How do I use Eulers Explicit method to solve a system of first-order ODEs

Hello, I have created a system of first order ODEs from the higher order initial value problem, but now I cannot figure out how to use Matlab to find the solution using Eulers explicit method. I have already used Eulers (implicit I think?) and third order runge Kutta as you can see below but I am lost on how to incorporte the 4 initial values and 4 different first order ODEs. Thank you for your help!!! The picture of the problem statement will be on top followed by the system of first-order ODEs I derived, then my Eulers implicit and runge kutta code and finally the code I am trying to work on right now.

if true
%EULER IMPLICIT METHOD
h = 0.1;
x0 = 0;
u0 = 1;
x_end = 2;
x(1)=x0;
u(1)=u0;
%number of steps
n = (x_end-x0)/h;
for k=1:n
  uprime(k+1) = -x(k)*u(k) + exp(-(x(k)^2)/2);
  x(k+1) = x(k) + h;
  u(k+1) = u(k) + uprime(k+1)*h;
end
%transpose x and u into column vectors
xvalues = x';
uvalues = u';
%%THIRD ORDER RANGE KUTTA METHOD
h = 0.1;
x0 = 0;
u0 = 1;
x_end = 2;
x1(1)=x0;
u1(1)=u0;
x1= 0:h:2;
y1= zeros(1,length(x1));
F = @(x,u) -x.*u + exp(-(x.^2)/2);
for i=1:(length(x1)-1)
  k1 = F(x1(i),u1(i));
  k2 = F(x1(i)+0.5*h,u1(i)+k1);
  k3 = F((x1(i)+0.5*h),(u1(i)-k1+2*k2));
  u1(i+1) = u1(i) + (1/6)*(k1+4*k2+k3)*h;
end
u1values = u1';
%%PLOT ALL THREE GRAPHS
exact = @(x) exp(-(x.^2)/2).*(x+1);
plot(x,u, '--b');
xlim([0 2])
hold on
fplot(exact,[x0,x_end], 'r');
hold on
plot(x1,u1, '--g');
legend('euler','exact','runge-kutta')
step = xvalues;
implicit = uvalues;
RungeKutta = u1values;
exact = exact(xvalues);
table(step,implicit,RungeKutta,exact)
end

First try:

if true
%%EULER EXPLICIT METHOD
h = 0.1;
x_end = 2;
x0 = 0;
x0_1 = 0;
x0_2 = 0;
x0_3 = 0;
u0 = 1.1;
u0_1 = -4.7;
u0_2 = 7.9;
u0_3 = -14.3;
u1(1)=u0;
u2(1)=u0_1;
u3(1)=u0_2;
u4(1)=u0_3;
x1(1)=x0;
x2(1)=x0_1;
x3(1)=x0_2;
x4(1)=x0_3;
%number of steps
n = (x_end-x0)/h;
for k=1:n
  up1(k+1) = -x1(k)*u1(k) + exp(-(x1(k)^2)/2);
  up2(k+1) = -x2(k)*u2(k) + exp(-(x2(k)^2)/2);
  up3(k+1) = -x3(k)*u3(k) + exp(-(x3(k)^2)/2);
  up4(k+1) = -x4(k)*u4(k) + exp(-(x4(k)^2)/2);
  x(k+1) = x1(k) + x2(k) + x3(k) + x4(k) + h;
  u(k+1) = u1(k) + u2(k) + u3(k) + u4(k) + up1(k+1)*h + up2(k+1)*h + up3(k+1)*h + up4(k+1)*h;
end 
%transpose x and u into column vectors
xvalues = x';
uvalues = u';
end

MOST RECENT CODE:

if true
%%EULER EXPLICIT METHOD
f = @(x,u) -x.*u + exp(-(x.^2)/2);
exact = @(x) exp(-2.*x).*(2-(4.81097.*10^(-16)).*(exp(x))-(exp(3.*x))+(0.1.*exp(5.*x)));
h = 0.1;
xRange = [0,1.5];
initial = [1.1, -4.7, 7.9, -14.3];
steps = xRange(2)/h;
x=zeros(steps+1,1);
x(1) = xRange(1);
h = ( xRange(2) - xRange(1) ) / steps;
u = (initial);
for k = 1 : steps
x(k+1) = x(k) + h;
u(k+1,:) = u(k,:) + h * (feval( f, x(k),(u(k,:))'))';
end 
xvalues= x'
uvalues= u'
plot(x,u, '--b');
xlim([0 1.5])
hold on
fplot(exact,[0,1.5], '--r');
legend('euler 0.1','exact');
step = xvalues;
implicit = uvalues;
exact = exact(xvalues);
table(step,implicit,exact)
end

[KSSV]

Don't attach code as image.....copy and paste the code here.....

[Erick Jablonski]

good point, fixed it

Answer 1

[Torsten]

This should help:

% function definition 
F=@(t,u)[u(2); u(3); u(4); -6*u(1)-u(2)+7*u(3)+u(4)];
% initial values
U(1:4,1) = [1.1; -4.7; 7.9; -14.3];
% Loop update
U(1:4,n+1) = U(1:4,n) + dx*F(t(n),U(1:4,n))

Best wishes

Torsten.

[Erick Jablonski]

Should their be a lowercase u and an uppercase U? sorry im just confused on how and where to put these updates into my code.

[Torsten]

u is just a formal function parameter ; you can replace it by whatever you like.

Best wishes

Torsten.