       Re: Help w/ Mathematica: simple stuff

• To: mathgroup at smc.vnet.net
• Subject: [mg30792] Re: Help w/ Mathematica: simple stuff
• From: "Orestis Vantzos" <atelesforos at hotmail.com>
• Date: Wed, 19 Sep 2001 00:16:31 -0400 (EDT)
• Organization: National Technical University of Athens, Greece
• References: <9nh2a1\$iu\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```To begin with, your ODE is not of the type y'[x]==r*y[x], because of the
x^(-1).
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
t].
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
there..)
Orestis

"Daita Mizohu" <sophtwarez at yahoo.com> wrote in message
news:9nh2a1\$iu\$1 at smc.vnet.net...
> Hello all,
>
> I'm a mathematica novice and only doing basic calculus.  We have a
> question that is as follows:
>
> y=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=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.
>

```

• Prev by Date: Re: Combinations
• Next by Date: Two plots, two ranges with a Stress-Strain diagram
• Previous by thread: Re: Help w/ Mathematica: simple stuff
• Next by thread: Re: Help w/ Mathematica: simple stuff