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Re: Help w/ Mathematica: simple stuff

  • To: mathgroup at
  • Subject: [mg30792] Re: Help w/ Mathematica: simple stuff
  • From: "Orestis Vantzos" <atelesforos at>
  • Date: Wed, 19 Sep 2001 00:16:31 -0400 (EDT)
  • Organization: National Technical University of Athens, Greece
  • References: <9nh2a1$iu$>
  • Sender: owner-wri-mathgroup at

To begin with, your ODE is not of the type y'[x]==r*y[x], because of the
So the solution is not y[x]==k Exp[r x].
Assume that y[t] solves y'[t]==r*y[t] (equ1) and is thus equal to k Exp[r
Now define g[t]:=y[Log[t]].
g'[t]==y'[Log[t]]/t (chain rule)== r*y[Log[t]]/t (by equ1)==r g[t]/t
So g solves your ODE... g[t]==y[Log[t]]==k Exp[r Log[t]]==k t^r
Which is exactly the solution that Mathematica produced (non surprises

"Daita Mizohu" <sophtwarez at> wrote in message
news:9nh2a1$iu$1 at
> Hello all,
> I'm a mathematica novice and only doing basic calculus.  We have a
> question that is as follows:
> y[1]=a
> y'[x]=(r*y[x])/x
> Explain Mathematica output of...
> y[x]=ax^r
> I guess I don't understand what x is doing in the denominator.  The
> y[1]=a part is straight forward; when x is 1 then no matter what
> constant r is it will be 1.  I also know from my material that when
> y[x]=kE^rx then y'[x]=r*y[x].
> Thanks much,
> Daita, M.

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