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Homework Help: Solve this equation

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve for x and y:
    x2 + (√8)x*sin[(√2)xy] +2 = 0

    2. Relevant equations



    3. The attempt at a solution
    Other than decomposing the root 8 i don't know what else to do. any hints? thanks.
     
  2. jcsd
  3. Feb 19, 2012 #2

    ehild

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    Hi kscplay,

    Transform the equation into the form (x+a)2+b=0

    ehild
     
  4. Feb 20, 2012 #3

    Mark44

    Staff: Mentor

     
  5. Feb 20, 2012 #4

    ehild

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    I meant (x+√2sin[√2xy])2=-2+2sin2[√2xy]).

    The left hand side can not be negative. The right hand side ?...


    ehild
     
  6. Feb 20, 2012 #5
    this might be a silly question, but wouldn't you first isolate y?
     
  7. Feb 20, 2012 #6

    ehild

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    That could be also a way to find the solution. But I would look at the right hand side of my equation and see if it can be positive.


    ehild
     
    Last edited: Feb 20, 2012
  8. Feb 20, 2012 #7

    ehild

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    If we isolate y we get [tex]y=\frac{1}{\sqrt{2}x}\arcsin(-\frac{x^2+2}{2\sqrt{2}x})[/tex] The argument of arcsin has to be in the range [-1,1]. What does it mean for x?

    My method of transforming the equation to [itex](x+\sqrt{2}\sin(\sqrt{2}xy)^2=-2+2sin^2(\sqrt{2}xy)[/itex] shows that the left hand side can not be negative, the right-hand side can not be positive so both have to be zero.

    [tex]x+\sqrt{2}\sin(\sqrt{2}xy)=0[/tex]
    [tex]\sin^2(\sqrt{2}xy)=1 [/tex]

    two equations, two unknowns...

    ehild
     
    Last edited: Feb 21, 2012
  9. Feb 21, 2012 #8

    Mentallic

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    That's genius! I'm glad I didn't miss out on reading this :smile:
     
  10. Feb 21, 2012 #9

    ehild

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    :smile:
    There was a typo , I corrected it in post #7. The square should not be there in my first equation. Correctly it is [tex]x+\sqrt{2}\sin(\sqrt{2}xy)=0[/tex]
    ehild
     
    Last edited: Feb 21, 2012
  11. Feb 21, 2012 #10
    That's a smart idea. Thanks ehild :)
     
  12. Feb 22, 2012 #11

    ehild

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    You are welcome, kscplay.:smile:

    The other way, jkristia suggested would work too.

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

    The magnitude of the argument of the arcsin function can not exceed 1:
    [tex] \left| \frac{x^2+2}{2 \sqrt{2} x} \right| \leq 1[/tex]

    The function

    [tex]
    \frac{x^2+2}{2\sqrt{2}x}[/tex] has extrema at x=±√2, minimum (1) for positive x and maximum (-1) for negative x. Only these values are allowed for x.

    ehild
     

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    Last edited: Feb 22, 2012
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