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>