Perhaps it would help to set a variable for the initial condition and
solve for that. Unfortunately, dsolve() doesn't presently handle the
case where the initial condition splits into two solutions.
>>> var('t0')
t0
>>> dsolve(ode, f(t), ics={f(0): t0})
Traceback (most recent call last):
File
On Fri, 10 Sept 2021 at 16:40, Nicolas Guarin wrote:
>
>
> 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 *
>
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))
