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Re: Help w/ Mathematica: simple stuff

  • To: mathgroup at
  • Subject: [mg30782] Re: Help w/ Mathematica: simple stuff
  • From: "Allan Hayes" <hay at>
  • Date: Wed, 19 Sep 2001 00:16:22 -0400 (EDT)
  • References: <9nh2a1$iu$>
  • Sender: owner-wri-mathgroup at


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]



Let's look at at your original code



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


Now we get



But still we get



Mathematica does not integrate without being told to do so.
Allan Hayes
Mathematica Training and Consulting
Leicester UK
hay at
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Daita Mizohu" <sophtwarez at> wrote in message
news:9nh2a1$iu$1 at
> 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.

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