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3. how do you plot these four codes on one set...

# Question: how do you plot these four codes on one set...

###### Question details

How do you plot these four codes on one set of axes? You might not need the codes to tell me how but I listed them anyway.

*****Codes:

Euler's Method:

function Euler()

% We will use Euler's method to investigate the first order DE dy/dt=cos(t-y) with initial condition y(0)=1

clear t % Clears all t inputs
clear y % Clears all y inputs

ti=0; % This is the left endpoint of our interval, t initial

h=1; % This is the step size, distance between t values

tf=100; % This is the right endpoint of our interval, t final

n=(tf-ti)/h; % This is the total amount of steps

t(1)=ti; % Declares the starting t value for the loop

y(1)=1; % This is the value of the initial condition at the first iteration of the loop

for j=1:n
t(j+1)=h+t(j);
y(j+1)=y(j)+h*cos(t(j)-y(j));
end

figure(1)

plot(t,y)

xlabel('t')
ylabel('y(t)')

title('Euler')

hold on % Allows plotting multiple solutions on the same set of axes
grid on % Outputs xy-axes for plot

Improved Euler:

function ImprovedEuler()

% We will use improved Euler's method, the one that corresponds to the midpoint method,
% to investigate the first order DE, dy/dt=cos(t-y) with initial condition y(0)=1

clear t % Clears all t inputs
clear y % Clears all y inputs

ti=0; % This is the left endpoint of our interval, t initial

h=1; % This is the step size, distance between t values

tf=100; % This is the right endpoint of our interval, t final

n=(tf-ti)/h; % This is the total amount of steps

t(1)=ti; % Declares the starting t value for the loop

y(1)=1; % This is the value of the initial condition at the first iteration of the loop

for j=1:n
t(j+1)=h+t(j);
f(j)=cos(t(j)-y(j));
y(j+1)=y(j) + h*f(j)/2 + h*cos(t(j)+h-y(j)-(h*f(j)))/2;
end

figure(2)

plot(t,y)

xlabel('t')
ylabel('y(t)')

title('Improved Euler')

hold on
grid on

Runge Kutta:

function RungeKutta()

% We will use Runge-Kutta, the method that corresponds to the trapezoidal weighted average,
% to investigate the first order DE dy/dt=cos(t-y) with initial condition y(0)=1

clear t % Clears all t inputs
clear y % Clears all y inputs

ti=0; % This is the left endpoint of our interval, t initial

h=1; % This is the step size, distance between t values

tf=100; % This is the right endpoint of our interval, t final

n=(tf-ti)/h; % This is the total amount of steps

t(1)=ti; % Declares the starting t value for the loop

y(1)=1; % This is the value of the initial condition at the first iteration of the loop

for j=1:n
t(j+1)=h+t(j);
f(j)=cos(t(j)-y(j));

k1(j)=f(j);
k2(j)=cos(t(j)+h/2-y(j)-(h*k1(j)/2));
k3(j)=cos(t(j)+h/2-y(j)-(h*k2(j)/2));
k4(j)=cos(t(j)+h-y(j)-(h*k3(j)));

y(j+1)=y(j)+(h*k1(j)/6)+(h*k2(j)/3)+(h*k3(j)/3)+(h*k4(j)/6);
end

figure(3)

plot(t,y)

xlabel('t')
ylabel('y(t)')

title('Runge-Kutta')

hold on
grid on

Plot Code:

% For loop to plot temperature vs time
for A=115:4:125
T=0:100
B=(98.6-A)*exp(-0.054*T)+A;
plot(T,B);
hold on
end

title('Body Temperature vs Time');
xlabel('Time (min)');
ylabel('Body Temperature (Fahenheit)');
grid on;