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E^y + e^-y =2x or e^2y - 2xe^y + 1 = 0 equation

  1. Sep 16, 2005 #1
    In this textbook i am looking at it says:

    "Thus e^y + e^-y =2x or

    e^2y - 2xe^y + 1 = 0"

    how did they go from the first to the second part?
     
  2. jcsd
  3. Sep 16, 2005 #2

    Hurkyl

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    Well, look at pieces of the equation and see if that gives you any clues.

    For example, their second equation has an e^(2y) in it1. Can you think of anything to do to the first equation so that the result will have an e^(2y) in it?

    1: I assume you meant e^(2y) and not e^2y (which is the same as (e^2)y)
     
  4. Sep 16, 2005 #3
    You should really try to figure this out yourself. What's the difference between the two equations?
     
  5. Sep 19, 2005 #4
    i still dont get how to go from

    [tex]e^y + exp(-y)=2x[/tex]

    to

    [tex]e^2y - 2xe^y + 1 = 0[/tex]

    help please
     
  6. Sep 19, 2005 #5
    I'm not sure, but that y should be raised too....e^(2y) not (e^2)y
     
  7. Sep 19, 2005 #6

    Hurkyl

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    Have you tried any of our hints?
     
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