       Re: How to simplify?

• To: mathgroup at smc.vnet.net
• Subject: [mg96218] Re: How to simplify?
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Mon, 9 Feb 2009 05:37:29 -0500 (EST)
• Organization: The Open University, Milton Keynes, UK
• References: <gmndih\$s09\$1@smc.vnet.net>

```In article <gmndih\$s09\$1 at smc.vnet.net>,
Aaron Fude <aaronfude at gmail.com> wrote:

> I'm sorry for totally belaboring this point, but I am having a hard
> time getting Mathematica be useful for me in this one respect. The
> following code shows that the linear ODE that I am trying to solve has
> 1/2 r BesselJ[1, r] as the particular solution. DSolve, however,
> returns an answer that I'm sure is correct. I tested it - numerically,
> the particular part is exactly 1/2 r BesselJ[1, r].
>
> But for someone who is looking for analytical insight, the answer is
> not useful. What can be done to simplify the expression so that it
> appears as 1/2 r BesselJ[1, r]
>
>
> S[r_] := 1/2 r BesselJ[1, r];
> D[S[r], {r, 2}] + 1/r D[S[r], r] + S[r] // FullSimplify
> DSolve[y''[r] + 1/r y'[r] + S[r] == BesselJ[0, r], y[r],
>   r] // FullSimplify

[snip]

What output did you get on your system, which is...? Here is what I get
on mine:

In:= S[r_] := 1/2 r BesselJ[1, r];
D[S[r], {r, 2}] + 1/r D[S[r], r] + S[r] // FullSimplify
DSolve[y''[r] + 1/r y'[r] + S[r] == BesselJ[0, r], y[r],
r] // FullSimplify

Out= BesselJ[0, r]

Out= {{y[r] -> -1 + 1/2 r BesselJ[1, r] + C + C Log[r]}}

In:= {\$Version, \$ReleaseNumber}

Out= {"6.0 for Mac OS X x86 (64-bit) (May 21, 2008)", 3}

Regards,
--Jean-Marc

```

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