Runge Kutta for system of eqs
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
I would like to approxiamte and plot the following system of equations: { u' = v ; v' = -4*u' - 5 u} using the Runge Kutta method.
I am able to plot it for the second equation, but having difficulties incorporating the u' = v to reflect the full system.
the exact solution I am comparing to is: x(t) = 3.*exp(-2.*t) .* cos(t) + exp(-2.*t) .* sin(t) corresponding to equation: ??′′(??)+4??′(??)+5??(??)=0 with ??(0)=3 and ??′(0)=−5. as initial conditions.
clear all
close all
clc
h = .1; % set the step size
x = 0:h:5; % set the interval of x
y = zeros(1,length(x));
y(1) = 3; % set the intial value for y
n = length(x)-1;
y_dot =@(x,y)(-4*x-5*y); %insert function to be solved
for i = 1:n
k1 = y_dot(x(i),y(i));
k2 = y_dot(x(i)+.5*h,y(i)+.5*k1*h);
k3 = y_dot(x(i)+.5*h,y(i)+.5*k2*h);
k4 = y_dot(x(i)+h,y(i)+k3*h);
y(i+1) = y(i)+((k1+2*k2+2*k3+k4)/6)*h;
end
Please explain the any updates yu make, since I am new to ML and need to conceptually understand the fundamentals.
Thanks.
3 Kommentare
Antworten (1)
darova
am 25 Nov. 2019
I wrote function in the way similar to ode45
y_dot = @(x,dx)[dx; -4*dx-5*x]; %insert function to be solved
for i = 1:n
k1 = y_dot(x(i),dx(i));
k2 = y_dot(x(i)+h/2*k1(1), dx(i)+h/2*k1(2));
k3 = y_dot(x(i)+h/2*k2(1), dx(i)+h/2*k2(2));
k4 = y_dot(x(i)+k3(1), dx(i)+k3(2));
u = (k1+2*k2+2*k3+k4)*h/6;
% shorter form
% k1 = h/2*y_dot(x(i),dx(i));
% k2 = h/2*y_dot(x(i)+k1(1), dx(i)+k1(2));
% k3 = h* y_dot(x(i)+k2(1), dx(i)+k2(2));
% k4 = h* y_dot(x(i)+k3(1), dx(i)+k3(2));
% u = (2*k1+4*k2+2*k3+k4)/6;
x(i+1) = x(i) + u(1);
dx(i+1) = dx(i) + u(2);
end
0 Kommentare
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!