Question

Please solve all parts

Consider the following IVP

y′ = 2t − y + 1, y(0) = 1.

Find an approximation of y(1) using

(a) the Euler’s method with N = 4,

(b) the Euler’s method with N = 8,

(c) the improved Euler’s method with N = 4,

(d) the improved Euler’s method with N = 8.

Solve the IVP, find y(1), and compare the accuracy |y(1)−yN | of the approximations.

Answer 1

Approximating y(1) using two methods with 4 different n values will take much time by hand. So I am providing MATLAB CODE for the same with all output.

part a

%%%%%%%%%%%% MATLAB CODE %%%%%%%%%%

% Euler Method
clear all
clc;
format short
f=@(t,y)(2*t-y+1); % dy/dt=f(t,y)=y-t^2
n=4;
a=0;
b=1;
h=(b-a)/n;
t = a:h:b;
y = zeros(1,length(t));
y(1)=1; % index has been taken as i instead of 0

for n=1:(length(t)-1)
y(n+1) = y(n)+h*f(t(n),y(n));
end
t_y=[t' y'] % Solution of ODE at t=0 to 1
yN=y(n+1) % Solution by Euler method
y1=1.7358 % Analytical solution
Accuracy=abs(y1-yN)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

OUTPUT

t_y =

0 1.0000
0.2500 1.0000
0.5000 1.1250
0.7500 1.3438
1.0000 1.6328

yN = 1.6328

y1 = 1.7358
Accuracy = 0.1030

Part (b)

%%%%%%%%%%%%%%% MATLAB CODE %%%%%%%%%%%%%%%%%

% Euler Method
clear all
clc;
format short
f=@(t,y)(2*t-y+1); % dy/dt=f(t,y)=y-t^2
n=8;
a=0;
b=1;
h=(b-a)/n;
t = a:h:b;
y = zeros(1,length(t));
y(1)=1; % index has been taken as i instead of 0

for n=1:(length(t)-1)
y(n+1) = y(n)+h*f(t(n),y(n));
end
t_y=[t' y'] % Solution of ODE at t=0 to 1
yN=y(n+1) % Solution by Euler method
y1=1.7358 % Analytical solution
Accuracy=abs(y1-yN)

%%%%%%%%%%%%%%%%%%%%%%%%%

OUTPUT

t_y =

0 1.0000
0.1250 1.0000
0.2500 1.0313
0.3750 1.0898
0.5000 1.1724
0.6250 1.2758
0.7500 1.3976
0.8750 1.5354
1.0000 1.6872

yN = 1.6872
y1 = 1.7358
Accuracy = 0.0486

Part (c)

%%%%%%%%%%%%%%%%%%%%%%%% MATLAB CODE %%%%%%%%%

% Modified (Improved) Euler method
clear all
clc;
format short
f=@(t,y)(2*t-y+1); % dy/dt=f(t,y)=y-t^2
n=4;
a=0;
b=1;
h=(b-a)/n;
t = a:h:b;
y = zeros(1,length(t));
y(1)=1; % index has been taken as i instead of 0

for n=1:(length(t)-1)
y(n+1) = y(n)+h*f(t(n)+0.5*h,y(n)+0.5*h*f(t(n),y(n)));
end
t_y=[t' y'] % Solution of ODE at t=0 to 1
yN=y(n+1) % Solution by Modified Euler method
y1=1.7358 % Analytical solution
Accuracy=abs(y1-yN)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

OUTPUT

t_y =

0 1.0000
0.2500 1.0625
0.5000 1.2207
0.7500 1.4537
1.0000 1.7451

yN = 1.7451
y1 = 1.7358
Accuracy = 0.0093

Part(d)

%%%%%%%%%%%%%%% MATLAB CODE

% Improved Euler Method
clear all
clc;
format short
f=@(t,y)(2*t-y+1); % dy/dt=f(t,y)=y-t^2
n=8;
a=0;
b=1;
h=(b-a)/n;
t = a:h:b;
y = zeros(1,length(t));
y(1)=1; % index has been taken as i instead of 0

for n=1:(length(t)-1)
y(n+1) = y(n)+h*f(t(n)+0.5*h,y(n)+0.5*h*f(t(n),y(n)));
end
t_y=[t' y'] % Solution of ODE at t=0 to 1
yN=y(n+1) % Solution by Euler method
y1=1.7358 % Analytical solution
Accuracy=abs(y1-yN)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

OUTPUT

t_y =

0 1.0000
0.1250 1.0156
0.2500 1.0587
0.3750 1.1261
0.5000 1.2148
0.6250 1.3224
0.7500 1.4468
0.8750 1.5858
1.0000 1.7379

yN = 1.7379
y1 = 1.7358
Accuracy = 0.0021