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 | 1 | initial version |
[ 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
Your proposed solution:
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: Px.plot((0,10))
Launched png viewer for Graphics object consisting of 1 graphics primitive
sage: limit(Px(t),t=oo)
15
HTH,
| | 2 | No.2 Revision |
[ 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
Your proposed solution:
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: Px.plot((0,10))
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
sage: limit(Px(t),t=oo)
15
In other words, your proposed solution is a sigmoïd...
HTH,
| | 3 | No.3 Revision |
[ 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
Your proposed solution:
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 :
sage: E1=(lambda a,b:a==b)(*map(exp, (S0*30).simplify_log().operands())) ; E1
x(t)/(x(t) - 15) == e^(30*_C + 30*t)
which can now be solved for x(t) :
sage: S1=E1.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.
HTH,
| | 4 | No.4 Revision |
[ 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
Your proposed solution:
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 :
sage: E1=(lambda a,b:a==b)(*map(exp, (S0*30).simplify_log().operands())) ; E1
x(t)/(x(t) - 15) == e^(30*_C + 30*t)
which can now be solved and solve it explicitly for x(t) :
sage: S1=E1.solve(x(t)) 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,
