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MathGroup Archive 1997

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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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