Re: DSolve with DiracDelta [CORRECTION]
- To: mathgroup at smc.vnet.net
- Subject: [mg75807] Re: [mg75764] DSolve with DiracDelta [CORRECTION]
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 11 May 2007 05:22:43 -0400 (EDT)
- Reply-to: hanlonr at cox.net
Actually, the problem was that the original boundary condition should have been
y[0]==1
rather than
y[0]==0
Bob Hanlon
---- Bob Hanlon <hanlonr at cox.net> wrote:
> It appears that you need a boundary condition at a time other than zero.
>
> Clear[soln];
> soln[t_]=y[t]/.
> DSolve[{y'[t] + a y[t] == DiracDelta[t],
> y[1] == Exp[-a]}, y[t], t][[1]]
>
> UnitStep[t]/E^(a*t)
>
>
> Bob Hanlon
>
> ---- Steffen Paul <steffen.paul at item.uni-bremen.de> wrote:
> > Hi
> > I tried to solve
> > DSolve[{y'[t] + \[Alpha] y[t] == DiracDelta[t], y[0] == 0}, y, t]
> >
> > and got
> >
> > -\[ExponentialE]^(-t \[Alpha]) (1 - HeavisideTheta[t])
> >
> > which is zero for t >0.
> >
> > The solution is correct but I expected somthing else:
> >
> > exp( - alpha t) UnitStep(t)
> >
> > which is zero for t <0 and which is also a solution.
> >
> > In engineering, these solutions are called impulse responses.
> >
> > The last solution is physically more usefull , because the system responds
> > after the excitation (DiracDelta).
> >
> >
> >
> > How can I force Mathematica to give only solutions with nonzero values for t
> > >0 ?
> >
> >
> >
> > Regards,
> >
> > Steffen
> >
> >
> >
> >
> >