showing the fourth-order Runge-Kutta Method results using a step size of 0.01.
   • Attach a third direction field graph showing the fourth-order Runge-Kutta Method results using a step size of 0.005.
   • Type or write a brief explanation of what happens with the plots as the step size is reduced.
5. For the initial-value problem in paragraphs 2 and 3 above, find the approximate the value of y when x = 4 using the fourth-order Runge-Kutta Method for each of the three different step sizes you used in paragraph 4.

Type or write:
“For the fourth-order Runge-Kutta Method with a step size of 0.1, y(4) =”, followed by

the answer accurate to five decimal places.
“For the fourth-order Runge-Kutta Method with a step size of 0.01, y(4) =”, followed by

the answer accurate to five decimal places.
“For the fourth-order Runge-Kutta Method with a step size of 0.005, y(4) =”, followed by

the answer accurate to five decimal places.

Answer 1

%Matlab code for numerical solution of ode using RK4 method
clear all
close all
%function for which Solution have to do
fun=@(x,y) exp(y).*sin(5*x);
%initial condition
xinit=0; yinit=0.5;
xend=4;
%all step sizes
hh=[0.1 0.01 0.005];
for i=1:3
    h=hh(i);
    [x_elr,y_elr]=euler_method(fun,yinit,xinit,xend,h);
    fprintf('\tUsing Euler, for h=%1.3f value of x(%2.2f) is %f\n',h,x_elr(end),y_elr(end))
    [x_rk4,y_rk4]=RK4(fun,yinit,xinit,xend,h);
    fprintf('\tUsing RK4, for h=%1.3f value of x(%2.2f) is %f\n',h,x_rk4(end),y_rk4(end))
  
end

plot(x_rk4,y_rk4,'linewidth',2)
hold on
plot(x_elr,y_elr,'--')

ylabel('x')
xlabel('y(x)')
title('x vs. y(x) plot')
legend('RK4 Solution','Euler Solution','location','best')

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Matlab code for Euler's forward
function [t_result,y1_result] = euler_method(f,y0,t0,tend,h)
%function for Euler equation solution
  
    %all step size
    N=(tend-t0)/h;
    %Initial values

    %t end values
    tn=t0:h:tend;
    % Euler steps
    y1_result(1)=y0;
    t_result(1)=t0;
    fprintf('\n\tFor step size %2.2f\n',h)
    for i=1:length(tn)-1
        t_result(i+1)= t_result(i)+h;
        y1_result(i+1)=y1_result(i)+h*double(f(t_result(i),y1_result(i)));
        %fprintf('At x=%2.2f y(%2.2f)=%f\n',t_result(i+1),t_result(i+1),y1_result(i+1))
    end
end

%%Matlab function for Runge Kutta Method
function [t_rk,y_rk]=RK4(f,yinit,tinit,tend,h)
    % RK4 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 %%%%%%%%%%%%%%%%%%%%%%%