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Math Help - First Order Question

  1. #1
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    Oct 2009
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    14

    First Order Question

    If i have:

    (3y^2+2xy)dx-(2xy+x^2)dy=0

    I firstly tried to solve this with Exact equations method. The partials I found were not equal therefore i assumed an integrating factor as a function of x and y. For both i did not come up with anything, as I got the factors as functions of both and x and y.

    Then I proceeded to use the Homogenous Substituion v= y/x.
    This lead me to getting a solution 2y+x=C.

    I'm not sure if my approach is correct. I know for homogenous whenever the form y/x is recognizable, we proceed with this approach. But when is it correct to use the Exact Equations method?

    Thanks.
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  2. #2
    Member
    Joined
    May 2009
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    New Delhi
    Posts
    153
    you are in right direction .
     put\ y= vx \quad \Rightarrow \frac {dy}{dx}=v+x \frac {dv}{dx}
    then DE  (3y^2+2xy)dx=(2xy+x^2)dy  \Rightarrow \frac{dy}{dx}=\frac{3y^2+2xy}{2xy+x^2}
    reduce to
     \frac {2v+1}{v(v+1)}dv= \frac{1}{x} dx
    resolve   \frac {2v+1}{v(v+1)} into partial fraction
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