Question

Problem #3: The Ralston method is a second-order method that can be used to solve an initial-value, first-order ordinary differential equation. The algorithm is given below: 2 Yi+1 = yi + k +k2)h Where kı = f(ti,y;) 3 k2 = ft;+ -h, y; +-kih You are asked to do the following: 3.1 Following that given in Inclass activity #10a, develop a MATLAB function to implement the algorithm for any given function, the time span, and the initial value. 3.2 Use your code to solve the following first-order ordinary differential equation from t = 0 tot = 1.5 with the initial condition y = 3 at t=0 with a time step of h = 0.3: dy --1.5y +7e64 3.3 dt Make a plot of your numerical solution (with symbols) of a time step of h = 0.3 and compare it with the solution by the MATLAB built-in ODE solver, "ode 45". Test the effect of the time step h on the accuracy of your numerical solutions. 3.4

Answer 1

The detailed solution is given below.

clear;
clc;

f = @(t,y) -1.5.*y + 7*exp(-0.4.*t);

H = [0.3 0.1 0.05];
nh = numel(H);
   
for j = 1:nh
    h = H(j); 
    t = 0:h:1.3;
    nt = numel(t);
    % Initial Condition
    y = zeros(1,nt);
    y(1) = 3;

    figure(j)
    hold on
    xlabel('t')
    ylabel('y')
    m = sprintf('h = %.2f\n',h);
    % Ralston Method
    for i = 1:nt-1
        k1 = f(t(i),y(i));
        k2 = f(t(i)+3/4*h,y(i)+3/4*k1*h);

        y(i+1) = y(i) + (1/3*k1 + 2/3*k2)*h;  
    end
    
    % ode45
    [to,yo]=ode45(@(t,y)f(t,y),t,y(1));
    plot(to,yo,'linewidth',2,'DisplayName','Ralston Method')
    plot(t,y,'-.','linewidth',2,'DisplayName','ode45')
    title(m);
    grid on
    legend show
end

h = 0.30 3.7 Ralston Method - ode45 3.6 3. 5 3.4 у 3. ديا 3.2 3.1 3 0.2 0.4 0.6 0.8 1 1.2 t

h = 0.10 3.7 Ralston Method -ode45 3.6 3 on 3.4 у 3.3 3.2 3.1 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 t

h = 0.05 3.7 Ralston Method -- ode45 3.6 3 on 3.4 у 3.3 3.2 3.1 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 t

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