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First Order ODE

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Find all solutions

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

    2. Relevant equations

    3. The attempt at a solution

    I know I have to use separation of variables because it isn't linear.

    so I get

    [tex] \frac{ydy}{(y^{2} - 1)^{\frac{3}{2}}} = \frac{1}{dxx^{2}}[/tex]

    Now I'm kinda stuck at how to integrate this. I think I'm supposed to use partial fraction expansion, but since there's the 3/2 exponent, I'm confused as to how to go about doing that. Tips???
  2. jcsd
  3. Oct 8, 2007 #2
    Check the right hand side of your equation to make sure the dx is in the right spot. Try substitution on the left
  4. Oct 8, 2007 #3
    Thanks, that helped. So now I substituted take the integral of both sides and I get

    [tex] x = (y^{2}-1)^{\frac{1}{2}} + c [/tex]

    This isn't matching up with the correct answer though. I took the integral and got -1/x on the left and 1 / - (y^2 - 1)^1/2 + c on the right. So I just took out the negative signs and changed to the reciprocal on the both sides, whatcha think?
  5. Oct 8, 2007 #4


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    Well that would be correct if the constant wasn't there but because it is...you just can't simply invert both sides

    if you brought [tex]\frac{1}{\sqrt{y^2-1}} +c [/tex] to the same base and then invert...that would be correct
  6. Oct 8, 2007 #5
    ahh I see. I got the correct answer the only thing that confuses me is that in the book the answer says [tex] y \equiv \pm 1 [/tex]... Actually I still don't really know what "equivalent" represents in this class.
  7. Oct 8, 2007 #6


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    well normally for the function to be continuous i thought it would be y not equal to +/- 1
  8. Oct 8, 2007 #7


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    In addition to the solutions you get by integrating, the equation has solutions where y' is identically equal to zero. These are them.
  9. Oct 8, 2007 #8
    So if if it didn't say " all solutions " I could just say that y=1, but what's the point of having the weird three line equal sign?
  10. Oct 8, 2007 #9


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    It just means y(x)=1 for all x. I.e. y is IDENTICALLY equal to one. Sure, if you don't need all solutions, then you could just pick an easy one.
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