Separable differential equation

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Homework Help Overview

The discussion revolves around solving a separable differential equation of the form dN/dt = k(10,000 - N(t)). The original poster attempts to integrate the equation but encounters confusion regarding the absolute value and the interpretation of the results, particularly in relation to the correct form of the solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration process and the handling of absolute values. Questions are raised about the interpretation of e^-ln(x) and the implications of the absolute value in the context of the solution. There is also exploration of the initial conditions and how they affect the constant A.

Discussion Status

Some participants provide clarifications regarding the mathematical properties involved, such as the behavior of absolute values and the relationship between N(t) and 10,000. There is an ongoing exploration of different interpretations of the solution and the implications of the initial conditions, with no explicit consensus reached.

Contextual Notes

The original poster's confusion stems from the integration process and the subsequent handling of absolute values, as well as the interpretation of the solution in relation to the initial condition N(0) = 200. The discussion reflects the complexity of the problem and the nuances in mathematical reasoning required to navigate it.

M55ikael
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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:
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You're almost there, when you took exponentials you should have obtianed:
<br /> 10^{4}-N(t)=Ae^{-kt}<br />
I don't know what you said e^{-C}=-C and hence:
<br /> N(t)=10^{4}-Ae^{-kt}<br />
and you would use the initial conditions to find A.
To answer your questions though:
1)e^{-\ln X}=e^{-1\times\ln X}=e^{\ln X^{-1}}=X^{-1}
2) No, if N(t)<10^{4} then |10^{4}-N(t)|=10^{4}-N(t)
3) See above
 
hunt_mat said:
You're almost there, when you took exponentials you should have obtianed:
<br /> 10^{4}-N(t)=Ae^{-kt}<br />
I don't know what you said e^{-C}=-C and hence:
<br /> N(t)=10^{4}-Ae^{-kt}<br />
and you would use the initial conditions to find A.
To answer your questions though:
1)e^{-\ln X}=e^{-1\times\ln X}=e^{\ln X^{-1}}=X^{-1}
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 e^{-kt}*e^{-C}. 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
 
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A=e^{-C}, you find this by setting t=0 and using N(0)=200. The abolsute value is defined as:
<br /> |x|=x\quad x\geqslant 0,|x|=-x\quad x&lt;0<br />
 
hunt_mat said:
A=e^{-C}, you find this by setting t=0 and using N(0)=200. The abolsute value is defined as:
<br /> |x|=x\quad x\geqslant 0,|x|=-x\quad x&lt;0<br />

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 x\geqslant correctly.
hunt_mat said:
<br /> |x|=x\quad x\geqslant 0<br />
 
I didn't, I stated that:
<br /> 10^{4}-N(t)\geqslant 0\Rightarrow |10^{4}-N(t)|=10^{4}-N(t)<br />
Mat
 
hunt_mat said:
I didn't, I stated that:
<br /> 10^{4}-N(t)\geqslant 0\Rightarrow |10^{4}-N(t)|=10^{4}-N(t)<br />
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:

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

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

|x|=x\quad x\geqslant 0

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

I think you meantN(t)-10^{4}&lt; 0\Rightarrow |-10^{4}+N(t)|=-10^{4}+N(t)
 
  • #11
I don't think I did, I think that 10^4-N(t)>0.
 
  • #12
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
We're saying the same thing, try to convince youself of that fact.
 
  • #14
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
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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