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Re: Simple ODE with time-dep BC

  • To: mathgroup at
  • Subject: [mg78015] Re: Simple ODE with time-dep BC
  • From: "Kevin J. McCann" <Kevin.McCann at>
  • Date: Thu, 21 Jun 2007 05:48:39 -0400 (EDT)
  • Organization: University of Maryland, Baltimore County
  • References: <f5as8d$9kj$>

Well, first off you did not enter the first DE correctly. It should read

DSolve[{h'[t]==-a*h[t]+c/h[t], h[0]==0}, h[t], t]

Note the h[t] in the c/h[t].

As for the time dependent one you put u[t]==v^2. Mathematica thinks the RHS
is a constant, and you can see why it doesn't make sense then. I.e. 
u[t]==constant implies the u'[t]=0. So, I guess that I don't understand 
your question.


Apostolos E. A. S. Evangelopoulos wrote:
> Hello all,
> I'm having 2 problems with solving the following ordinary differential equation:
> h*h'[t]==-a*h^2+c
> Problem 1:
> Mathematica doesn't like the form of the above:
> DSolve[{h'[t]\[Equal]-a*h[t]+c/h, h[0]\[Equal]0}, h[t], t]
> returns
> DSolve::dvnoarg : The function h appears with no arguments. More...
> If, now, I reduce the above equation to
> DSolve[{1/2*u'[t]\[Equal]-a*u[t]+c,u[0]\[Equal]0},u[t],t]
> (by substitution of u=h^2)
> then Mathematica solves for u[t] without complaints. The thing is, though, I could solve the latter by hand, really, so what's the point? Is Mathematica not supposed to be able to solve the equation in the first form, really (it appears to be non-linear, but intrinsically it's not)?
> Problem 2:
> I 'd like to impose a time dependent boundary condition, so, instead of h[0]==0, something like h[t]==v (some constant), or, the more complicated, h[t]==t. How do I solve that? I have tried as follows, with the reduced u[t] form:
> DSolve[{u'[t]/2\[Equal]-a*u[t]+c,u[t]\[Equal]v^2},u[t],t]
> resulting in
> DSolve::overdet : The system has fewer dependent variables than equations, so is overdetermined. More...
> Thank you, all, for your help, in advance.
> Apostolos

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