# Solution of Coupled Differential equation for a 2d linear flow using RK4 method in python 3

I want to study the dynamics of a 2d linear flow, whose dynamical equation is- $$\begin{pmatrix} \dot{x_1}\\ \dot{x_2}\\ \end{pmatrix}=\begin{pmatrix} 1 & 1\\ 4 & -2\\ \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}$$. Now I have tried to solve and plot y vs. x of this coupled differential equation using RK4 in python for the initial condition $$(y_0=2, x_0=-1)$$. My code is following, but the graph is not correct [In the graph origin should be a saddle point, $$\begin{pmatrix} -0.25\\ 1\\ \end{pmatrix}$$ axis should be stable manifold and $$\begin{pmatrix} 1\\ 1\\ \end{pmatrix}$$ axis should be unstable manifold]-

import numpy as np
from math import sqrt

import matplotlib.pyplot as plt

# Equations:

def V(u,t):
x1, x2, v1, v2 = u
return np.array([ v1, v2, (x1+x2), -(4*x1-2*x2)])

def rk4(f, u0, t0, tf , n):
t = np.linspace(t0, tf, n+1)
u = np.array((n+1)*[u0])
h = t[1]-t[0]
for i in range(n):
k1 = h * f(u[i], t[i])
k2 = h * f(u[i] + 0.5 * k1, t[i] + 0.5*h)
k3 = h * f(u[i] + 0.5 * k2, t[i] + 0.5*h)
k4 = h * f(u[i] + k3, t[i] + h)
u[i+1] = u[i] + (k1 + 2*(k2 + k3) + k4) / 6
return u, t

u, t  = rk4(V, np.array([2.0, 0., -1, 1.]) , 0. , 1. , 1000)
x1, x2, v1, v2 = u.T
plt.plot(x1,x2)
plt.show()


Can anyone helps me to write this code.

You are implementing the system $$\pmatrix{\ddot x_1\\\ddot x_2} = \pmatrix{1&1\\-4&2} \pmatrix{x_1\\x_2}$$ in its first-order version with $$v_k=\dot x_k$$. It is not surprising that this different system results in different solutions.
def V(u,t):