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Re: simple question on DSolve
*To*: mathgroup at smc.vnet.net
*Subject*: [mg125012] Re: simple question on DSolve
*From*: Bob Hanlon <hanlonr357 at gmail.com>
*Date*: Fri, 17 Feb 2012 06:28:26 -0500 (EST)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201202141137.GAA17715@smc.vnet.net>
You are missing a comma in your equation list between the two boundary
conditions. This eliminates the condition y'[1] == 0.
Bob Hanlon
On Thu, Feb 16, 2012 at 3:26 AM, Christoph Lhotka
<christoph.lhotka at fundp.ac.be> wrote:
> Hi,
>
> if I am using DSolve:
>
> DSolve[{y''[t] + k y[t] == 0,
> y'[0] == 0
> y'[1] == 0}, {y[t]}, t]
>
> I get:
>
> {{y[t] -> C[1] Cos[Sqrt[k] t]}}
>
> but
>
> D[sol, t] /. {{t -> 0}, {t -> 1}}
>
> gives
>
> {0, -Sqrt[k] C[1] Sin[Sqrt[k]]}.
>
> does it mean that C[1]=0 or any other idea?
>
> best,
>
> christoph
>
>
>
>
> On 02/15/2012 10:44 AM, Bob Hanlon wrote:
>> eqns = {y''[x] + k*y[x] == 0,
>> y'[0] == yp0, y'[1] == yp1};
>>
>> sol = DSolve[eqns, y, x][[1]] // Simplify
>>
>> {y -> Function[{x}, (1/Sqrt[
>> k])(yp0 Cos[Sqrt[k] x] Cot[Sqrt[k]] -
>> yp1 Cos[Sqrt[k] x] Csc[Sqrt[k]] + yp0 Sin[Sqrt[k] x])]}
>>
>> eqns /. sol // Simplify
>>
>> {True, True, True}
>>
>> y[x] /. sol /. {yp0 -> 0, yp1 -> 0}
>>
>> 0
>>
>> Appears to be forced by boundary conditions.
>>
>>
>> Bob Hanlon
>>
>>
>> On Tue, Feb 14, 2012 at 6:37 AM, Gualtiero Badin
>> <gualtiero.badin at gmail.com> wrote:
>>> Hello,
>>> if I try to solve the simple boundary value problem
>>>
>>> y''+ky=0
>>> y'(0)=0
>>> y'(1)=0
>>>
>>> mathematica returns me y=0, that is correct but that is not the
>>> complete answer... Does anyone know how to get the complete answer?
>>> (of course i know the complete answer, but I would like to solve some
>>> uglier versions of the same problem...)
>>> Thanks
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