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Math Help - 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; April 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,
    \frac{dy}{dx}-(.2)y=-(.02)x
    This is a Linear-First Order Differencial Equation
    Thus,
    P(x)=-.2 and Q(x)=-.02x
    Thus,
    I(x)=\exp \left( \int -.2dx \right)=e^{-.2x}
    Thus, the solutions are,
    y=\frac{1}{I(x)} \left( \int I(x)Q(x) dx \right)
    Substitute,
    y=e^{.2x}\int (-.02)e^{-.2x}x dx
    Thus,
    y=-(.02)e^{.2x}\int xe^{-.2x}dx
    Integrate by parts,
    y=-(.02)e^{.2x}\left( -5xe^{-.2x}+5\int e^{-.2x}dx \right)
    Thus,
    =(-.02)e^{.2x} \left( -5xe^{-.2x}-25e^{-.2x} \right)
    Thus,
    y=.1x+.5+C
    Thus all solutions are,
    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,
    f(x)=.1x+C
    Then initially ya have, f(0)=2000
    Thus,
    f(0)=.1(0)+C=2000
    Thus,
    C=2000

    Thus, the unique equation that satisfies this OED is,
    y=.1x+2000
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    You have,
    \frac{dy}{dx}-(.2)y=-(.02)x
    This is a Linear-First Order Differencial Equation
    Thus,
    P(x)=-.2 and Q(x)=-.02x
    Thus,
    I(x)=\exp \left( \int -.2dx \right)=e^{-.2x}
    Thus, the solutions are,
    y=\frac{1}{I(x)} \left( \int I(x)Q(x) dx \right)
    Substitute,
    y=e^{.2x}\int (-.02)e^{-.2x}x dx
    Thus,
    y=-(.02)e^{.2x}\int xe^{-.2x}dx
    Integrate by parts,
    y=-(.02)e^{.2x}\left( -5xe^{-.2x}+5\int e^{-.2x}dx \right)
    Thus,
    =(-.02)e^{.2x} \left( -5xe^{-.2x}-25e^{-.2x} \right)
    Thus,
    y=.1x+.5+C
    Thus all solutions are,
    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 C=1000, and
    substituting the solution back into the equation gives:

    <br />
0.1=200<br />
.



    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack
    Which can't be right, as the initial condition gives C=1000, and
    substituting the solution back into the equation gives:

    <br />
0.1=200<br />
.



    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,
    \frac{dy}{dx}+P(x)y=Q(x)
    Then,
    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.

    <br />
\frac{dy}{dx}=0.2y-0.02x<br />

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

    <br />
\frac{dy_1}{dx}=0.2y_1<br />

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

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

    <br />
k\ Ae^{kx}=0.2e^{kx}<br />

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

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

    <br />
a=0.2ax+0.2c-0.02x<br />
,

    which means that a=0.1 and c=0.5

    So the general solution to the DE is:

    <br />
y(x)=Ae^{0.2x}+0.1x+0.5<br />
.

    Now when x=0 y=1000, so:

    <br />
1000=A+0.5<br />
,

    hence A=999.5, and the solution is:

    <br />
y(x)=999.5e^{0.2x}+0.1x+0.5<br />
.

    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; April 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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