Re: Help w/ Mathematica: simple stuff
- To: mathgroup at smc.vnet.net
- Subject: [mg30782] Re: Help w/ Mathematica: simple stuff
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Wed, 19 Sep 2001 00:16:22 -0400 (EDT)
- References: <9nh2a1$iu$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Daita, Perhaps what you want is DSolve[{y'[x]==(r*y[x])/x, y[1]==a},y,x] {{y -> Function[{x}, a*x^r]}} y = y/.%[[1]] Function[{x}, a*x^r] y[z] a*z^r Let's look at at your original code Clear[y] y[1]=a; y'[x]=(r*y[x])/x; This instructs Mathematica that y[1] should be replaced by a and that y'[x] should be replaced by (r*y[x])/x. It does not follow that y'[z] will be replaced by (r*y[z])/z. (here you can replace z with anything other than x). To get this you need y'[x_]=(r*y[x])/x; Now we get y'[z] (r*y[z])/z But still we get y[x] y[x] Mathematica does not integrate without being told to do so. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Daita Mizohu" <sophtwarez at yahoo.com> wrote in message news:9nh2a1$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. >