Too tricky, but maybe this helps:
sage: c,x = var("c,x")
sage: y = function('y',x)
sage: soln = desolve(diff(y,x)+sin(x)*y^6==0,y); soln
1/(5*y(x)^5) == c - cos(x)
sage: soln.subs(x = pi)
1/(5*y(pi)^5) == c + 1
sage: ssoln = soln.__repr__(); ssoln
'1/(5*y(x)^5) == c - cos(x)'
sage: ssoln = ssoln.replace("y(x)","9"); ssoln
'1/(5*9^5) == c - cos(x)'
sage: ssoln = ssoln.replace("x","pi"); ssoln
'1/(5*9^5) == c - cos(pi)'
sage: nsoln = eval(ssoln)
sage: nsoln
c + 1 == 0
Hopefully there are better ways.
Using SymPy:
sage: x = Symbol('x')
sage: C1 = Symbol('C1')
sage: y = Function('y')
sage: soln = dsolve(Derivative(y(x),x)+sin(x)*y(x)^6,y(x)); soln
(C1 - 5*cos(x))**(-1/5)
sage: ypi = soln.subs(x,pi); ypi
(5 + C1)**(-1/5)
But now
sage: solve(ypi - 9, C1)
yields an error which I don't understand.
On Tue, Apr 28, 2009 at 1:59 PM, ma...@mendelu.cz <ma...@mendelu.cz> wrote:
>
> Hello, I just want to test, if I can do with my students in computer
> lab what they usualy do on the paper
>
> 1. find general solution
> 2. substitute from initail conditions
> 3. find c
> 4. use this c in general solution
>
> I agree that this may be strange to do this in the case, we have a
> command for solving IVP.
> I just wanted to know, if it is possible to substitute some value for y
> (x), if y(x) is declared as a function.
>
> Thanks. Robert
>
> On 28 Dub, 16:45, David Joyner <wdjoy...@gmail.com> wrote:
>> I don't know if you are asking about how to use ICs in
>> desolve or if you are asking about how to do substitutions.
>> Anyway, I get this:
>>
>> sage: y=function('y',x)
>> sage: desolve(diff(y,x)+sin(x)*y^6==0,y)
>> 1/(5*y(x)^5) == c - cos(x)
>> sage: desolve(diff(y,x)+sin(x)*y^6==0,y,[pi,9])
>> 1/(5*y(x)^5) == (-295245*cos(x) - 295244)/295245
>>
>> Does that help?
>>
>> On Tue, Apr 28, 2009 at 10:36 AM, ma...@mendelu.cz <ma...@mendelu.cz> wrote:
>>
>> > Dear memebers of SAGE-support
>>
>> > I wonder if it is possible to substitute initial conditions into an
>> > equation produced by desolve. I tried something like
>>
>> > y=function('y',x)
>> > desolve(diff(y,x)+sin(x)*y^6==0,y)
>> > sol({x:pi,y:9})
>>
>> > and got
>>
>> > 1/(5*y(pi)^5) == c + 1
>>
>> > but I would like to see
>>
>> > 1/(5*9^5) == c + 1
>>
>> > many thanks
>>
>> > Robert
> >
>
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