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Simple Differential Equation

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data
    "Solve this differential equation algebraically, subject to the initial condition that [tex]y=10[/tex] at [tex]t=0[/tex]


    2. Relevant equations
    [tex]\frac{dy}{dt} = 2y*\frac{1000-y}{1000}[/tex]


    3. The attempt at a solution
    I first reduced the right side to [tex]\frac{-y^2}{500} + 2y[/tex]
    After that I separated the variables, but I don't know how to integrate that function.
    Can anyone point me in the right step?
    [tex]\int \frac {dy}{\frac{-y^2}{500}+2y}[/tex]
     
    Last edited: Mar 13, 2010
  2. jcsd
  3. Mar 13, 2010 #2

    Dick

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    Have you done any partial fractions recently? You can write the integrand as A/(y)+B/(1000-y). You just need to find A and B.
     
  4. Mar 13, 2010 #3
    I am in fact working on the section over partial fractions. Thank you very much Dick. I see how partial fractions plays into this. Might I ask how you got [tex]\frac{A}{y} + \frac{B}{1000-y}[/tex]

    I'm having trouble seeing the partial fractions. Most of them I have worked with I use the shortcut to Heaviside's Method. Many have been in the form of [tex]\int \frac {Linear}{Quadratic}[/tex]
     
    Last edited: Mar 13, 2010
  5. Mar 13, 2010 #4

    Dick

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    y and (1000-y) were the factors of your original expression before you multiplied it out. You didn't really want to do that.
     
  6. Mar 13, 2010 #5
    OK. So what would be on the left-hand side of the equal sign?
    [tex]x=\frac{A}{y}+\frac{B}{1000-y}[/tex]

    I understand you multiply the left side by one of the denominators on the right side and then plug in the x value that would have zeroed the denominator. I just don't know what goes on the left side. Is it the [tex]\frac{dy}{dt}[/tex]?
     
  7. Mar 13, 2010 #6
    Nevermind. My poor algebra skills were my downfall this time. I figured out that it's [tex]\int \frac{-500}{y(y-1000)} [/tex]
     
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