dsolve produces an incorrect result?

[Dan Lewis]

Hi folks,

Looks like sympy/dsolve produces an incorrect solution to an ODE:

from sympy import *

​

mass, g, b, t = symbols('mass g b t')

v, x = symbols('v x', cls=Function)

​

paraChute = mass*v(t).diff(t)-mass*g+b*(v(t))**2

​

solution = dsolve(paraChute,v(t),hint='lie_group')

solution

Out[1]:

Eq(v(t), sqrt(g)*sqrt(mass)/(sqrt(b)*tanh(sqrt(b)*sqrt(g)*(C1*mass + t)/sqrt(mass))))

Sorry in advance if I'm doing something silly - the solution shouldn't be 1/tanh.  Should be tanh.

[Aaron Meurer]

According to checkodesol(paraChute, solution), the solution is correct.

Aaron Meurer

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[Aaron Meurer]

The reason this is happening is that both tanh(x) and coth(x) satisfy
the same differential equation, namely f'(x) = 1 - f(x)**2, because
d/dx tanh(x) = 1 - tanh(x)**2 and d/dx coth(x) = 1 - coth(x)**2. This
is your ODE (up to a change in variable).

Aaron Meurer

[Aaron Meurer]

In particular, tanh(x + I*pi/2) = coth(x) and coth(x + I*pi/2) =
tanh(x). You can verify this with SymPy:

In [22]: x = Symbol('x', real=True)

In [23]: tanh(x + I*pi/2).expand(complex=True).simplify()
Out[23]:
1
───────
tanh(x)

In [24]: 1/tanh(x + I*pi/2).expand(complex=True).simplify()
Out[24]: tanh(x)

But note that the arbitrary constant in the solution of your ODE is
like coth(a*t + C1). So the solution represents both tanh() and
coth(), depending on the value of C1.

Aaron Meurer

[Dan Lewis]

Yikes.  Indeed - being silly and not thinking through my complex numbers.  Thanks for the detailed response(s).