code of euler's method

Hi, i follow every protocol steps for euler's method, but my results are too increased and they are not correct. Anyone could see if i´m doing anything wrong? i think it happens because my derivatives are floating too much.

[Sara]

What's the expected result? What are the functions you're trying to solve?

Answer 1

[George Papazafeiropoulos]

A simple application of Euler method:

Define the function:

function E=euler(f,a,b,ya,M)
h=(b-a)/M;
Y=zeros(1,M+1);
T=a:h:b;
Y(1)=ya;
for j=1:M
    Y(j+1)=Y(j)+h*f(T(j));
end
E=[T' Y'];
end

where - f is the function entered as function handle

- a and b are the left and right endpoints

- ya is the initial condition E(a)

- M is the number of steps

- E=[T' Y'] where T is the vector of abscissas and Y is the vector of ordinates

Then run the code:

f=@(x) x^2;
a=0;
b=10;
ya=0;
M=200;
YY=euler(f,a,b,ya,M)

You can adjust your problem according to the above algorithm.

[Rachel Lee]

How would you find the error between Euler's Method and the Exact Soln using truncation? I think we need the derivative but nothing I do seems to work.

[Rachel Lee]

%------------------------------------Functions

function [E] = odeEuler(f,a,b,ya,M)

%M is the no of steps taken

h=(b-a)/M;

Y=zeros(1,M+1);

T=a:h:b;

Y(1)=ya; %this value is 4 for this problem

for j=1:M

Y(j+1)=Y(j)+h*f(T(j));

end

E=[T' Y'];

end

%------------------------------------Executable

%goal print out three iterations of this soln

y0 = 4; %initial y value

t = [0 2 4]'; %this is our specific system of t

size = length(t);

fn = @(t)(4/1.3)*(exp(0.8*t) - exp(-0.5*t))+2*exp(-0.5*t);

dfn = @(t) 4*exp(0.8*t) - 0.5 * fn;

h = 2; err = 0; %initial conditions

a = t(1,:);%0

b = t(size,:);%4

[Soln] = odeEuler(fn,a,b,y0,h);

A = t;

B = Soln(:,2);

C = fn(t);

%producing the graph

plot(t,B,t,C);

title('Comparing Linearization Methods')

legend('Eulers Method','Exact Soln: 4/1.3)*(exp(0.8*t) - exp(-0.5*t))+2*exp(-0.5*t)')

%producing a Table with M iterations

Data = [A B C];

VarNames = {'time domain','Eulers Method','Exact Soln'};

T = table(Data(:,1),Data(:,2),Data(:,3),'VariableNames',VarNames)

Answer 2

[SkyRazor]

hello, could you please post your equation and give us some explanations?

Answer 3

[ahmed abdelmageed]

function E=euler(f,a,b,ya,M)

h=(b-a)/M;

Y=zeros(1,M+1);

T=a:h:b;

Y(1)=ya;

for j=1:M

Y(j+1)=Y(j)+h*f(T(j));

end

E=[T' Y'];

end

Answer 4

[Sandip Das]

%Published in 19th july 2021

%Sandip Das

clc

clear all

dydt=input('\n Enter the function : ');

x0=input('\n Enter initial value of x : ');

y0=input('\n Enter initial value of y : ');

xn=input('\n Enter the final value of x: ');

h=input('\n Enter the step length h: ');

i=0;

while i<xn

tempy=y0+h*dydt(x0,y0);

tempx=x0+h;

x0=tempx;

y0=tempy;

i=i+h;

end

fprintf('The value of y at t=%f is %f \n',x0,y0);