Use the Runge–Kutta algorithm for systems with h= 0.1 to approximate the solution to the initial value problem.

$$\begin{gathered} \mathrm{x}\text{'}=\mathrm{yz};\mathrm{x}\left(0\right)=0, \\ \mathrm{y}\text{'}=-\mathrm{xz};\mathrm{y}\left(0\right)=1, \\ \mathrm{z}\text{'}=\frac{-\mathrm{xy}}{2};\mathrm{z}\left(0\right)=1, \end{gathered}$$

At t=1.

Given initial value problem is:

$$\begin{gathered} x\text{'}=yz \\ y\text{'}=-xz \\ z\text{'}=\frac{-xy}{2} \end{gathered}$$

And the initial conditions are:

$$\begin{gathered} x\left(0\right)=0 \\ y\left(0\right)=1 \\ z\left(0\right)=1 \end{gathered}$$

To get the solution apply the Runge-Kutta method in mat lab for t=1.

Function n[t,x] =Runge_Kutta(f,t0,t_end,init_cond,h)
%we begin at time t0 and end when we reach t_end
%init_cond(i) contains the initial value of x_i
%f contains functions such that x_i'=f_i(t,x1,x2,...)

t(:,1)=t0; % t0 is the initial value of t
x(:,1)=init_cond; % initial conditions are set

i=1;
while t(:,i) <t_end

k1=f(t(i),x(:,i));
k2=f(t(i)+0.5*h,x(:,i)+0.5*h*k1);
k3=f(t(i)+0.5*h,x(:,i)+0.5*h*k2);
k4=f(t(i)+h,x(:,i)+h*k3);

x(:,i+1)=x(:,i)+(h/6)*(k1+2*k2+2*k3+k4);

t(:,i+1)=t(:,i)+ h;
i=i+1;

end

clear all

init_cond=[0;1;1];

f=@(t,X) [X(2)*X(3);-X(1)*X(3);-X(1)*X(2)/2];

[t,x] = Runge_Kutta(f,0,1,init_cond,0.1);

table(t(:,end),x(1,end),x(2,end),x(3,end),'VariableNames',{'t','x','y','z'})

Then the results are;

Therefore,$\mathrm{X}\left(1\right)=0.803,\mathrm{Y}\left(1\right)=0.59598,\mathrm{Z}\left(1\right)=0.82316$.