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Math Help - Exact differential equation

  1. #1
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    Exact differential equation

    hello all

    I am having problem understanding this:
    i solved (a) :
    (4x^3 y^2 + 3x^2 y^4)dx+(2x^4 y + 4x^3 y^3)dy=0
    and got this:
    x^4 y^2 +x^3 Y^4 = c
    the question is (b) to solve for IC :
    y(0)=0
    and give all the solution. what is there beside x=0 ; y=0 ??
    i think maybe x=-y^2 but I'm not sure
    thanks for the help
    mormor83
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  2. #2
    MHF Contributor Calculus26's Avatar
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    From the IC:

    c= 0

    Therfore you would think

    x^4 y^2 + x^3 y^4 = 0 is the solution

    However

    dy/dx = (4x^3 y^2 + 3x^2 y^4)/2x^4 y + 4x^3 y^3)

    reduces to:

    dy/dx = (4x^3 y + 3x^2 y^3)/2x^4 + 4x^3 y^2)

    which has the equiilibrium solution y = 0 as well

    Another way of looking at this is

    x^4 y^2 + x^3 y^4 = 0

    y^2( x^4 +x^3y^2) =0

    which yields y = 0 or x = -y^2 as you suggested.

    Note

    dy/dx = (4x^3 y + 3x^2 y^3)/2x^4 + 4x^3 y^2)

    does not satisfy the conditions of the uniqueness theorem.
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  3. #3
    Senior Member Peritus's Avatar
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    Nov 2007
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    This is the resulting equation after applying the IC:

    <br />
x^3 y^2 \left( {x + y^2 } \right) = 0<br />

    so all the possible solutions are on the following curve:

    <br />
y = \left\{ {\left. { \pm \sqrt { - x} } \right|x \geqslant 0} \right\}<br />

    which is a canonical parabola rotated 90 degrees to the left.
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  4. #4
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    thank you very much for that. ( just as i thought but i was not sure)
    by the way what do you use to write in "math font"?
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