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Re: Partial Derivatives
- To: mathgroup at smc.vnet.net
- Subject: [mg7927] Re: [mg7774] Partial Derivatives
- From: jpk at max.mpae.gwdg.de
- Date: Wed, 23 Jul 1997 15:46:07 -0400
- Sender: owner-wri-mathgroup at wolfram.com
Hi
> I am trying to find partial derivatives of
> an equation, eg y''[x] + y'[x] + x^2 ==0.
> What I would like is to say something like pD[eqn,y[x]] and get back 0
> while pD[eqn,y''[x]] will give me 1. Certainly D does this.
What do You want ?
Coefficient[eqn[[1]],y''[x]] gives 1
and
Coefficient[eqn[[1]],y[x]] gives 0
It is a miss use of derivatives to find the coefficients.
> However,
> I also want pD[eqn,x] to give me 2x and not y'''[x] + y''[x] + 2 x.
> Can anyone please suggest a way to get around this?
Yes, invent a new kind of differential calculus. Newton and Leibniz have
it done -- just try it again.
D[ y''[x] + y'[x] + x^2,x] == y'''[x] + y''[x] + 2 x
is correct in the common sense of differntial calculus. How ever it is up
to You to invent a new one.
>
> Note that entering the equation as y'' + y' + x^2 does not work. Here
> D[eqn,y] gives me the strange result y''' + y'' which I really cannot
> understand.
If the input is strange the result of any operation on it will probably
not become better. What is y' ?
d y
--- (??)
d ?
You can not made a derivative on a symbol with respect to nothing. You can
look for the total derivative with Dt[y] of a symbol.
Hope that helps
Jens
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