Hard differential equation problem

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Homework Help Overview

The discussion revolves around a differential equation of the form dm/dt = -km(t), where participants are exploring the steps involved in solving it. The subject area includes ordinary differential equations and the method of separation of variables.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the transformation of the equation into the form dm/m = -k(t)dt and question the reasoning behind this step. There are inquiries about the nature of k, whether it is a constant or a function of time. Some express confusion about the integration process and the meaning of various mathematical notations.

Discussion Status

The conversation is ongoing, with participants seeking clarification on the steps to solve the equation. Some have offered guidance on the method of separation of variables, while others are questioning their understanding of derivatives and integrals, indicating a mix of knowledge levels among participants.

Contextual Notes

There is a noted lack of familiarity with calculus concepts among some participants, which may be impacting their ability to engage with the problem effectively. Additionally, there is discussion about the appropriateness of the problem's difficulty level for the participants' current understanding.

whatisreality
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Homework Statement


I really need to know what to do with this to solve it.
dm/dt=-km(t)
I was given the second step, dm/m=-k(t)dt, and wondered if someone could explain why you do that? I know that after that step you have to integrate. If anyone could explain logm=-kt+c? I don't know where any of that comes from! And I know I probably shouldn't be trying this yet, it isn't homework, I'm just curious how it works...

Homework Equations





The Attempt at a Solution

 
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You write [tex]\frac{dm}{dt}=-km[/tex]
and then
[tex]\frac{dm}{m}=-k(t)dt[/tex]
Is k a function of t or a constant? If it is a function of t then it is unsolvable
By the way the problem is by far not hard, perhaps the easiest ode(ordinary differential equation) you will ever find! Are you familiar with the method of separation of variables?
 
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Um... nope, I don't know the separation of variables.
dm/dt=km(t), the t is there, it's time so I don't think it's constant? And I don't know how to tell if it's a function of t. K is proportionate flow rate of water (cubic metres per second)
I think it's hard because I've never seen anything like this before, and no one's told me how to do them! :S
 
whatisreality said:
Um... nope, I don't know the separation of variables.
dm/dt=km(t), the t is there, it's time so I don't think it's constant?
I think it's hard because I've never seen anything like this before, and no one's told me how to do them! :S

t is variable but what about k? Is it a number or a function of t?
 
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No, k has to be a number. This equation is solvable, I just don't understand the steps. K is proportionate flow rate of water; it's a number. Basically, I need to understand the steps that have to be followed to solve this?
 
whatisreality said:
No, k has to be a number. This equation is solvable, I just don't understand the steps. K is proportionate flow rate of water; it's a number. Basically, I need to understand the steps that have to be followed to solve this?

Very well. Here is my solution. Feel free to ask me anything you don't understand
The method is called separation of variables, you will find it in the very beginning of any ode book

[tex] \frac{dm(t)}{dt}=-km(t)\Leftrightarrow dm(t)=-km(t)dt\Leftrightarrow \frac{dm(t)}{m(t)}=-kdt\Leftrightarrow \int\frac{dm(t)}{m(t)}= \int -kdt\Leftrightarrow \ln(m(t))=-kt+c\Leftrightarrow m(t)=e^{-kt+c}[/tex]
 
What do the arrows mean? Why is it dm(t) divided by dt? Sorry! I know this is going to start getting annoying, but I really don't understand most of it! And wait - why is there an extra c at the end?? What does e mean?
 
Have you now or ever studied differential or integral calculus?
 
whatisreality said:
What do the arrows mean? Why is it dm(t) divided by dt? Sorry! I know this is going to start getting annoying, but I really don't understand most of it!

We obviously don't use the same mathematical notation. The arrows mean that if the equality before them is true then the equality after them is also true and if the equality after them is true then the equality before them is also true.

dm(t) divided by dt is the derivative of th function m(t) with respect to t.
 
  • #10
Integral yes, a little, with raising powers by one so on non-linear things but not with fractions? Differential no. Never. Which is why I am stuck, and trying to learn, presumably why I was given the easiest type?

Thanks for trying, but I'm starting to think this may be a lost cause! Thanks, 3.1415... Sorry I'm a bit of a hopeless student, the explanation was good, it's just me!
 
  • #11
whatisreality said:
Integral yes, a little, with raising powers by one so on non-linear things but not with fractions? Differential no. Never. Which is why I am stuck, and trying to learn, presumably why I was given the easiest type?

You can't learn to solve odes without knowledge of derivatives and integrals.
 
  • #12
I thought a derivative was from differentiation? So is integration at least the thing I was describing?

Was that a stupid question?

Okay, so can someone please tell me how to do derivatives? Let's start with the basics. :)
 
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  • #13
No, no, no, wait. dm/dt=km(t), so dm/m=-k(t)dt. So can't I integrate that to make
logm=-kt+c so m=mo, logm/mo=kt
m=m(o)e^kt??
 
  • #14
That is almost perfect! (You forgot the - sign in front of k, and mo is m at t=0, is not it?) You know how to solve a separable differential equation!

ehild
 
  • #15
As echild said it is almost correct. I suggest ,however, not to continue reading your book regarding differential equations. Instead you should read a book regarding limits, derivatives and integrals
 
  • #16
Maybe. Thanks for the advice - and I did forget the minus sign (oops!), and mo is m at t=0, yes... well, I almost got it! :D Oh, it wasn't a book I was reading, it was just curiousity. It's okay, my maths teacher said he'd teach me things like limits, derivatives and integrals - we already did two of those, just not limits, and I didn't know how to apply those things. It's fine, all sorted - thanks for all the help! :D
 
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