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Math Help - bernoulli equation/partial fraction

  1. #1
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    bernoulli equation/partial fraction

    so i know this question can be done both way

    y' = 3y - y^2

    <=>

    y' = y(3-y)


    alright so i dont think i can do it by bernoulli because i dont have two variables so i dont think thats gonna work but
    I need to solve the initial value problem y(0) = 2

    i get up to

    dy
    -- = 3dt
    y(1-y/3)

    i just don't know how to to setup and do the partial fraction
    Last edited by treetheta; October 14th 2009 at 07:36 PM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Quote Originally Posted by treetheta View Post
    so i know this question can be done both way

    y' = 3y - y^2

    <=>

    y' = y(3-y)


    alright so i dont think i can do it by bernoulli because i dont have two variables so i dont think thats gonna work but
    I need to solve the initial value problem y(0) = 2

    i get up to

    dy
    -- = 3dt
    y(1-y/3)

    i just don't know how to to setup and do the partial fraction
    \frac{1}{y(3-y)}=\frac{A}{y}+\frac{B}{3-y}\implies A(3-y)+By=1\implies (B-A)y+3A=1\implies A=\frac{1}{3},B=\frac{1}{3}

    So,

    \int\frac{y'}{y(3-y)}\,dx=\frac{1}{3}\int\frac{y'}{y}\,dx+\frac{1}{3  }\int\frac{y'}{3-y}\,dx=\int\,dx

    Letting k=e^{3c_0}, You end up with

    \frac{1}{3}(\ln(y)-\ln(3-y))=x+c_0\implies\ln\left(\frac{y}{3-y}\right)=3x+3c_0\implies\frac{y}{3-y}=e^{3x}\cdot \underbrace{e^{3c_0}}_{k}\implies y=3ke^{3x}-kye^{3x}\implies y(1+ke^{3x})=3ke^{3x}\implies y=\frac{3ke^{3x}}{1+ke^{3x}}

    At the point (0,2), we have 2=\frac{3k}{1+k}\implies k=2.

    So \boxed{y=\frac{6e^{3x}}{1+2e^{3x}}}
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