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Re: Help w/ Mathematica: simple stuff

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  • Subject: [mg30876] Re: Help w/ Mathematica: simple stuff
  • From: "Warren Wolfe" <wwolfe18 at>
  • Date: Fri, 21 Sep 2001 04:04:14 -0400 (EDT)
  • References: <9nh2a1$iu$>
  • Sender: owner-wri-mathgroup at

"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.
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[1]==a},y[x],x] and all is well.

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