# Solution of ODE expressed with hyperbolic trigonometric function With Sage 7.3, consider

t=var('t')
v=function('v')(t)
m, g, h = var('m g h')
assume(m > 0)
assume(g > 0)
assume(h > 0)
sol = desolve(m*diff(v,t) == m*g - h*v**2, v, ivar=t)
show(sol)
sol.solve(v)


The outcomes are

$$\newcommand{\Bold}{\mathbf{#1}}-\frac{m \log\left(\frac{h v\left(t\right) - \sqrt{g h m}}{h v\left(t\right) + \sqrt{g h m}}\right)}{2 \, \sqrt{g h m}} = C + t$$

$$\newcommand{\Bold}{\mathbf{#1}}\left[v\left(t\right) = \frac{\sqrt{g h m} {\left(e^{\left(\frac{2 \, \sqrt{g h m} C}{m} + \frac{2 \, \sqrt{g h m} t}{m}\right)} + 1\right)}}{h {\left(e^{\left(\frac{2 \, \sqrt{g h m} C}{m} + \frac{2 \, \sqrt{g h m} t}{m}\right)} - 1\right)}}\right]$$

Nevermind desolve came with an implicit solution I had to further solve by hand but then solve missed that the last expression for v(t) is a mere hyperbolic tangent. How would I get the solution in its simpler form?

$$v(t) = \sqrt{\frac{gm}{h}}\ \text{tanh}\left(\sqrt{\frac{gh}{m}}(t + m C)\right)$$

Thanks for any hint!

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## Comments

It doesn't help for the hyperbolic ones, but maxima does have some routines to go back and forth between exponential and trigonometric expressions:

sage: tan(x)._maxima_().exponentialize().sage()
(-I*e^(I*x) + I*e^(-I*x))/(e^(I*x) + e^(-I*x))
sage: tan(x)._maxima_().exponentialize().demoivre().sage()
sin(x)/cos(x)


I don't think these routines are directly exposed in sage and "demoivre" doesn't know about hyperbolic routines.

In general, I would expect computer algebra system would give preference to exponential expressions, because it's easy to define a "normal form" for them, whereas it's not so clear what that should mean for hyperbolic/trigonometric functions in general: there's too much redundancy.

Thanks a lot for those hints: I had no idea I could get access to Maxima in such a way. As for my problem, from a human point of view, it is natural to want the "shorter" answer. I have just checked that Mathematica does actually produce the formula with tanh. Besides, the math behind exponentialize and demoivre must have an equivalent in hyperbolic trigonometry.