Question

Using MATLAB_R2017a, solve #3 using the differential equation in question #2 using Simulink, present the model and result.

2. Differential Equation (5 points) Using (i) Euler's method and (ii) modified Euler's method, both with step size h-0.01, to construct an approximate solution from F0 to F2 for xt 2, 42 with initial condition x(0)=1. Compare both results by plotting them in the same figure. 3. Simulink (5 points) Solve the above differential equation using simplink. Present the model and result.

Answer 1

Text :-

dxdt = @(t,x) sqrt(x*t/(x.^2+t.^2));
h = 0.01;
t = 0:h:2;      % time vector (t0,t1,...tn)
x_euler = zeros(1,length(t));      % Initializing Euler solution vector having same size as time vector
x_mod_euler = zeros(1,length(t));  % Initializing modified Euler solution

x_euler(1) = 1;             % Initial Conditions
x_mod_euler(1)=1;

for k = 2:length(t)
x_euler(k) = x_euler(k-1) + h*dxdt(t(k-1),x_euler(k-1));
x_mod_euler(k) = x_mod_euler(k-1) + (h/2)*(dxdt(t(k-1),x_mod_euler(k-1)) + dxdt(t(k),x_euler(k)));
end

% Comparing Results in table (Intermediate Values)
fprintf("\n %8s %16s %20s\n","t","Euler","Modified Euler")
fprintf(" %8.2f %16.6f %20.6f\n",[t(1:20:end)',x_euler(1:20:end)',x_mod_euler(1:20:end)'].')
% Ploting Both on Same Graph
p = plot(t,x_euler,'-.',t,x_mod_euler,'--');
p(1).LineWidth = 1.6;       p(2).LineWidth = 1.6;
legend(["Euler Method","Modified Euler Method"],'Location','Best')
xlabel('t');    ylabel("x(t)");      title("Euler Method vs Modifed Euler Method")

Simulink :-

1 Scope Integrator Clock u(2) sqrt(u(1) u(2)/(u(1)2+ u(2)A2)) Fcn Bus Creator

Euler Method vs Modifed Euler Method 2.4 r 2.2 1.8 1.6 1.4 Euler Method Modified Euler Method 1.2 0 0.2 0.40.6 0.8 1 12 1.416 1.8 2