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I have a system of coupled differential equations, one of which is second-order. I am looking for a way to solve them in Python. I would be extremely grateful for any advice on how can I do that!

$k$ is just a constant

$$ \left(\frac{dr}{dT}\right)^2=k-\left(1-\frac{5}{r}\right)\left(3+\frac{2}{r^2}\right)\\ \frac{d\varphi}{dT}=\frac{1}{r^2} $$

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    $\begingroup$ Is that $(\frac{dr}{dt})^2$ the square of $\frac{dr}{dt}$, or a second derivative $\frac{d^2}{dt^2}r$? That notation is typically used for the first derivative squared, but you mention "second order" twice, so it is not clear. $\endgroup$ – Federico Poloni Jan 28 at 9:18
  • $\begingroup$ Seems like a homework... It is really bad written though... $\endgroup$ – Alone Programmer Jan 28 at 19:21
  • $\begingroup$ @Лада, I've edited the formulas in your question (btw, we have LaTeX enabled!). Please, make sure that I did not miss anything. $\endgroup$ – Anton Menshov Jan 29 at 1:54
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The first step is to transform the second order equation to a set of two coupled first order equations. Define an auxiliary function $u(T) = \frac{dr(T)}{dT}$. This results in the system

$$\begin{align} \frac{du}{dT} &= k-(1-\frac{5}{r})(3+\frac{2}{r^2}) \\ \frac{dr}{dT} &= u\\ \frac{d\phi}{dT} & = \frac{1}{r^2} \end{align} $$

Now you have a set of three coupled first order equations in the form fit for solving with solve_ivp. See SciPy documentation for solve_ivp.

import numpy as np  
from scipy.integrate import solve_ivp 
import matplotlib.pyplot as plt

def rhs(t,Y):
    dY = np.zeros_like(Y)
    k = 1.0
    dY[0] = k - (1 - 5/Y[1])*(3+2/Y[1])
    dY[1] = Y[0]
    dY[2] = 1/Y[1]**2
    return dY

Y0 = np.array([0,1,0]) 
sol = solve_ivp(rhs, [0,10], Y0, method='Radau', dense_output=True)

t = np.linspace(0, 10, 1001) 
Y = sol.sol(t) 

plt.plot(t, Y[1],'-', label='r(t)') 
plt.plot(t, Y[2],'-', label='phi(t)')
plt.legend(loc='best')
plt.xlabel('T')

resulting plot

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