Question

using matlab thank you

3 MARKS QUESTION 3 Background The van der Pol equation is a 2nd-order ODE that describes self-sustaining oscillations in which energy is withdrawn from large oscillations and fed into the small oscillations. This equation typically models electronic circuits containing vacuum tubes. The van der Pol equation is: dt2 dt where y represents the position coordinate, t is time, and u is a damping coefficient The 2nd-order ODE can be solved as a set of 1st-order ODEs, as shown below. Here, z is a 'dummy' variable The two 1st-order ODEs above are coupled and must be solved simultaneously. i.e. the solutions are dependent on each other and therefore cannot be solved individually. The first iteration will solve for y1 and z1 using initial conditions yo and zo. The second iteration will solve for y2 and z2 using y1 and z1, and so forth You are strongly encouraged to complete a few iterations by hand to fully understand the process. Q3a In the euler2.m file, complete the function file to perform Euler's method to solve two 1st-order ODE equations simultaneously You should still have three figure windows by the end of this task. Q3b In the Q3b.m file, consider initial conditions y (0) 1 and z(0) 1, and 1. Solve the van der Pol equation using the euler2() function written in Q2a with time steps of 0.25, 0.125 and 0.0625 fromt 0 to 30s In figure(4)1, plot the following in 2-by-1 subplot arrangement using the 'colours' variable provided in the m file to represent time steps of 0.25, 0.125 and 0.0625, respectively Top panel] y against t solved using Euler's method for each time step dy [Bottom panell x against t solved using Euler's method for each time step It is suspected that the solutions obtained are not highly accurate. Use fprintf() to print a statement describing how the accuracy of the solutions can be improved

Answer 1

function dYdT = f(T,Y)
mew =1;

y = Y(1); z = Y(2);
dydt = z;
dzdt = mew*(1-y^2)*z-y;

dYdT = [dydt dzdt];
end

------------------------------------------------------------------------

clear
close all
clc

a = 0; b = 30; % time interval [0 30]
H = [0.25 0.125 0.0625]; % values of h

for j=1:numel(H)
y = [1 1]; %initial conditions [y(0) z(0)] or [y(0) y'(0)]
h = H(j);
t = a:h:b;
n = (b-a)/h;

% Euler's method
for i=1:n
y(i+1,:) = y(i,:) + h.*f(t(i),y(i,:));
end

% y(t) is stored in 1st column, dydt is stored in 2nd column of y

% plotting y(t)
figure
plot(t,y(:,1))
xlabel('t')
ylabel('y(t)')
title(['h = ' num2str(h)])
grid on

% plotting subplots showing y(t) and corresponding dydt
figure
subplot(2,1,1)
plot(t,y(:,1)) %y(t)
title(['h = ' num2str(h)])
ylabel('y(t)')
xlabel('t')
subplot(2,1,2)
plot(t,y(:,2)) % dydt
ylabel('dydt')
xlabel('t')
end

fprintf('The accuracy can be improved by using a smaller step size h \n')

h 0.25 3 -1 -2 -3 20 10 15 25 30

h = 0.125 2.5r 1.5 0.5 0.5 -1 1.5 -2 -2.5 10 15 20 25 30

h 0.0625 2.5 1.5 0.5 0.5 -1 1.5 -2 -2.5 10 15 20 25 30

h 0.25 -2 10 15 20 25 30 -2 10 20 15 25 30

h = 0.125 -2 30 25 20 15 10 -2 30 25 20 15 10

h 0.0625 -2 10 15 20 25 30 -2 10 20 15 25 30

The accuracy can be improved by using a smaller step size h

COMMENT DOWN FOR ANY QUERY RELATED TO THIS ANSWER,

IF YOU'RE SATISFIED, GIVE A THUMBS UP
~yc~