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Math Help - Differential equation / Seperable

  1. #1
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    Differential equation / Seperable

    Hello!

    I have been given an equation as following;

    \frac{dy}{dt}=ay(A-y)

    where
    y>0
    A and a are positive constants


    1)
    I am then to find the constants \alpha and \beta so the equation can be given as this:

    \frac{1}{y(A-y)}= \frac{\alpha}{y}+ \frac{\beta}{A-y)}


    2)
    And at last find a general solution when y(0)= \frac{A}{2}

    Don't know where to start actually...

    Any help is greatly appreciated!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    This is just the partial fraction decomposition you learned in integral calculus.

    For


    Multiplty both sides by y(A-y)

    Then let y = A to find beta =1/A and y = 0 to find alpha = 1/A
    To solve separate the variables and integrate
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  3. #3
    Eater of Worlds
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    If we separate variables and integrate, we get:

    y=\frac{Ae^{aAt}}{e^{aAt}+AC}

    Using the initial condition, y(0)=A/2, gives C=1/A.

    Which results in:

    y=\frac{Ae^{aAt}}{e^{aAt}+1}
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  4. #4
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    A bit heavy for me atm, but I think I see how you two did it. Struggling a bit with DE...


    If I could trouble you with an addition to this calculation, as for what happens to y when t\rightarrow \infty?

    How to explain/show is my concern...
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  5. #5
    Moo
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    Hello,
    Quote Originally Posted by jokke22 View Post
    A bit heavy for me atm, but I think I see how you two did it. Struggling a bit with DE...


    If I could trouble you with an addition to this calculation, as for what happens to y when t\rightarrow \infty?

    How to explain/show is my concern...
    y=\frac{Ae^{aAt}}{e^{aAt}+1}=\frac{A}{1+e^{-aAt}}, by dividing both numerator and denominator by e^{aAt}

    And since aA is positive, -aA is negative.

    We know that \lim_{x\to\infty} e^{-x}=0

    Hence the limit is...
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  6. #6
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    This is the logistic equation, correct?. Here is a general tutorial to discuss the solutions of the logistic equation.

    Let's start from scratch. Okey-doke.

    \frac{dy}{dt}=a(A-y)y for y>0. A and a are constants.....[1]

    Separate variables:

    \frac{dy}{(A-y)y}=adt

    Integrate:

    \int\frac{1}{(A-y)y}dy=a\int dt+C_{1}.....[2]

    Where C_{1} is a constant of integration. By use of partial

    fractions as Calculus26 showed:

    \frac{1}{(A-y)y}=\frac{1}{A}\left(\frac{1}{A-y}+\frac{1}{y}\right)

    Therefore, [2] becomes:

    \frac{1}{A}\left(\int \frac{dy}{A-y}+\int\frac{dy}{y}\right)=a\int dt+C_{1}

    or \frac{1}{A}\left(-ln|A-y|+ln|y|\right)+at+C_{1}

    when we integrate.

    Therefore, ln\left|\frac{A-y}{y}\right|=-Aat-AC_{1}.

    Exponentiating gives:

    \left|\frac{A-y}{y}\right|=Ce^{-Aat}

    where C=e^{-AC_{1}}>0 is now our arbitrary constant.

    Since y>0 by our assumption, if we also assume y<A, the absolute

    value signs are done with.

    \left|\frac{A-y}{y}\right|=-\left(\frac{A-y}{y}\right)

    We can also consider either case by simply letting the arbitrary cnstant C

    be positive or negative.

    Therefore, \frac{A-y}{y}=Ce^{-Aat}.

    Solving for y:

    y(t)=\frac{A}{1+Ce^{-Aat}}.....[3]

    which is the general solution of [1].

    To evaluate C we need the initial conditions:

    y(0)=a_{0}=\frac{A}{2}

    Sub into [3]:

    y(0)=\frac{A}{1+Ce^{0}}

    \frac{A}{2}=\frac{A}{1+C}

    C=1

    Hence, thus, and therefore,

    y(t)=\frac{A}{1+e^{-Aat}}

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Now, you asked what happens to y when t-->infinity.

    Suppose a>0 and A>0. If y(t) represents population then certain models

    say that a(A-y)=\text{birth rate-death rate}

    If we have a small population(small y), then a(A-y)>0

    and the birth rate exceeds the death rate. Thus, the population is growing.

    We take note that:

    \lim_{t\to \infty}y(t)=\frac{A}{1+(\frac{A}{a_{0}}+1)\display  style\lim_{t\to \infty}e^{-Aat}}=A

    Since \lim_{t\to \infty}e^{-Aat}=0

    This is a curious result since it says the population approaches a limiting

    value A independent of the initial population size we call

    a_{0}

    A is called the limiting population that the syatem can support. If

    a_{0}>A, the population decreases.
    \
    If a_{0}<A the population increases.

    We can also take note that when y=A, [1] becomes y'=0 and we are at

    equilibrium(birth rate=death rate).

    Here is a haphazard 'paint' graph and I hope this little tutorial helps some to understand it.
    Attached Thumbnails Attached Thumbnails Differential equation / Seperable-logistic.gif  
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  7. #7
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    That was some heavy explaining! Appreciate the effort GREATLY and will look it over asap!
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