Separable differential equation

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  • #1
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Homework Statement



I'm trying to solve this equation:

dN/dt = k(10 000 - N(t))

The attempt at a solution

dN/10 000 - N(t) = k dt

I integrate, and I'm left with:

- ln l10 000 - N(t)l = kt +C

I raise both sides with e. I don't know what e^- ln(xxx) is so I multiply both sides with -1

l10 000 - N(t)l = e^(-kt) * e^(-C) = -Ce^(-kt)

I now get the following:
N(t) = -10 000 - Ce^(-kt)

However the correct solution according to my book is:
N(t) = (+)10 000 - Ce^(-kt)

I am left with three questions:

1. What is e^- ln(...)?
2. Was I correct in making letting l10 000 - N(t)l = 10 000 + N(t) ?
3. What did I do wrong?


Thanks!
 
Last edited:

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,745
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You're almost there, when you took exponentials you should have obtianed:
[tex]
10^{4}-N(t)=Ae^{-kt}
[/tex]
I don't know what you said [tex]e^{-C}=-C[/tex] and hence:
[tex]
N(t)=10^{4}-Ae^{-kt}
[/tex]
and you would use the initial conditions to find A.
To answer your questions though:
1)[tex]e^{-\ln X}=e^{-1\times\ln X}=e^{\ln X^{-1}}=X^{-1}[/tex]
2) No, if N(t)<10^{4} then |10^{4}-N(t)|=10^{4}-N(t)
3) See above
 
  • #3
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You're almost there, when you took exponentials you should have obtianed:
[tex]
10^{4}-N(t)=Ae^{-kt}
[/tex]
I don't know what you said [tex]e^{-C}=-C[/tex] and hence:
[tex]
N(t)=10^{4}-Ae^{-kt}
[/tex]
and you would use the initial conditions to find A.
To answer your questions though:
1)[tex]e^{-\ln X}=e^{-1\times\ln X}=e^{\ln X^{-1}}=X^{-1}[/tex]
2) No, if N(t)<10^{4} then |10^{4}-N(t)|=10^{4}-N(t)
3) See above

Wow, that was quick!
Could you explain what A is equal to? I raise - kt - C with e and get [tex]e^{-kt}*e^{-C}[/tex]. How did you get A?

Also, it seems I have missed something when it comes to absolute values, could you explain why N(t) becomes negative and 10^4 becomes positive?

The initial conditions are that N(0)=200
 
Last edited:
  • #4
hunt_mat
Homework Helper
1,745
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[tex]A=e^{-C}[/tex], you find this by setting t=0 and using N(0)=200. The abolsute value is defined as:
[tex]
|x|=x\quad x\geqslant 0,|x|=-x\quad x<0
[/tex]
 
  • #5
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[tex]A=e^{-C}[/tex], you find this by setting t=0 and using N(0)=200. The abolsute value is defined as:
[tex]
|x|=x\quad x\geqslant 0,|x|=-x\quad x<0
[/tex]

Thanks so much!
How do you determine N(t) is less than zero? Should I enter a value for t or something?

Furthermore, my book has taken a slightly different approach ending up with:

l N(t) - 10 000) l = Ce^(-kt)

They then conclude that N(t) = 10 000 - Ce^(-kt) because N(t) is equal to or less than 10 000. That would seem to contradict the previous statement if I interpreted the symbol [tex]x\geqslant[/tex] correctly.
[tex]
|x|=x\quad x\geqslant 0
[/tex]
 
  • #6
hunt_mat
Homework Helper
1,745
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I didn't, I stated that:
[tex]
10^{4}-N(t)\geqslant 0\Rightarrow |10^{4}-N(t)|=10^{4}-N(t)
[/tex]
Mat
 
  • #7
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I didn't, I stated that:
[tex]
10^{4}-N(t)\geqslant 0\Rightarrow |10^{4}-N(t)|=10^{4}-N(t)
[/tex]
Mat

I wasn't accusing you of anything, I'm only trying to figure out how this works.

How do I determine the value of N(t) if at all? I am still clueless.

Why is this true:

[tex]N(t)-10^{4}\geqslant 0\Rightarrow |-10^{4}+N(t)|=-(-10^{4}-N(t))[/tex]

[tex]10^{4}-N(t)/geqslant 0\Rightarrow |10^{4}-N(t)|=10^{4}-N(t)[/tex]

[tex]|x|=x\quad x\geqslant 0[/tex]

It doesn't make sense to me. The first two statements seem to contradict each other IMHO.
 
Last edited:
  • #8
hunt_mat
Homework Helper
1,745
26
I know you weren't. I was clarifying.
[tex]
N(t)-10^{4}\geqslant 0\Rightarrow |-10^{4}+N(t)|=-10^{4}+N(t)
[/tex]
Is true be definition as |x|=x if x>0.
Once you get rid of the modulus sign, you can just re-arrange.
 
  • #9
hunt_mat
Homework Helper
1,745
26
You know this from solving the diffential equation:
[tex]
10^{4}-N(t)=Ae^{-kt}
[/tex]
Rearrange to obtain:
[tex]
N(t)=10^{4}-Ae^{-kt}
[/tex]
We find A by using N(0)=200, hence:
[tex]
200=10^{4}-A
[/tex]
Does this help at all?
 
  • #10
17
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Yes it does :-)

I think you meant[tex]N(t)-10^{4}< 0\Rightarrow |-10^{4}+N(t)|=-10^{4}+N(t)[/tex]
 
  • #11
hunt_mat
Homework Helper
1,745
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I don't think I did, I think that 10^4-N(t)>0.
 
  • #12
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My book states the following:
N(t) ≤ 10 000 => N(t) - 10 000 ≤ 0 => l N(t) - 10 000 l = -(N(t) - 10 000)

EDIT:
You are misunderstanding, please note that I have reversed the values. Like I said, this is the approach the book took, not me. They multiplied by -1 before integrating and got a different modulus.
 
Last edited:
  • #13
hunt_mat
Homework Helper
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We're saying the same thing, try to convince youself of that fact.
 
  • #14
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Yup. But back to the N(t) value question. Do you take into account all the values of N(t) when solving the inequality and getting rid of the modulus?

I guess you solve the inequality for N(0) = 200 up to N(x)= 10 000
 
  • #15
hunt_mat
Homework Helper
1,745
26
You don't solve any inequalities. All I am saying that saying is that N(t)<10000, and so |10000-N(t)|=10000-N(t), from here on in the derivation goes as I said it did.
 

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