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What is the interval for the particular solution?

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data
    dy/dx = (9x²)/(9y²-11) , y(1)=0

    2. Relevant equations


    3. The attempt at a solution
    I've found the particular solution:

    3y³-11y = 3x³-3

    The solution is defined for -⅓√11<y<⅓√11,
    What about the x values? Plugging the y values into the equation doesn't work of course..
     
  2. jcsd
  3. Feb 2, 2016 #2

    RUber

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    It looks to me that your solution is defined everywhere...how did you conclude the interval above?
     
  4. Feb 2, 2016 #3

    RUber

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    I see now...since you start with y=1, there is no way to continuously make it outside that interval -- where ##9y^2 - 11 = 0##.
    To solve for the domain of x, I would recommend finding the range (max, min) of the function of y over the interval of definition. This would give you a domain defined by:
    ## 3x^3 - 3 \in \left[ \min_{y \in ( -\sqrt{11}/3, \sqrt{11}/3)} \{3y^3 -11\} , \max_{y \in ( -\sqrt{11}/3, \sqrt{11}/3)} \{3y^3 -11\}\right].##
     
  5. Feb 2, 2016 #4
    Yes I understand all of this, however my assignment wants me to show the interval of existence like this:

    A < x < B

    That is, to find the domain, not the range. How can I do that?
     
  6. Feb 2, 2016 #5
    I don't think actual values are possible here? Only approximations of "x" for values greater by closest to -⅓√11 and smaller but closest to ⅓√11...
     
  7. Feb 2, 2016 #6

    RUber

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    The function of y, ##f(y) = 3y^3-11y## has its own domain and range. The domain is ##-\sqrt{11}/3 < y < \sqrt{11}/3. ## What is its range?
    If ##g(x)=3x^3 - 3##, then the range of f(y) must equal the range of g(x). Then you find the domain of x that restricts g to the range you have found.
    For example if the minimum of f(y) is -10, you would say:
    ## g(x) > -10 \\ 3x^3 - 3 > -10 \\ x^3 > -7/3 \\ x > (-7/3)^{1/3}##
     
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