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