Re: simple question on DSolve
- To: mathgroup at smc.vnet.net
- Subject: [mg124984] Re: simple question on DSolve
- From: Christoph Lhotka <christoph.lhotka at fundp.ac.be>
- Date: Thu, 16 Feb 2012 03:26:22 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201202141137.GAA17715@smc.vnet.net> <201202150944.EAA29401@smc.vnet.net>
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
- References:
- simple question on DSolve
- From: Gualtiero Badin <gualtiero.badin@gmail.com>
- Re: simple question on DSolve
- From: Bob Hanlon <hanlonr357@gmail.com>
- simple question on DSolve