Solve a system of equations using Runge Kutta of order 4 in python

Question

I want to solve the following system using the numerical method Runge Kutta of order 4:

y'=-xy x'=-x+sin(t)

with initial conditions x(0)=0.75 and y(0)=0

Using RungeKutta of order code 4 I tried this:

def GlucoseTT(x, t, params):

a = params["a"]
b= params["b"]
c= params["c"]

xdot= np.array([a*x[0]-b*x[1], -c*x[0]+np.sin(t)])

return xdot



def RK4(f, x0, t0, tf, dt):
   t=np.arange(t0,tf,dt)
   nt=t.size

nx=x0.size
x=np.zeros((nx,nt))

x[:,0]=x0

for k in range(nt-1):
    k1= dt*f(t[k], x[:,k])
    k2= dt*f(t[k]+dt/2, x[:,k]+k1/2)
    k3= dt*f(t[k]+dt/2, x[:,k]+k2/2)
    k4= dt*f(t[k]+dt, x[:,k]+k3)
    
    dx=(k1+2*k2+2*k3+k4)/6
    
    x[:,k+1]=x[:,k]+dx
return x, t

#Define Problem

params = {"a":1, "b":1, "c":1}
f= lambda t, x : GlucoseTT(x, t, params)

x0= np.array([0.75,0])


#Solve ODE
t0=0
tf= 100
dt= 0.1

x,t =RK4(f, x0, t0, tf, dt)

plt.plot(t,x[0,:],"r")

I dont' know if this code works....

Answer 1

Not an answer, but the solution of the equations, so that you can compare:

X = C e^-t + (sin(t) - cos(t)) / 2
Y = D e^-t - C t e^-t + cos(t) / 2

The initial conditions give the system

0.75 = C - 0.5
0    = D + 0.5