# How to solve a Bernoulli Diff EQ?

Hi!

I've been trying to figure out how to solve a Bernoulli Diff EQ in Sage for 2 hours now and no matter what I do, it always gives me this long and/or complicated answer when the actual answer isn't anywhere near what sage says

My problem is

dx/dt = 2x(15-x),  P(0) = 10


And my sage code is

t=var('t')
x=function('x')(t)
de=diff(x,t)==2*x(15-x)
h=desolve(de,x,ics=[0,10],algorithm="fricas",contrib_ode=true)
z=desolve_laplace(de,dvar=x,ics=[0,10])
show(expand(h))
show(expand(z))


It gives me this:

-1/30*log(x(t) - 15) + 1/30*log(x(t)) == -1/30*I*pi + t + 1/30*log(10) - 1/30*log(5)


But what I want is this

x=(30e^(30t))/(1+2e^(30t))


Any ideas on how to get this?

Thanks!

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[ I suppose that this is not homework. If it is, you should abstain from using my answer and find yourself your solution... ]

The x unknown of the D. E. is a function, so we'd better declare and use it as such :

sage: x=function("x")
sage: var("t")
t
sage: de=diff(x(t),t)==2*x(t)*(15-x(t))


Maxima (Sage's default DE solver) doesn't give an explicit answer :

sage: SM=desolve(de, x(t), ivar=t, contrib_ode=True) ; SM
-1/30*log(x(t) - 15) + 1/30*log(x(t)) == _C + t


Sympy does give an implicit answer :

sage: from sympy import dsolve, sympify
sage: SS=dsolve(*map(sympify, (de, x(t))))._sage_() ; SS
x(t) == 15/(C1*e^(-30*t) + 1)


Declare the integration constant and check the solution :

sage: var("C1")
C1
sage: bool(de.substitute_function(x, SS.rhs().function(t)))
True


Solve C1 for your boundary conditions :

sage: SC1=C1.subs(t==0).subs(x(0)==0).solve(C1) ; SC1
[C1 == 0]


The solution of your problem is therefore :

sage: SS.subs(SC1)
x(t) == 15


sage: Px=(30*e^(30*t))/(1+2*e^(30*t)).function(t) ; Px
t |--> 30*e^(30*t)/(2*e^(30*t) + 1)


does not fulfill your boundary condition :

sage: Px(0)
10


BTW :

sage: limit(Px(t),t=oo)
15
sage: limit(Px(t),t=-oo)
0
sage: Px.plot((-5, 5))
Launched png viewer for Graphics object consisting of 1 graphics primitive


In other words, your proposed solution is a sigmoïd...

HTH,

EDIT : Maxima's solution can be exploited with a little "manual" work :

sage: S0=desolve(de, x(t)) ; S0
-1/30*log(x(t) - 15) + 1/30*log(x(t)) == _C + t


Let's declare the arbitrary constant :

sage: var("_C")
_C


log properties allow us to reshape this solution and solve it explicitly for x(t) :

sage: S1=(S0*30).simplify_log().solve(x(t)) ; S1
[x(t) == 15*e^(30*_C + 30*t)/(e^(30*_C + 30*t) - 1)]


which can be itself rewritten :

sage: var("a, b")
(a, b)
sage: w0,w1=(SR.wild(u) for u in range(2))
sage: S1t=[u.subs((b*(a/(a-1))==b*(1+1/(a-1))).subs(a==w0, b==w1)) for u in E1.solve(x(t))] ; S1t
[x(t) == 15/(e^(30*_C + 30*t) - 1) + 15]


This singleton solution checks the differential equation :

sage: [bool(de.substitute_function(x,u.rhs().function(t))) for u in S1t]
[True]


Is there a value of _C satisfying the boundary condition ?

sage: S2=S1[0].subs(t==0).subs(x(0)==15) ; S2
15 == 15*e^(-30*_C) + 15


Nope : e^(-30*_C) is nozero for all _C complexes.

It should be noted that Maxima's solution does not describe the "special case x(t)=15, which is a solution of de :

sage: de.substitute_function(x,SR(15).function(t))
0 == 0


HTH,

more

Be careful and make multiplication explicit.

The differential equation studied here is $$x'(t) = 2 · x(t) · (15 - x(t))$$ and not $$x'(t) = 2 · x(15 - x(t))$$ so instead of

de = diff(x, t) == 2 * x(15 - x)


you want

de = diff(x, t) == 2 * x * (15 - x)

more

I did, but that didn't change the answer

( 2021-04-30 02:59:31 +0200 )edit