# Revision history [back]

### Solvingan ODE and simplifying the result

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 allocate to the solution (i.e. h_sol) the RHS of the above? without the h(t) == part.

I had to set by hand (it is ease, but it would be nice to automatize the allocation)

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?

 2 None slelievre 10223 ●9 ●107 ●205 http://carva.org/samue...

### Solvingan ODE and simplifying the result

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 allocate to the solution (i.e. h_sol) the RHS of the above? without the h(t) == part.

I had to set by hand (it is ease, but it would be nice to automatize the allocation)

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?