Runge Kutta for system of eqs

Question

jojo 2019-11-24

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/492904-runge-kutta-for-system-of-eqs

编辑： jojo 2019-11-26

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.

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jojo 2019-11-25

Yes. ?′′(?)+4?′(?)+5?(?)=0 this is the full equation.

jojo 2019-11-25

编辑：jojo 2019-11-25

Just having difficulty breaking it into a system of equations and incorporate it into the code

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Answer 1

darova 2019-11-25

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/492904-runge-kutta-for-system-of-eqs#answer_403316

在 MATLAB Online 中打开

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