How do you isolate for y when 0 = 2y + e^y

  1. How do you isolate for y when you have the equation 0 = 2y + e^y - 4x + 3?

    Any tips, useful links or solutions and an explanation would be greatly appreciated!

    Thanks!
     
  2. jcsd
  3. Numerics seems to be the only way. Usually when you have to isolate a variable that's acted on by different types of functions, it's very difficult or impossible to do analytically.
     
  4. :(

    10char
     
  5. In a lot of situations, it's not necessary to get analytic solutions. What is this for?
     
  6. HallsofIvy

    HallsofIvy 40,391
    Staff Emeritus
    Science Advisor

    The "Wolfram Alpha" solution that johnqwertyful links to use the "Lambert W function" which is defined as the inverse function to [itex]f(x)= xe^x[/itex]. It cannot be written in terms of any simpler function.
     
  7. Well, when we have these mixed exponential equations, we try to put it in Lambert-W form. That is, in the form:

    [tex]g(x,y)e^{g(x,y)}=h(x)[/tex]

    Then by definition of the Lambert W function which you can look up, we take the W function of both sides and obtain:

    [tex]g(x,y)=W(h)[/tex]

    Now, doing a little moving around of your equation:

    [tex]1/2 e^y=2x-3/2-y[/tex]
    [tex]1/2 e^{2x-3/2}=e^{-y} e^{2x-3/2}(2x-3/2-y)[/tex]

    or:

    [tex](2x-3/2-y)e^{2x-3/2-y}=1/2 e^{2/x-3/2}[/tex]

    I'll let you finish it to isolate y in terms of the perfectly valid (multi-valued) function of x in terms of the Lambert W-function.
     
    Last edited: Sep 13, 2013
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