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[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. >