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Mathematical Modeling

  1. Apr 5, 2010 #1

    jgens

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    1. The problem statement, all variables and given/known data

    When certain viral particles enter the bdoy, the replicate to 160% every four hours and the immune system eliminates these particular viral particles at the rate of 50000 viral particles per hour. Find an equation modeling this viral growth.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I could really just use some help getting this problem off the gound, so any information would be appreciated. Thanks!
     
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  3. Apr 5, 2010 #2

    Dick

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    Any thoughts? I'm not sure what you are confused about. They want you to set up an ODE with terms representing growth rate and elimination rate, right?
     
  4. Apr 5, 2010 #3

    jgens

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    I'm not exactly sure what they want me to do. My orignial thought was [itex]n = n_0(1.6^{t/4}) - 50000t[/itex] but this doesn't seem right since it assumes that every particle has the opportunity to replicate before being eliminated. I haven't done much with ODEs but if you don't mind helping me set one up, I think that I might be able to solve the problem.
     
  5. Apr 5, 2010 #4

    Dick

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    I don't think that's right either. Your replication rate is independent of the number viral particles present. And it's not an ODE. Want to try again? You want an expression for dn/dt.
     
  6. Apr 5, 2010 #5

    jgens

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    I'm sorry, but do you think that you could give me a little bigger hint? I'm really struggling to do anything with dn/dt.
     
  7. Apr 5, 2010 #6

    Dick

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    Here's a really big hint. How about an equation of the form dn/dt=c*n-b where c and b are constants? Can you use your replication data to find c?
     
  8. Apr 5, 2010 #7

    jgens

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    So, using that hint I'm probably looking for something like . . .

    [tex]\frac{dn}{dt} = (1.6)n - b[/itex]
     
  9. Apr 5, 2010 #8

    Dick

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    Well, no. Let's go back a step. dn/dt=(growth rate)-(elimination rate), right? Growth rate should be proportional to n, also right? So you can find growth rate by solving dn/dt=c*n. What's the solution? Now if n grows by 160% in 4 hours, what's c?
     
  10. Apr 5, 2010 #9

    jgens

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    Okay, so would c = ln(1.6)/4 then?
     
  11. Apr 5, 2010 #10

    Dick

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    There you go. That's it.
     
  12. Apr 5, 2010 #11

    jgens

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    Okay, so . . .

    [tex]\frac{dn}{dt} = \frac{\ln{1.6}}{4}n - 50000[/tex]

    And this is a differential equation of the form

    [tex]n' + P(x)n = Q(x)[/tex]

    Is this right so far?
     
  13. Apr 5, 2010 #12

    Dick

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    It is. But that's a little harder than it needs to be. Your equation is separable.
     
  14. Apr 5, 2010 #13

    jgens

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    How is the above DE seperable?
     
  15. Apr 5, 2010 #14

    Dick

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    In the usual way. dn/(ln(1.6)*n/4-50000)=dt. All of the n's on the left, all of the t's on right. Separable.
     
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