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Thread: First Order Question

  1. #1
    Oct 2009

    First Order Question

    If i have:


    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?

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