This post was prepared to be upload to ask.sagemath.org, but I got a
warning "Spam was detected on your post, sorry for if this is a mistake"
that forbids me to post the question.
-----------------------------
I'm interested in solving the differential equation $$3 h' + 3 h^2 = c_1,$$
where $c_1$ is a positive real number.
var('t')
var('c1', latex_name=r'c_1')
h = function('h')(t)
eq = -3*h^2 + c1 - 3*diff(h, t)
eq_sol = desolve(eq, h, ivar=t, contrib_ode=True)
The above code works, but it's not solved explicitly for $h$, so
h_sol = solve(eq_sol, h)
h_sol = h_sol[0]
h_sol
This gives something like $$h\left(t\right) = \frac{\sqrt{3} \sqrt{c_{1}}
{\left(e^{\left(\frac{2}{3} \, \sqrt{3} C \sqrt{c_{1}} + \frac{2}{3} \,
\sqrt{3} \sqrt{c_{1}} t\right)} + 1\right)}}{3 \,
{\left(e^{\left(\frac{2}{3} \, \sqrt{3} C \sqrt{c_{1}} + \frac{2}{3} \,
\sqrt{3} \sqrt{c_{1}} t\right)} - 1\right)}},$$
in sage notation (non-LaTeX) it starts like
h(t) == 1/3*sqrt(3)*sqrt(c1)* ...
**Question 1:** Is there a way to assing to the solution (i.e. `h_sol`) the
RHS of the above? without the `h(t) == ` part.
I had to assign by hand (it is ease, but it would be nice to automatize the
assignation)
var('C') # the integration constant introduced above
h_sol = 1/3*sqrt(3)*sqrt(c1)* ...
Then, by simply looking at the solution it is clear that it can be
simplified. I tried things like
h_sol = h_sol.canonicalize_radical()
h_sol = h_sol.collect_common_factors()
h_sol = h_sol.simplify_rectform(complexity_measure = None)
but none of them returns the expected result, which could be obtained from
Mathematica's kernel
mathematica("DSolve[3*h'[t] + 3*h[t]^2 == C[1], h[t], t]//FullSimplify")
$$ \sqrt{\frac{c_1}{3}} \tanh\left( \sqrt{\frac{c_1}{3}} (t - 3 c_2)
\right) $$
**Question 2:** How could the expression `h_sol` be manipulated to obtain
the hyperbolic tangent?
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