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Math Help - First Order Differential Equation (Homogeneous?)

  1. #1
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    First Order Differential Equation (Homogeneous?)

    So I'm stuck on:

    (1-x)y' = 1-y

    I think it's homogeneous but even that I'm not sure of. (I'm pretty new to first and second order ODEs so I'm finding it really hard at the moment!)

    I simply took the (1-x) to the other side to make it an equation in y'(x) but I have no idea on what to do now.

    I know for homogeneous equations you can say that y=xv and let
    y'(x) = xdv/dx + v for example but I can't get that to work here (suggesting that I might be wrong in thinking that it's a homogeneous equation....).

    Any help is really appreciated!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    It is not homogeneus, it is a separated variables equation.
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  3. #3
    Super Member girdav's Avatar
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    It's not a homogeneous equation because y=0 is not a solution.
    You can first solve the equation (1-x^2)y'=y^2 (which is homogeneous) then try to find a particular solution, which is not difficult.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by girdav View Post
    It's not a homogeneous equation because y=0 is not a solution.
    You can first solve the equation (1-x^2)y'=y^2 (which is homogeneous) then try to find a particular solution, which is not difficult.

    More directly:

    (1-x^2)dy=(1-y^2)dx\Leftrightarrow \dfrac{dy}{1-y^2}=\dfrac{dx}{1-x^2}\Leftrightarrow\displaystyle\int \dfrac{dy}{1-y^2}=\displaystyle\int\dfrac{dx}{1-x^2}\Leftrightarrow \ldots
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