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Math Help - Bernoulli's Equation

  1. #1
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    Bernoulli's Equation

    I also have a bernouli's equation problem that I wonder if you guys would check the work on. Here goes:
    Bernouli's equation:
    y'+p(x)y=Q(x)y^n

    dv/du=(u-v)^2-2(u-v)-2 w=u-v dw/du=1-dv/du dv/du=1-dw/du
    1-dw/du=w^2-2w-2
    dw/du=-w^2+2w+3
    dw/du-2w=-w^2+3
    p=-2, Q=-W^2+3
    Linear form:
    dz/dw+(1-n)PZ=(1-n)Q
    dz/dw=(1-2)(-2)(Z)=(1-2)(-W^2+3)
    dz/dw+2z=w^2-3
    u=e∫2dw=e^2w
    z=1/u(
    ∫uq+c)
    Z=1/e^2w(
    ∫(e^2w)(w^2-3) + C Now where I go from here is a mystery to me. Anyone have some idea of how to intergrate that function?

    Oh yeah, Answer is (u-v-3)e^4u=c(u-v+1)
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  2. #2
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    Quote Originally Posted by Sebastian View Post

    dv/du=(u-v)^2-2(u-v)-2 w=u-v dw/du=1-dv/du dv/du=1-dw/du
    1-dw/du=w^2-2w-2
    dw/du=-w^2+2w+3
    dw/du-2w=-w^2+3
    p=-2, Q=-W^2+3
    Wait.
    Why is Q(u)=-W^2+3

    Remember it must be of the form,
    Say for example,
    Q(u)w^2
    Now if you say that,
    Q(u)=-w^2+2
    Then you imply that,
    Q(u)w^2=-w^2+2
    Which is clearly false.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Wait.
    Why is Q(u)=-W^2+3

    Remember it must be of the form,
    Say for example,
    Q(u)w^2
    Now if you say that,
    Q(u)=-w^2+2
    Then you imply that,
    Q(u)w^2=-w^2+2
    Which is clearly false.
    I don't understand what to do from here then. Does my w become my z function? If not, how do I make it so?
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  4. #4
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    Quote Originally Posted by Sebastian View Post
    I don't understand what to do from here then. Does my w become my z function? If not, how do I make it so?
    Here are two ways of writing it (assuming it is non-zero),

    -w^2+2=(-w+2/w)w^1 thus Q=-w+2/w

    Or,

    -w^2+2=(-1+2/w^2)w^2 thus Q=-1+2/w^2
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