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Math Help - [SOLVED] Tricky First Order ODE

  1. #1
    Junior Member
    Joined
    Mar 2010
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    51

    [SOLVED] Tricky First Order ODE

    y\left( y^{3}\; -\; x \right)dx\; +\; x\left( y^{3}\; +\; x \right)dy\; =\; 0

    Can't seem to find the family of solutions...

    Any ideas?
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  2. #2
    Junior Member
    Joined
    Mar 2010
    Posts
    51
    Finally got it!

    Regroup the original equation to get:

    y^{3}\left( ydx\; +\; xdy \right)\; +\; x^{2}dy\; -\; xydx\; =\; 0
    or:
    y^{3}\left( xy \right)'\; +\; x^{2}dy\; -\; xydx\; =\; 0

    Divide through by [Math]y^{3}[/tex] and eventually get it into the form:

    2\left( xy \right)'\; -\; \left( \frac{x^{2}}{y^{2}} \right)'\; =\; 0

    And a final family of solutions in the form of:

    2xy^{3}\; -\; x^{2}\; =\; cy^{2}
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