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Re: Re: PDE's with DSolve

  • To: mathgroup at
  • Subject: [mg6055] Re: [mg6045] Re: [mg5983] PDE's with DSolve
  • From: sherod at boussinesq.Colorado.EDU (Scott Herod)
  • Date: Sun, 16 Feb 1997 01:11:57 -0500
  • Organization: /usr/local/lib/rn/organization
  • Sender: owner-wri-mathgroup at

Perhaps I am misunderstanding your comments.  If so please forgive my

When I say "Solve the differential equation F(x,y(x),y'(x)) = 0"
I mean find the function y(x) such that F(x,y,y') is always zero on some
set (interval) of x's.  For many functions F there are standard ways
to find such functions y(x).  

In the case of the problem that I asked Mathematica to solve (in easy

  ---- y(x,t) = 0   (or   y_{tt} = 0}

the true solution is y(x,t) = a(x) + b(x) * t
where a and b are arbitrary functions of x.

Mathematica on the otherhand returns  y(x,t) = a(x) + b(t)  which
is wrong.  

If you ask Mma to solve y_{xt} = 0 it appears to return
y = k x t + a(x) + b(t).  I say appears because the actual result is

{{y[x[1],x[2]]\[Rule]-DSolve`DSolveDump`b$12 x[1] x[2]+C[1][x[1]]+C[2][x[2]]}}
                     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^  (??)

At any rate this is also not correct.

Scott Herod
sherod at


In article <5e0vt8$dlr at> "Christopher R. Carlen" <crobc at> writes:
>Scott Herod wrote:
>> I was happy to see that DSolve can solve some quasi-linear partial
>> differential equations and was happily playing with it.  I then tried
>> the following:
>> DSolve[D[y[x,t], {t,2}] == 0, y[x,t], {x,t}].
>> Rather distressing.
>> Scott Herod
>Wouldn't it be necessary to have previously defined y[x,t] to be some
>explicit function?
>In other words, I cannot possibly solve even
>F(x,y,y')=0  without having knowledge of the function y(x) .
>Once y(x) is defined, then I can consider the means to find a solution.
>Christopher R. Carlen
>crobc at
>carlenc at

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