Differential equations with initial condition in Python (change a working code)

[Euler2]

I am pretty new in Python. I have a working code which looks like

import numpy as np
import scipy.integrate as sp
import sys
g, l, k = 9.81, 10, float(input())
PTS = 6
def rhs( y, t, g, l, k ): return [ k*t/l * np.cos(y[1]) - g/l * np.sin(y[1]),y[0] ]
y0 = [ 0, 0 ]
t = np.linspace( 0, 6, PTS+1 )
y = sp.odeint( rhs, y0, t, args = (g, l, k) )
for i in range(PTS): print( t[i], np.arctan(k*t[i]/g) *180/np.pi )

where you have to write a value 0.5 (it is value of "k"). The output of this code is:

time value of y
0.0 0.0
1.0 2.91774937823
2.0 5.82044363971
3.0 8.69348861202
4.0 11.5231772897
5.0 14.297051396

Now I want to use a value in time t = 5 (y = 14.297051396) as my initial condition and find values of y in time t = 6, ..., 16 for k=1.1. How should I change my code?

I tried this code

import numpy as np
   import scipy.integrate as sp
   import sys
   g, l, k = 9.81, 10, float(input())
   PTS = 12
   def rhs( y, t, g, l, k ): return [ k*t/l * np.cos(y[1]) - g/l * np.sin(y[1]), y[0] ]
   y0 = [ 14.297051396, 5 ]
   t = np.linspace( 5, 17, PTS+1 )
   y = sp.odeint( rhs, y0, t, args = (g, l, k) )
   for i in range(PTS): print( t[i], np.arctan(k*t[i]/g) *180/np.pi )

and write 1.1, but the output is the same as in the case that I write y0 = [ 0, 0 ].

What did I do wrong? Could somebody explain to me, how to change it? (I am self-learner in Python and I do not study programming).

[Euler2]

I tried to write it as a code for t=0:1:12 (and hence my condition is in time t=0), but it also did not work....

import numpy as np
import scipy.integrate as sp
import sys
g, l, k = 9.81, 10, float(input())
PTS = 12
def rhs( y, t, g, l, k ): return [ k*t/l * np.cos(y[1]) - g/l * np.sin(y[1]),
y[0] ]
y0 = [14.297051396, 0 ]
t = np.linspace( 0, 12, PTS+1 )
y = sp.odeint( rhs, y0, t, args = (g, l, k) )
for i in range(PTS): print( t[i], np.arctan(k*t[i]/g) *180/np.pi )