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MathGroup Archive 2007

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Re: New to mathematica: Question about solving

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74694] Re: [mg74664] New to mathematica: Question about solving
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sat, 31 Mar 2007 01:39:19 -0500 (EST)
  • Reply-to: hanlonr at cox.net

I must have evaluated out of order. Change this to read

soln=DSolve[{y'[x] == y[x] + y[x]^3}, y[x], x]

{{y[x] -> -((I*E^(x + C[1]))/
      Sqrt[-1 + E^(2*x + 2*C[1])])}, 
  {y[x] -> (I*E^(x + C[1]))/Sqrt[-1 + E^(2*x + 2*C[1])]}}

soln=soln/.
        expr1_*(expr2_)^(-1/
        2):>(expr2/(expr1^2))^(-1/2)/.C[1]:>-Log[C]/2//ExpandAll

{{y[x] -> 1/Sqrt[C/E^(2*x) - 1]}, 
  {y[x] -> 1/Sqrt[C/E^(2*x) - 1]}}


Bob Hanlon

---- Bob Hanlon <hanlonr at cox.net> wrote: 
> soln=DSolve[{y'[x] == y[x] + y[x]^3}, y[x], x]
> 
> {{y[x] -> -((I*E^(x + C[1]))/
>       Sqrt[-1 + E^(2*x + 2*C[1])])}, 
>   {y[x] -> (I*E^(x + C[1]))/Sqrt[-1 + E^(2*x + 2*C[1])]}}
> 
> soln=soln/.{
>         expr1_*(expr2_)^(-1/2):>(expr2/(expr1^2))^(-1/2),
>         C[1]:>-Log[C]/2}//ExpandAll
> 
> {{y[x] -> 1/Sqrt[C/E^(2*x) - 1]}, 
>   {y[x] -> 1/Sqrt[C/E^(2*x) - 1]}}
> 
> 
> Bob Hanlon
> 
> ---- traz <t_raz at yahoo.com> wrote: 
> > Whenever I try to solve differential equations in mathematica, I get a solution with an imaginary part different from the solution in a text book. For example:
> > 
> > DSolve[{y'[x] == y[x] + y[x]^3}, y, x]
> > 
> > will give me a solution that has an imaginary part and not the one I expect here from the text book:
> > 
> > {+(Ce^(-2x)-1)^(-1/2), -(Ce^(-2x)-1)^(-1/2)}
> > 
> > Can anyone give me a tip on how to do this? Also does anyone know of an online tuttorial that goes into details a little bit?
> > 



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