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MathGroup Archive 2002

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Re: Why these graphs differ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg33959] Re: Why these graphs differ?
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Wed, 24 Apr 2002 01:22:31 -0400 (EDT)
  • References: <aa3gm6$80q$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Vladimir Bondarenko" <vvb at mail.strace.net> wrote in message
news:aa3gm6$80q$1 at smc.vnet.net...
> These solutions, naturally, coincide.
>
>          DSolve[{y'[z] == z, y[0] == 1}, y[z], z][[1, 1, 2]]
> Evaluate[DSolve[{y'[z] == z, y[0] == 1}, y[z], z][[1, 1, 2]]]
>
> (2 + z^2)/2
> (2 + z^2)/2
>
>
> But, surprisingly, the corresponding graphs are not identical:
>
> Plot[         DSolve[{y'[z] == z, y[0] == 1}, y[z], z][[1, 1, 2]],  {z, 0,
1}]
> Plot[Evaluate[DSolve[{y'[z] == z, y[0] == 1}, y[z], z][[1, 1, 2]]], {z, 0,
1}]
>
>
> Is it a feature or a problem?
>
>
> Vladimir Bondarenko
>

Vladimir,
With
    Plot[ DSolve[{y'[z] == z, y[0] == 1}, y[z], z][[1, 1, 2]],  {z, 0, 1}]
numerical values are successively assigned to z and Mathematica tries to
solve; however
DSolve cannot work with, for example z=0.5, hence the warning messages that
you get.
You are in fact plotting z.
For example

    z=.5;

    DSolve[{y'[z]==z,y[0]==1},y[z],z][[1,1,2]]

        DSolve::dsvar: 0.5` cannot be used as a variable.

        0.5

--
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





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