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