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.
>