Re: Surprising DSolve problem
- To: mathgroup at smc.vnet.net
- Subject: [mg123969] Re: Surprising DSolve problem
- From: "Nasser M. Abbasi" <nma at 12000.org>
- Date: Tue, 3 Jan 2012 05:26:18 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <jdrn85$8sr$1@smc.vnet.net>
- Reply-to: nma at 12000.org
On 1/2/2012 1:44 AM, Sam Takoy wrote: > Hi, > > I find that my Mathematica (8.0.1) > > can do > > DSolve[ y''[x] + (2 y[x])/(Cosh[x]^2) == 0, y, x] > > but can't do > > DSolve[ y''[x] + (2 y[x])/(a^2 Cosh[x/a]^2) == 0, y, x] (*Additional > constant "a") > > If I am not mistaken, the equations are equivalent by a trivial change > of variables. > > What's going on here? > > Thanks, > > Sam > Not sure why but when 'a' is an integer, it is not happy. So this is a way around it, just add a 1.0*a in place of a in the DSolve, and now it goes through: ----------------------------------- DSolve[y''[x] + (2 y[x])/(a^2 Cosh[x/(1.0*a)]^2) == 0, y, x] -------------------------------------------- To make sure I had the solution as DSolve[ y''[x] + (2 y[x])/(Cosh[x]^2) == 0, y, x] (Which you say should be the same), I did gave both the some initial conditions, and plotted the solution for some random values of 'a', like this: --------------------------------- ClearAll[y, x, a]; sol1 = y[x] /. First@DSolve[{y''[x] + (2 y[x])/(a^2 Cosh[x/(1.0*a)]^2) == 0, y[0] == 0, y'[0] == 1}, y[x], x]; ----------------------------------- which gave: (1. Sqrt[1. -1. Sech[x/a]^2])/Sqrt[1/a^2] Now plot it ------------------------------ Plot[sol1 /. a -> RandomInteger[10], {x, -.02, .02}, PlotRange -> All] ------------------------------ But when I plot the second solution, for the same range, they are not the same. (Are sure that adding an 'a' will make no difference?) ------------------------------------ sol2 = y[x] /. First@DSolve[{y''[x] + (2 y[x])/(Cosh[x]^2) == 0, y[0] == 0, y'[0] == 1}, y[x], x] ----------------------------------- Which gave Tanh[x] Plot[sol2, {x, -.01, .01}] does not give the same solution shape. Tanh[x] =!= (1. Sqrt[1. -1. Sech[x/a]^2])/Sqrt[1/a^2] for any 'a'. so, The 2 differential equations do not appear to be the same? What change of variables will make them the same for any 'a' ? --Nasser