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MathGroup Archive 2007

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Re: Substitutions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77940] Re: Substitutions
  • From: Albert <awnl at arcor.net>
  • Date: Wed, 20 Jun 2007 05:31:17 -0400 (EDT)
  • References: <f58d00$886$1@smc.vnet.net>

Hi,

your problem description is quite vague, it is usually better to show 
mathematica code in input form, what you have tried and what is wrong 
with the result you achieved...

anyway, in situations like yours, I usually try to solve for one of the 
variables and replace only that. Then usually everything cancels 
automatically, and if not I massage the result with Simplify and other 
commands until I'm happy, I am using a very simple example here, but it 
shows the idea:
In[39]:= expr = a*(b/c)*Derivative[1][y][x] + (b/c)*y[x] == 0

Out[39]= (b*y[x])/c + (a*b*Derivative[1][y][x])/c == 0

In[43]:= expr /. {a*(b/c) -> d}

Out[43]= (b*y[x])/c + d*Derivative[1][y][x] == 0

In[44]:= Solve[a*(b/c) == d, b]

Out[44]= {{b -> (c*d)/a}}

In[45]:= expr /. {b -> (c*d)/a}

hth,

albert


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