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Lambert W function

  1. Jul 8, 2014 #1
    Dear scholars,

    I am working on the following equation and wonder whether my solution is correct. The actual problem is more complex but the example below captures the main features. I have not a lot of experience with the Lambert W function. Thanks in advance for your comments!

    Equation:
    $$Y=x \exp(x^2)$$

    Substitute: $$u=x^2$$

    Then:
    $$Y=\sqrt{u} \exp(u)$$
    $$Y^2=u \exp(2u)$$
    $$2Y^2=2u \exp(2u)$$

    Then using W function:
    $$2u=W(2Y^2)$$
    From which $$x$$ follows.
     
  2. jcsd
  3. Jul 8, 2014 #2
    Correct.
     
  4. Jul 8, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, very good!
     
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