Question

I DESPERATELY NEED HELP WITH THIS DIFFERENTIAL EQUATIONS MATLAB ASSIGNMENT IM SUPPOSED TO BE LEARNING BUT WE HAVE A SUB AND HE DIDN'T TEACH IT! ITS EULER AND IMPROVED EULER IN MATLAB! HERE IS THE LINK FOR THE IMAGE FILE THAT SHOWS THE FULL INSTRUCTIONS FOR THE CODE.

https://imgur.com/a/gjmypLs

Also, here is my code so far that I borrowed form an old assignment but the data is all wrong and the application of the code is slightly different so either create new code or modify mine so that it matches up with this assignment! I WILL RATE ASAP! Thanks!

%% Exercise 1

clc
clear all

t= linspace(0, 0.5, 100);
y= 3*exp(2*t);
%Inline function
f= inline('2*y', 't', 'y');
[t5, y5]= euler(f, [0, 0.5], 3, 5);
y5(end);
[t50, y50]= euler(f, [0, 0.5], 3, 50);
y50(end);
[t500, y500]= euler(f, [0, 0.5], 3, 500);
y500(end);
[t5000, y5000]= euler(f, [0, 0.5], 3, 5000);
y5000(end);
% Error of Euler's Method = Exact - Approximation
e5 = y(end) - y5(end); % error at N=5
e50 = y(end) - y50(end);
e500 = y(end) - y500(end);
e5000 = y(end) - y5000(end);
%Ratio of errors
ratio1= e5/e50; % error between 5 and 50
ratio2= e50/e500; % 5 and 500
ratio3= e500/e5000; % 5 and 5000
%Compile the data
N= [5;50;500;5000];
Approximation= [y5(end);y50(end);y500(end);y5000(end)];
Error= [e5;e50;e500;e5000];
Ratio= [0;ratio1;ratio2;ratio3];
T=table(N,Approximation,Error,Ratio)
%% Exercise 2
figure(1)
t = 0:.45:10; y = -30:6:42 ; % define grid of values in t and y direction
[T,Y]=meshgrid(t,y); % creates 2d matrices of points in the ty-plane
dT = ones(size(T)); % dt=1 for all points
dY = -2*Y; % dy = -2*y; this is the ODE
quiver(T,Y,dT,dY) % draw arrows (t,y)->(t+dt, t+dy)
axis tight % adjust look
hold on
hold off

figure(2)
t = 0:.45:10; y = -30:6:42 ; % define grid of values in t and y direction
[T,Y]=meshgrid(t,y); % creates 2d matrices of points in the ty-plane
dT = ones(size(T)); % dt=1 for all points
dY = -2*Y; % dy = -2*y; this is the ODE
quiver(T,Y,dT,dY) % draw arrows (t,y)->(t+dt, t+dy)
axis tight % adjust look
hold on
% Exact solution
t = linspace(0,10,200);
y = 3*exp(-2*t);
plot(t,y,'k-', 'linewidth',2)
hold off

%% Part c

figure(3)
t = 0:.45:10; y = -30:6:42 ; % define grid of values in t and y direction
[T,Y]=meshgrid(t,y); % creates 2d matrices of points in the ty-plane
dT = ones(size(T)); % dt=1 for all points
dY = -2*Y; % dy = -2*y; this is the ODE
quiver(T,Y,dT,dY) % draw arrows (t,y)->(t+dt, t+dy)
axis tight % adjust look
hold on
% Exact solution
t = linspace(0,10,200);
y = 3*exp(-2*t);
plot(t,y,'k-', 'linewidth',2)
% Euler's Approximation for N=8
f= inline('-2*y', 't', 'y'); %inline function
[t8, y8]= euler(f, [0, 10], 3, 8); % N=8
plot(t8,y8,'ro-', 'linewidth',2)
hold off
axis([0 10 -30 40])

%% Part d

figure(4)
t = 0:.4:10; y = -1:0.4:3; % define grid of values in t and y direction
[T,Y]=meshgrid(t,y); % creates 2d matrices of points in the ty-plane
dT = ones(size(T)); % dt=1 for all points
dY = -2*Y; % dy = -2*y; this is the ODE
quiver(T,Y,dT,dY) % draw arrows (t,y)->(t+dt, t+dy)
axis tight % adjust look
hold on
% Exact solution
t = linspace(0,10,200);
y = 3*exp(-2*t);
plot(t,y,'k-','linewidth',2)
% Euler's Approximation for N=16
f= inline('-2*y', 't', 'y'); %inline function
[t16, y16]= euler(f, [0, 10], 3, 16); % N=16
plot(t16,y16,'ro-','linewidth',2)
hold off
axis([0 10 -1 3])   
% The step size of the function is larger than expected due to the
% oscillation of the beginning values, but geometrically then follows the
% field direction vectors.
%% Excerise 3

% Display contents of impeuler M-file.
type 'impeuler.m'
% Define ODE function for dy/dt = f(t,y) = 2y
f= inline('2*y', 't', 'y');
% Compute numerical solution to IVP with 5 timesteps using "improved Euler."
[t5, y5] = impeuler(f, [0 0.5],3,5)
function [t,y] = impeuler(f,tspan,y0,N)
% Solves the IVP y' = f(t,y), y(t0) = y0 in the time interval tspan = [t0,tf]
% using Euler's method with N time steps.
% Input:
% f = name of inline function or function M-file that evaluates the ODE
% (if not an inline function, use: euler(@f,tspan,y0,N))
% For a system, the f must be given as column vector.
% tspan = [t0, tf] where t0 = initial time value and tf = final time value
% y0 = initial value of the dependent variable. If solving a system,
% initial conditions must be given as a vector.
% N = number of steps used.
% Output:
% t = vector of time values where the solution was computed
% y = vector of computed solution values.
m = length(y0);
t0 = tspan(1);
tf = tspan(2);
h = (tf-t0)/N; % evaluate the time step size
t = linspace(t0,tf,N+1); % create the vector of t values
y = zeros(m,N+1); % allocate memory for the output y
y(:,1) = y0'; % set initial condition
for n=1:N
f1= f(t(n), y(:,n));
f2= f(t(n+1), y(:,n)+h*f1);
y(:,n+1)= y(:,n) +h*(f1+f2)/2;
end
t = t'; y = y'; % change t and y from row to column vectors
end

% The output of the values versus the given match the data correctly.
%% Exercise 4

%Euler's method
t= linspace(0, 0.5, 100);
y= 3*exp(2*t);
f= inline('2*y', 't', 'y'); %inline function
[t5, y5]= impeuler(f, [0, 0.5], 3, 5); % N=5
y5(end);
[t50, y50]= impeuler(f, [0, 0.5], 3, 50); % N=50
y50(end);
[t500, y500]= impeuler(f, [0, 0.5], 3, 500); % N=500
y500(end);
[t5000, y5000]= impeuler(f, [0, 0.5], 3, 5000); % N=5000
y5000(end);
% Error of Euler’s method
e5 = y(end) - y5(end); % error at N=5
e50 = y(end) - y50(end); % error at N=50
e500 = y(end) - y500(end); % error at N=500
e5000 = y(end) - y5000(end); % error at N=5000
% Ratio Error
ratio1= e5/e50; % ratio of error between 5 and 50
ratio2= e50/e500; % 5 and 50
ratio3= e500/e5000; % 5 and 50
% Compile the data into a table
N= [5;50;500;5000];
Approximation= [y5(end);y50(end);y500(end);y5000(end)];
Error= [e5;e50;e500;e5000];
Ratio= [0;ratio1;ratio2;ratio3];
T=table(N,Approximation,Error,Ratio)

% The step size of this example is 10, and the ratio of error is 100.
% This follows the k^2 decrease factor.
%% Exercise 5

% Part a same as in Exercise 2 (see Exercise 2a)
% Pand b same as in Exercise 2 (see Exercise 2b)
%% Part C
figure(5)
t = 0:.45:10; y = -30:6:42 ; % define grid of values in t and y direction
[T,Y]=meshgrid(t,y); % creates 2d matrices of points in the ty-plane
dT = ones(size(T)); % dt=1 for all points
dY = -2*Y; % dy = -2*y; this is the ODE
quiver(T,Y,dT,dY) % draw arrows (t,y)->(t+dt, t+dy)
axis tight % adjust look
hold on
% Exact solution for Euler
t = linspace(0,10,200);
y = 3*exp(-2*t);
plot(t,y,'k-', 'linewidth',2)
% Improved Euler's Approximation for N=8
f= inline('-2*y', 't', 'y'); %inline function
[t8, y8]= impeuler(f, [0, 10], 3, 8); % N=8
plot(t8,y8,'ro-', 'linewidth',2)
hold off
axis([0 10 -30 40])
%% Part D

figure(6)
t = 0:.4:10; y = -1:0.4:3; % define grid of values in t and y direction
[T,Y]=meshgrid(t,y); % creates 2d matrices of points in the ty-plane
dT = ones(size(T)); % dt=1 for all points
dY = -2*Y; % dy = -2*y; this is the ODE
quiver(T,Y,dT,dY) % draw arrows (t,y)->(t+dt, t+dy)
axis tight % adjust look
hold on
% Exact solution
t = linspace(0,10,200);
y = 3*exp(-2*t);
plot(t,y,'k-','linewidth',2)
% Improved Euler's Approximation for N=16
f= inline('-2*y', 't', 'y'); %inline function
[t16, y16]= impeuler(f, [0, 10], 3, 16); % N=16
plot(t16,y16,'ro-','linewidth',2)
hold off
axis([0 10 -1 3])

% In comparison to the N = 8 Euler approximation,
% the N = 16 approximation more closely follows the function resulting
% in larger accuracy using Euler’s method.

Answer 1

clc
clear all

t= linspace(0, 0.5, 100);
y= 3*exp(2*t);
%Inline function
f= inline('2*y', 't', 'y');
[t5, y5]= euler(f, [0, 0.5], 3, 5);
y5(end);
[t50, y50]= euler(f, [0, 0.5], 3, 50);
y50(end);
[t500, y500]= euler(f, [0, 0.5], 3, 500);
y500(end);
[t5000, y5000]= euler(f, [0, 0.5], 3, 5000);
y5000(end);
% Error of Euler's Method = Exact - Approximation
e5 = y(end) - y5(end); % error at N=5
e50 = y(end) - y50(end);
e500 = y(end) - y500(end);
e5000 = y(end) - y5000(end);
%Ratio of errors
ratio1= e5/e50; % error between 5 and 50
ratio2= e50/e500; % 5 and 500
ratio3= e500/e5000; % 5 and 5000
%Compile the data
N= [5;50;500;5000];
Approximation= [y5(end);y50(end);y500(end);y5000(end)];
Error= [e5;e50;e500;e5000];
Ratio= [0;ratio1;ratio2;ratio3];
T=table(N,Approximation,Error,Ratio)
%% Exercise 2
figure(1)
t = 0:.45:10; y = -30:6:42 ; % define grid of values in t and y direction
[T,Y]=meshgrid(t,y); % creates 2d matrices of points in the ty-plane
dT = ones(size(T)); % dt=1 for all points
dY = -2*Y; % dy = -2*y; this is the ODE
quiver(T,Y,dT,dY) % draw arrows (t,y)->(t+dt, t+dy)
axis tight % adjust look
hold on
hold off

figure(2)
t = 0:.45:10; y = -30:6:42 ; % define grid of values in t and y direction
[T,Y]=meshgrid(t,y); % creates 2d matrices of points in the ty-plane
dT = ones(size(T)); % dt=1 for all points
dY = -2*Y; % dy = -2*y; this is the ODE
quiver(T,Y,dT,dY) % draw arrows (t,y)->(t+dt, t+dy)
axis tight % adjust look
hold on
% Exact solution
t = linspace(0,10,200);
y = 3*exp(-2*t);
plot(t,y,'k-', 'linewidth',2)
hold off