On Fri, 10 Sept 2021 at 16:40, Nicolas Guarin <[email protected]> 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 *
> 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?
Are the different solutions not equivalent for different values of the
constants (bearing in mind the possibility for the constants to be
non-real)? This suggests that one solution can be transformed to the other
for appropriate values of the constants:
In [*41*]: C1, C2 = symbols('C1, C2')
In [*42*]: s1, s2 = solve(Eq(tanh(C1 + t), coth(C2 + t)), C1)
In [*43*]: s1.rewrite(exp).simplify()
Out[*43*]:
⎛ 2⋅C₂⎞
log⎝-ℯ ⎠
───────────
2
In [*44*]: s2.rewrite(exp).simplify()
Out[*44*]:
⎛ ________⎞
⎜ ╱ 2⋅C₂ ⎟
log⎝-╲╱ -ℯ ⎠
In other words tanh(C1 + t) can be equivalent to coth(C2 + t) if C1 =
log(-exp(2*C2))/2. Of course if C2 is real then this implies that C1 is not.
Oscar
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