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 > >