# Runge Kutta for system of eqs

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jojo on 24 Nov 2019
Edited: jojo on 26 Nov 2019
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.
jojo on 25 Nov 2019
Just having difficulty breaking it into a system of equations and incorporate it into the code

darova on 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