I have the following ODE
z'' = 1 - z²
That has as solutions z=tanh(C + t) and z=coth(C + t), depending on the
initial condition being greater or less than 1. When I use dsolve I get the
latter
from sympy import *
init_session()
ode = Eq(f(t).diff(t), 1 - f(t)**2)
sol = dsolve(ode, f(t))
And sol is
-1
f(t) = ──────────── ,
tanh(C₁ - t)
That is equivalent to coth(C + t).
Is there any reason for SymPy to give this answer and not the other or is
this a bug?
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