       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==a},y,x]

{{y -> Function[{x}, a*x^r]}}

y = y/.%[]

Function[{x}, a*x^r]

y[z]

a*z^r

Let's look at at your original code

Clear[y]

y=a;
y'[x]=(r*y[x])/x;

This instructs Mathematica that y 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=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.
>

```

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