Re: simple question on DSolve

```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
>> 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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