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Re: DSolve
*To*: mathgroup at smc.vnet.net
*Subject*: [mg83177] Re: DSolve
*From*: dh <dh at metrohm.ch>
*Date*: Wed, 14 Nov 2007 04:42:13 -0500 (EST)
*References*: <fhc3n2$4vv$1@smc.vnet.net>
Hi Ray,
there is no doubt that u[x]==0 is a solution.
But are there other solutions? To answer this consider: u''=-\[Lambda]^2
(0.25 - x^2) u[x]. This means that between [-0.5,0.5] where (0.25 - x^2)
is positive, the right side of the equation is - pos u[x] where pos is a
positive expression. This means that for u!=0 the curve bends towards
the x-axis. If \[Lambda] is larger, this effect is stronger. Therefore,
starting at x=-1/2 with a given slope, the curve hits u[1/2]==0 only for
very special values of \[Lambda].
Shortly, the generic solution is u[x]==0, other solutions are only
possible for special values of \[Lambda].
hope this helps, Daniel
Raj wrote:
> eqn = D[u[x], {x, 2}] + \[Lambda]^2 (0.25 - x^2) u[x] == 0
> DSolve[{eqn, u[-1/2] == 0, u[1/2] == 0}, u[x], x]
>
> This returns {{u[x] -> 0}} while another CAS system returns a solution
> in terms of WhittakerW function.
>
> Am I doing something wrong or is Mathematica not able to solve this
> equation symbolically?
>
> Thanks,
>
> Raj
>
>
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