       Re: Help w/ Mathematica: simple stuff

• To: mathgroup at smc.vnet.net
• Subject: [mg30876] Re: Help w/ Mathematica: simple stuff
• From: "Warren Wolfe" <wwolfe18 at home.com>
• Date: Fri, 21 Sep 2001 04:04:14 -0400 (EDT)
• References: <9nh2a1\$iu\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```"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.
>
y[x] = a*x^r IS the correct solution. To verify this calculate y'[x] and you
get r*a*x^(r-1) which is
the same as (r*a*x^r)/x  which is r*y[x]/x   Note that E^x plays no roll in
the solution of this diffeq.

Just execute  DSolve[{y'[x]==(r*y[x])/x,y==a},y[x],x] and all is well.

```

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