Euler's method for two first order differential equations?

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Yuhong Jin
Yuhong Jin 2019-11-27
回答： Eng 2023-4-23
I was trying to solve two first order differential equations like below using the Euler's method and plot two graphs with x and y as a function of t.
The differential equations are:
dxdt = @(x,t) -1.*y-0.1.*x;
dydt = @(y,t) x-0.5.*y;
I tried this script below:
a=0; %initial time
b=2000; %final time
h = 0.01; % time step
N = (b-a)./h;
t=a:h:b;
t(1)=a;
x(1)=0;
y(1)=0;
dxdt = @(x,t) -1.*y-0.1.*x;
dydt = @(y,t) x-0.5.*y;
for n=1:N
x(n+1)=x(n)+h.*dxdt(x(n),t(n));
y(n+1)=y(n)+h.*dydt(y(n),t(n));
t(n+1)=a+n*h;
end
plot(t,x,'-',t,y,'--')
But there was an error saying
Unable to perform assignment because the left and right sides have a different
number of elements.
Error in Q6 (line 15)
x(n+1)=x(n)+h.*dxdt(x(n),t(n));
I did not quite understand why.
Could anyone help me with this please?
Thanks in advance.

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采纳的回答

Jim Riggs
Jim Riggs 2019-11-27
编辑：Jim Riggs 2019-11-27
Try preallocating x and y
nsteps = (b-a)/h + 1; % this is the number of elements in t(a:h:b) (It is = N+1)
x=zeros(1,nsteps);
y=zeros(1,nsteps);
Also, when you define t as
t=a:h:b;
This creates a vector with T(1)=a and t(end)=b, so the following line
t(1)=a
is not necessary.
Inside your for loop, t has already been defined, above, as t=a:h:b, so you don't need to re-define it.
x and y need to be preallocated to size "nsteps", which is 1 greater than N, in order for your loop to work.
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显示 8更早的评论
Chris Horne
Chris Horne 2022-3-29
I figured it out. The function definition delimiters typo. The code produces then 2D plot of the approximating functions.
Chris Horne
Chris Horne 2022-3-31
Is the term 'forward Euler' the same as 'Euler' in terms of the algorithm? Here is my method for solving 3 equaitons as a vector:
% This code solves u'(t) = F(t,u(t)) where u(t)= t, cos(t), sin(t)
% using the FORWARD EULER METHOD
% Initial conditions and setup
neqn = 3; % set a number of equations variable
h=input('Enter the step size: ') % step size will effect solution size
t=(0:h:4).';%(starting time value 0):h step size
nt = size(t,1); % size of time array
%(the ending value of t ); % the range of t
% define the function vector, F
F = @(t,u)[t,cos(t),sin(t)]; % define the function 'handle', F
% with hard coded vector functions of time
u = zeros(nt,neqn); % initialize the u vector with zeros
v=input('Enter the intial vector values of 3 components using brackets [u1(0),u2(0),u3(0)]: ')
u(1,:)= v; % the initial u value and the first column
%n=numel(u); % the number of u values
% The loop to solve the ODE (Forward Euler Algorithm)
for i = 1:nt-1
u(i+1,:) = u(i,:) + h*F(t(i),u(i,:)); % Euler's formula for a vector function F
end

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更多回答（1 个）

Eng
Eng 2023-4-23
% Define the differential equation dydx = @(x,y) x + y;
% Define the initial condition y0 = 1;
% Define the step size and interval h = 0.1; x = 0:h:1;
% Initialize the solution vector y = zeros(size(x)); y(1) = y0;
% Apply the Modified Euler Method for i = 1:length(x)-1 k1 = dydx(x(i), y(i)); k2 = dydx(x(i+1), y(i)+hk1); y(i+1) = y(i) + h/2(k1+k2); end
% Plot the solution plot(x,y) xlabel('x') ylabel('y')

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