Separable differential equation

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!

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hunt_mat
Homework Helper
You're almost there, when you took exponentials you should have obtianed:
$$10^{4}-N(t)=Ae^{-kt}$$
I don't know what you said $$e^{-C}=-C$$ and hence:
$$N(t)=10^{4}-Ae^{-kt}$$
and you would use the initial conditions to find A.
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

You're almost there, when you took exponentials you should have obtianed:
$$10^{4}-N(t)=Ae^{-kt}$$
I don't know what you said $$e^{-C}=-C$$ and hence:
$$N(t)=10^{4}-Ae^{-kt}$$
and you would use the initial conditions to find A.
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

Last edited:
hunt_mat
Homework Helper
$$A=e^{-C}$$, you find this by setting t=0 and using N(0)=200. The abolsute value is defined as:
$$|x|=x\quad x\geqslant 0,|x|=-x\quad x<0$$

$$A=e^{-C}$$, you find this by setting t=0 and using N(0)=200. The abolsute value is defined as:
$$|x|=x\quad x\geqslant 0,|x|=-x\quad x<0$$

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.
$$|x|=x\quad x\geqslant 0$$

hunt_mat
Homework Helper
I didn't, I stated that:
$$10^{4}-N(t)\geqslant 0\Rightarrow |10^{4}-N(t)|=10^{4}-N(t)$$
Mat

I didn't, I stated that:
$$10^{4}-N(t)\geqslant 0\Rightarrow |10^{4}-N(t)|=10^{4}-N(t)$$
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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hunt_mat
Homework Helper
I know you weren't. I was clarifying.
$$N(t)-10^{4}\geqslant 0\Rightarrow |-10^{4}+N(t)|=-10^{4}+N(t)$$
Is true be definition as |x|=x if x>0.
Once you get rid of the modulus sign, you can just re-arrange.

hunt_mat
Homework Helper
You know this from solving the diffential equation:
$$10^{4}-N(t)=Ae^{-kt}$$
Rearrange to obtain:
$$N(t)=10^{4}-Ae^{-kt}$$
We find A by using N(0)=200, hence:
$$200=10^{4}-A$$
Does this help at all?

Yes it does :-)

I think you meant$$N(t)-10^{4}< 0\Rightarrow |-10^{4}+N(t)|=-10^{4}+N(t)$$

hunt_mat
Homework Helper
I don't think I did, I think that 10^4-N(t)>0.

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.

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hunt_mat
Homework Helper
We're saying the same thing, try to convince youself of that fact.

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

hunt_mat
Homework Helper
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.