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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
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