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Mathematical Modeling

  • Thread starter jgens
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  • #1
jgens
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Homework Statement



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.

Homework Equations



N/A

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!
 

Answers and Replies

  • #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?
 
  • #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.
 
  • #4
Dick
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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.
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.
 
  • #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.
 
  • #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?
 
  • #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]
 
  • #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?
 
  • #9
jgens
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Okay, so would c = ln(1.6)/4 then?
 
  • #10
Dick
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Okay, so would c = ln(1.6)/4 then?
There you go. That's it.
 
  • #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?
 
  • #12
Dick
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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?
It is. But that's a little harder than it needs to be. Your equation is separable.
 
  • #13
jgens
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How is the above DE seperable?
 
  • #14
Dick
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How is the above DE seperable?
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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