# Solve the Doomsday equation

#### Laxer42

Problem Statement
Solve the “Doomsday equation”

dN/dt = N^α

and show that for α > 1, the solutions approach a vertical asymptote in finite time. What happens for α < 1?
Relevant Equations
n/a
dN/dt = N^α
int(N^-a)=int(dt)
[N^(1-a)]/[1-a]+c=t

I believe this is a solution, but it does not produce the results I am told to be looking for.

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#### Orodruin

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but it does not produce the results I am told to be looking for.
Did you try solving for N?

#### Laxer42

Did you try solving for N?
Yes. I didn't think that helped though.

#### Orodruin

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Yes. I didn't think that helped though.
It also does not help that you do not tell us what you got.

#### Laxer42

It also does not help that you do not tell us what you got.
Thanks for being nice about it.

N=(t-at)^(1/(1-a))

#### Orodruin

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N=(t-at)^(1/(1-a))
What happened to the integration constant?

Thanks for being nice about it.

#### Laxer42

What happened to the integration constant?

I assumed the constant was equal to zero.

If you want to be rude because I've made honest mistakes, I don't think your help will be productive for either of us.

#### Orodruin

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I assumed the constant was equal to zero.
You cannot arbitrarily set integration constants to zero. What do you get if you do not assume this?

If you want to be rude because I've made honest mistakes, I don't think your help will be productive for either of us.
Your problem here is your own arrogance, not me being rude. You are the one who wrongly interpreted a factual statement about the process (it does not help your learning process to keep people trying to help you in the dark regarding what you have done so far) for rudeness and decided to respond with a snarky remark.

Had you actually provided all the necessary information in your OP, this thread would have been over by now.

#### Borg

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If you want to be rude because I've made honest mistakes
How exactly should someone correct your mistakes without being seen as rude?

#### Laxer42

How exactly should someone correct your mistakes without being seen as rude?
"Did you solve for N?"
"Yes. I didn't think that helped though."
"Well show me what you got anyway."
...
"By the way, in the future you should make sure to show any and all work you have done regardless of whether you think it's right or even relevant. It usually speeds things along."

#### Laxer42

You cannot arbitrarily set integration constants to zero. What do you get if you do not assume this?
N=(t-at)^(1/(1-a))+c

#### Orodruin

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The constant appears correctly in your original post. If you solve for N, it will not be the same thing as first solving for N with the constant equal to zero and then adding a constant. I suggest you rename the constant $t_0$ and put it on the other side of the equation. If that does not help, please show us each step of when you solve for N.

#### Laxer42

The constant appears correctly in your original post. If you solve for N, it will not be the same thing as first solving for N with the constant equal to zero and then adding a constant. I suggest you rename the constant $t_0$ and put it on the other side of the equation. If that does not help, please show us each step of when you solve for N.
N=[(t-$t_0$)(1-α)]^(1/(1-α))

#### Orodruin

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Right, but for $\alpha > 1$ I would introduce a new constant $\beta = \alpha -1 > 0$ to tidy things up. What can you then observe as $t\to t_0$?

#### Laxer42

Right, but for $\alpha > 1$ I would introduce a new constant $\beta = \alpha -1 > 0$ to tidy things up. What can you then observe as $t\to t_0$?

N=[-(t-$t_0$)(B)]^(1/(1-α))

It would go to zero as $t\to t_0$

#### Orodruin

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So N=[-(t-$t_0$)($\beta )]^(1/(1-α))$

It would go to zero as $t\to t_0$
Would it? Introduce $\beta$ in the exponent too and remember that $\beta > 0$ so also $1/\beta > 0$. I also suggest writing $-(t-t_0)\beta$ as $\beta(t_0-t)$ (the relevant times for the finite solution are $t < t_0$).

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#### Orodruin

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"Did you solve for N?"
"Yes. I didn't think that helped though."
So let me tell you how that comes through when you do not provide your actual work:
"Did you solve for N?"
"Yes, of course I did, why are you not helping?"

"Did you solve for N?"
"Yes. I got X. I did not think it helped because Y"

Again, how you expect people to be able to help you without providing your actual work is beyond me and it is also against forum rules. You provided your actual work up until $N^{1-\alpha}/(1-\alpha) + t_0 = t$ so naturally the next question is about whether you tried to solve for $N$ or not. If you did you should then provide your work. It is not strange that people become irritated with you when they have to drag information out of you over the course of several posts rather than getting straight to the point.

"By the way, in the future you should make sure to show any and all work you have done regardless of whether you think it's right or even relevant. It usually speeds things along."
This should not even need pointing out to you. It is quite obvious that we cannot help you unless we do not undertand where your problem lies. Try to imagine how things seem from the other side.

#### Ray Vickson

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"Did you solve for N?"
"Yes. I didn't think that helped though."
"Well show me what you got anyway."
...
"By the way, in the future you should make sure to show any and all work you have done regardless of whether you think it's right or even relevant. It usually speeds things along."
None of those remarks are rude; they are meant to be helpful, and you would find them to be so if you were less defensive and more willing to work with helpers.

"Solve the Doomsday equation"

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