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How to solve this equation?

  1. Sep 13, 2011 #1
    Hello All,

    Could you please indicate me how to resolve this equation:

    ln(y) (x^4 - y^4)+ ln(x) (y^4 - x^4) + (y^2 - x^2)^2 = 0

    I am really struggling with it....

    Thanks in advance,

    C:)
     
  2. jcsd
  3. Sep 13, 2011 #2

    Mentallic

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    What exactly do you mean by solving it? What do you need to do?
     
  4. Sep 13, 2011 #3
    Mentallic,

    Thanks a lot for your reply.
    I mean, I want to know the value of "X" when "Y=5".

    Thanks in advance,
     
  5. Sep 13, 2011 #4

    chiro

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    Hey CIMP and welcome to the forums.

    Do you know about root finding algorithms?
     
  6. Sep 13, 2011 #5

    Mentallic

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    Then what you're looking for is to solve explicitly for x. However, you can't do that with the use of elementary functions (the usual - addition, multiplication, square roots, powers, trig, exponentials). You will need to find a numeric approximation to the answer.

    There are trivial solutions however such as y=x and y=-x, but notice that with logarithms present in the equation, this means you need to restrict x and y to be more than zero, so only y=x, x>0 is a trivial solution.

    So, if y=5 then x=5 as well, but that's not the only solution. The others can be found by numeric methods.
     
  7. Sep 13, 2011 #6
    How about this, to get going:

    x^4-y^4=-(y^4-x^4)

    So you could write (ln(y)-ln(x))(x^4-y^4)+(y^2-x^2)^2=0
    ln(y/x)(x^4-y^4)+(y^2-x^2)^2=0
    ln(y/x)(x^4-y^4)=-(y^2-x^2)^2
    ln(y/x)(x^2-y^2)(x^2+y^2)=-(y^2-x^2)^2
    ln(y/x)(x^2+y^2)=(y^2-x^2)
    ln(y/x)=(y^2-x^2)/R^2

    Where R is the distance from the origin.

    I don't know if this helps at all- it may give you a better idea of what the graph looks like though :uhh:
     
  8. Sep 14, 2011 #7
    Hello Mentallic,
    Thank you very much for your help. I really could advance more with your explanation. I made a plot as suggested by Jamma, and I could see that the function when X<5 it tend to -infinit and if X>5 it also tends to +infinit.

    Hello Jama,
    Thank you very much for your advice. It is vey useful to make a plot of the function before going to more sophisticated things!

    Hello Chiro,
    I use Maxima and Matlab and it was the first time that I was looking how to extract the root algoritms with them, but it was quite harder to use it... :bugeye:In fact, I pass all the day of yesterday dealing with that. Then I give up and I ploted the function and it was easy easier to see that it has just one solution when X=5. By lucky!
    But I need to learn how to use the software....

    Thanks everybody. I am so happy to be in this forum. Thansk again for your time and help...

    CIMP
     
  9. Sep 14, 2011 #8

    Mentallic

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    Not quite. Even though ln(y/x) is real for x,y<0, the original equation was of the form ln(y)-ln(x) in which case x,y<0 is NOT valid. You need to keep in mind that whatever transformations you do, you only consider the domain of the original function, not your manipulated function.
     
  10. Sep 14, 2011 #9

    Char. Limit

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    Actually, I'm going to have to disagree with you here. It looks to me as if x=5 IS the only solution corresponding to y=5.
     
  11. Sep 14, 2011 #10

    Mentallic

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    Ahh yes they're in fact double roots. Thanks for spotting it!

    [tex]\ln{y}(x^4-y^4)+\ln{x}(y^4-x^4)+(y^2-x^2)^2=0[/tex]

    [tex](y^2-x^2)\left(\ln{\frac{1}{y}}(x^2+y^2)+\ln{x}(x^2+y^2)+y^2-x^2\right)=0[/tex]

    The first factor corresponds to the roots of [itex]y=\pm x[/itex] thus only [itex]y=x, x>0[/itex] to satisfy the domain, and the second factor is

    [tex]\ln{\frac{x}{y}}(x^2+y^2)=x^2-y^2[/tex]

    Which gives us the same roots as the other factor, since ln(1)=0.
     
  12. Sep 15, 2011 #11
    Yes! :smile:
    Thanks!!!!

    "Mathematics is the language of nature..."
     
  13. Sep 15, 2011 #12
    thanks, i hat the same question
     
  14. Sep 15, 2011 #13
    So we've just solved all your homework for you guys, pfft.
     
  15. Sep 15, 2011 #14

    Mentallic

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    CIMP, in the future, please post your questions in the homework help section.
     
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