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Math Help - Identifying First Order Diff Equations

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    Identifying First Order Diff Equations

    So the equation is (xy+y)y' = x-xy
    It says to identify is the equation is Bernoulli, Separable, Linear and Homogeneous
    If these definitions are wrong on how to identify which is which, please let me know, because i researched this because my professor doesn't teach well.
    Bernoulli - y' + P(x)y = Q(x)y^n
    Separable - y' = p(x)h(y)where p and h are functions (I don't really understand this one too much.)
    Homogeneous - Look at the degrees of each grouping of coefficients, if they are all the same means homogeneous (Ex. xy^2 y' = x^3e^(y/x) - x^2y, so all the coefficient grouping ignoring y' are 3)
    Linear - (Q)y' + (P)y = Q(x) where P and Q are continuous functions


    So (xy+y)y' = x-xy
    Wouldn't I manipulate this to (xy+y)y' + xy = x and that would be linear

    Also
    x(1+y^2) + y(1+x^2)y' = 0
    x(1-y) + y(1+x^2)y' = 0

    How did you come up with the conclusion you did for each one?
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  2. #2
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    Re: Identifying First Order Diff Equations

    Quote Originally Posted by bustacap09 View Post
    So the equation is (xy+y)y' = x-xy
    It says to identify is the equation is Bernoulli, Separable, Linear and Homogeneous
    If these definitions are wrong on how to identify which is which, please let me know, because i researched this because my professor doesn't teach well.
    Bernoulli - y' + P(x)y = Q(x)y^n
    Separable - y' = p(x)h(y)where p and h are functions (I don't really understand this one too much.)
    Homogeneous - Look at the degrees of each grouping of coefficients, if they are all the same means homogeneous (Ex. xy^2 y' = x^3e^(y/x) - x^2y, so all the coefficient grouping ignoring y' are 3)
    Linear - (Q)y' + (P)y = Q(x) where P and Q are continuous functions


    So (xy+y)y' = x-xy
    Wouldn't I manipulate this to (xy+y)y' + xy = x and that would be linear

    Also
    x(1+y^2) + y(1+x^2)y' = 0


    x(1-y) + y(1+x^2)y' = 0

    How did you come up with the conclusion you did for each one?
    If you factor the original equation you get

    (x+1)y\frac{dy}{dx}=x(1-y) \iff \frac{ydy}{1-y}=\frac{xdx}{x+1}

    What does this tell you?
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