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Another Straight Forward DE

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Find a 1 parameter family of solutions for the differential equation yx^2 dy - y^3 dx = 2x^2 dy

    3. The attempt at a solution

    yx^2 dy - y^3 dx = 2x^2 dy
    x^2 dy - y^2 dx = 2x^2/y dy
    dy - y^2/x^2 dx =
    - y^2/x^2 dx = 2/y dy - dy
    Divide all by y^2
    dx/x^2 = (2dy/y - dy ) * -1/y^2
    dx/x^2 = -2dy/y^3 + dy/y^2
    Integrate
    -1/x = y^2 - y + C

    Where y ≠ 0 & x ≠ 0

    While the book gives an answer of
    (cx+1)y^2 = (y -1)x
    with the same domain
     
  2. jcsd
  3. Jul 9, 2012 #2

    BruceW

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    dx/x^2 = -2dy/y^3 + dy/y^2
    This line was correct, but then integrating this does not lead to your next line:
    -1/x = y^2 - y + C
    I think it was maybe a mistake? I've done similar mistakes loads of times.
     
  4. Jul 9, 2012 #3

    SammyS

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    Check the integration for y.

    What is [itex]\displaystyle \int (-2y^{-3}+y^{-2})\,dy\ ?[/itex]
     
  5. Jul 9, 2012 #4
    Thank you Bruce & Sammy. It seems mental integration isn't always reliable. :devil:

    So my new answer is -1/x = y^2 - y^-1 + C

    But I still don't have the answer in the book (cx+1)y^2 = (y -1)x

     
  6. Jul 9, 2012 #5

    SammyS

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    Hopefully, you mean -1/x = y-2 - y-1 + C.
     
  7. Jul 9, 2012 #6
    Hmm, forgot to fix that too!
     
  8. Jul 10, 2012 #7

    SammyS

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    Does that mean you now have the book's answer?
     
  9. Jul 10, 2012 #8
    Yes! By distributing the variables within parentheses and then dividing by x*y^2.

    Thank you. :approve: I am not sure why the book chose to put it in that form though.
     
  10. Jul 10, 2012 #9

    BruceW

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    Nice work! The book might have chosen that form to test rearranging skills. Also, there are no reciprocals which is always nice.
     
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