Re: A ODE I need to solve

*To*: mathgroup at smc.vnet.net*Subject*: [mg111124] Re: A ODE I need to solve*From*: Sam Takoy <sam.takoy at yahoo.com>*Date*: Wed, 21 Jul 2010 07:12:25 -0400 (EDT)*References*: <i20q77$kh9$1@smc.vnet.net> <i23k6l$f0d$1@smc.vnet.net>

That's quite nifty. Thanks! On 7/20/2010 3:45 AM, schochet123 wrote: > On Jul 19, 9:09 am, Sam Takoy<sam.ta... at yahoo.com> wrote: >> Hi, >> >> Mathematica has a problem with this: >> >> DSolve[y''[x] + (2/Cosh[x - h]^2 - 1) y[x] == 0, y, x] >> >> although the solution is not too difficult. One of the solutions is >> >> 1/(Cosh[2(x-h)]+1)^(1/2) >> >> Is there a way to help Mathematica along? >> >> Thanks! > > Define > > lhs[y_] := D[y, x, x] + (2/Cosh[x - h]^2 - 1) y > > Then when > > DSolve[lhs[y[x]] == 0, y, x] > > doesn't yield an answer, but you know the solution 1/(Cosh[2(x-h)] > +1)^(1/2) you can use the variation of parameters method > > DSolve[lhs[1/(Cosh[2 (x - h)] + 1)^(1/2) w[x]] == 0, w, x] > > which yields > > {{w -> Function[{x}, > C[2] + C[1] (1/2 (-h + x) - 1/4 Sinh[2 (h - x)])]}} > > C[2]->1, C[1]->0 corresponds to the solution you already knew, and > C[2]->0, C[1]->1 yields a second, linearly independent solution > y[x_]=1/(Cosh[2(x-h)]+1)^(1/2) (1/2 (-h + x) - 1/4 Sinh[2 (h - x)]) > > Steve >