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Thread: Epidemic (Diff Equation)

  1. #1
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    Epidemic (Diff Equation)

    The spread of a disease in a community is modelled by the following differential equation:

    dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

    1) Show the equation that for the family of 'curves' in the plane for which dy/dx is a constant, is the form y=mx+c.

    2) Solve the equation using the linear 1st order method, given that initially there are on thousand infected individuals.
    Last edited by Natasha1; Apr 30th 2006 at 09:58 AM.
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  2. #2
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    Quote Originally Posted by Natasha1
    The spread of a disease in a community is modelled by the following differential equation:

    dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

    1) Show the equation that for the family of 'curves' in the plane for which dy/dx is a constant, is the form y=mx+c.

    2) Solve the equation using the linear 1st order method, given that initially there are on thousand infected individuals.
    You have,
    $\displaystyle \frac{dy}{dx}-(.2)y=-(.02)x$
    This is a Linear-First Order Differencial Equation
    Thus,
    $\displaystyle P(x)=-.2$ and $\displaystyle Q(x)=-.02x$
    Thus,
    $\displaystyle I(x)=\exp \left( \int -.2dx \right)=e^{-.2x}$
    Thus, the solutions are,
    $\displaystyle y=\frac{1}{I(x)} \left( \int I(x)Q(x) dx \right)$
    Substitute,
    $\displaystyle y=e^{.2x}\int (-.02)e^{-.2x}x dx$
    Thus,
    $\displaystyle y=-(.02)e^{.2x}\int xe^{-.2x}dx$
    Integrate by parts,
    $\displaystyle y=-(.02)e^{.2x}\left( -5xe^{-.2x}+5\int e^{-.2x}dx \right)$
    Thus,
    $\displaystyle =(-.02)e^{.2x} \left( -5xe^{-.2x}-25e^{-.2x} \right)$
    Thus,
    $\displaystyle y=.1x+.5+C$
    Thus all solutions are,
    $\displaystyle y=.1x+C$

    Maybe I made a mistake I did not check my work, need to go soon.
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  3. #3
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    Assuming I did everything correctly and that,
    $\displaystyle f(x)=.1x+C$
    Then initially ya have, $\displaystyle f(0)=2000$
    Thus,
    $\displaystyle f(0)=.1(0)+C=2000$
    Thus,
    $\displaystyle C=2000$

    Thus, the unique equation that satisfies this OED is,
    $\displaystyle y=.1x+2000$
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    You have,
    $\displaystyle \frac{dy}{dx}-(.2)y=-(.02)x$
    This is a Linear-First Order Differencial Equation
    Thus,
    $\displaystyle P(x)=-.2$ and $\displaystyle Q(x)=-.02x$
    Thus,
    $\displaystyle I(x)=\exp \left( \int -.2dx \right)=e^{-.2x}$
    Thus, the solutions are,
    $\displaystyle y=\frac{1}{I(x)} \left( \int I(x)Q(x) dx \right)$
    Substitute,
    $\displaystyle y=e^{.2x}\int (-.02)e^{-.2x}x dx$
    Thus,
    $\displaystyle y=-(.02)e^{.2x}\int xe^{-.2x}dx$
    Integrate by parts,
    $\displaystyle y=-(.02)e^{.2x}\left( -5xe^{-.2x}+5\int e^{-.2x}dx \right)$
    Thus,
    $\displaystyle =(-.02)e^{.2x} \left( -5xe^{-.2x}-25e^{-.2x} \right)$
    Thus,
    $\displaystyle y=.1x+.5+C$
    Thus all solutions are,
    $\displaystyle y=.1x+C$

    Maybe I made a mistake I did not check my work, need to go soon.
    Which can't be right, as the initial condition gives $\displaystyle C=1000$, and
    substituting the solution back into the equation gives:

    $\displaystyle
    0.1=200
    $.



    RonL
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    Quote Originally Posted by CaptainBlack
    Which can't be right, as the initial condition gives $\displaystyle C=1000$, and
    substituting the solution back into the equation gives:

    $\displaystyle
    0.1=200
    $.



    RonL
    Where is the mistake?
    I hate making posts when I got to got in 20 minutes and they take like 30 minutes to write.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    Where is the mistake?
    I hate making posts when I got to got in 20 minutes and they take like 30 minutes to write.
    I will have to examine the solution carefully to find it, you are using a method
    I don't usually employ so it will need more carefully examination that it would
    normally take.

    RonL
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  7. #7
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    I don't use that method either, well at least I haven't been taught it like that so for me it's double dutch as they say in England! Which means I still don't get it

    RoL help
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  8. #8
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    Natasha, the method I used was,
    $\displaystyle \frac{dy}{dx}+P(x)y=Q(x)$
    Then,
    $\displaystyle y=\frac{1}{I(x)}\left( \int I(x)Q(x)dx \right)$
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  9. #9
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    Quote Originally Posted by Natasha1
    The spread of a disease in a community is modelled by the following differential equation:

    dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

    1) Show the equation that for the family of 'curves' in the plane for which dy/dx is a constant, is the form y=mx+c.

    2) Solve the equation using the linear 1st order method, given that initially there are on thousand infected individuals.
    We have a linear first order ODE with constant coefficients.

    $\displaystyle
    \frac{dy}{dx}=0.2y-0.02x
    $

    The method I normaly employ to solve these equations is to find the general
    solution $\displaystyle y_1(x)$ of the homeogeneous equation:

    $\displaystyle
    \frac{dy_1}{dx}=0.2y_1
    $

    Then if $\displaystyle y_2(x)$ is any solution (Particular Integral) of the original equation
    then the general solution of the original equation $\displaystyle y(x)=y_1(x)+y_2(x)$.

    Now to solve the homogeneous equation we try a solution of
    the form: $\displaystyle y_1(x)=Ae^{kx}$, so:

    $\displaystyle
    k\ Ae^{kx}=0.2e^{kx}
    $

    therefore $\displaystyle k=0.2$, and $\displaystyle y_1(x)=Ae^{0.2x}$.

    For the particular integral a reasonable guess would be of the form
    $\displaystyle y_2(x)=ax+c$. Substituting this into the equation gives:

    $\displaystyle
    a=0.2ax+0.2c-0.02x
    $,

    which means that $\displaystyle a=0.1$ and $\displaystyle c=0.5$

    So the general solution to the DE is:

    $\displaystyle
    y(x)=Ae^{0.2x}+0.1x+0.5
    $.

    Now when $\displaystyle x=0$ $\displaystyle y=1000$, so:

    $\displaystyle
    1000=A+0.5
    $,

    hence $\displaystyle A=999.5$, and the solution is:

    $\displaystyle
    y(x)=999.5e^{0.2x}+0.1x+0.5
    $.

    Given the wording of the problem I suspect that the intention is that you
    use the method that PH was proposing (as it applies to general first order
    ODEs)

    RonL
    Last edited by CaptainBlack; Apr 30th 2006 at 11:02 AM.
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  10. #10
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    I finally found my mistake
    It was a common one, notice I forgot the Constant after the derivative.
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