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Solution of a differential equation related to exponential decay

Hey. :smile:

I am learning about exponential decay and have come across this article:

http://en.wikipedia.org/wiki/Exponential_decay#Solution_of_the_differential_equation

When it gets to the section titled "Solution of the differential equation", I do not follow how the step from line 2 to 3 is made i.e. the integration. It appears to me as though the integral symbol has just been stuck in front of both sides of the equation and the with respect to the variable bit left out as d(the variable) is already there. Why is it acceptable to do this?
Original post by Magu1re
Hey. :smile:

I am learning about exponential decay and have come across this article:

http://en.wikipedia.org/wiki/Exponential_decay#Solution_of_the_differential_equation

When it gets to the section titled "Solution of the differential equation", I do not follow how the step from line 2 to 3 is made i.e. the integration. It appears to me as though the integral symbol has just been stuck in front of both sides of the equation and the with respect to the variable bit left out as d(the variable) is already there. Why is it acceptable to do this?

You have done separation of variables before, haven't you? I think it's somewhere in C4.

The rearrangement that they are using is a bit shaky, mathematically but it's fine to use it as a notation before performing another operation (such as integration).

EDIT: The bold part doesn't make sense to me. I'm not too sure what you mean by that but the "w.r.t. the variable" part is definitely in the LHS and RHS integrals.
(edited 13 years ago)
Reply 2
Avatar for Ben121
Ben121
4
What do you get if you integrate 1/x ?
Reply 3
Avatar for limetang
limetang
19
You have 1/N(t) x dN(t) =-(lambda) dt

You just have to integrate the left with respect to dN(t) and the right with respect to dt (lambda is just a constant).

So the integral of 1/N(t) =ln(t) + C and the integral of -lambda dt = -(lambda)t + c

Hopefully that helps if you can follow my pretty bad notation.
Original post by Ben121
What do you get if you integrate 1/x ?


Original post by limetang
You have 1/N(t) x dN(t) =-(lambda) dt

You just have to integrate the left with respect to dN(t) and the right with respect to dt (lambda is just a constant).

So the integral of 1/N(t) =ln(t) + C and the integral of -lambda dt = -(lambda)t + c

Hopefully that helps if you can follow my pretty bad notation.

I think you guys have misunderstood his question. He's asking why it's acceptable to separate the variables, rather than how to evaluate the integrals.
Reply 5
Avatar for Magu1re
Magu1re
OP
3
Original post by limetang
You have 1/N(t) x dN(t) =-(lambda) dt

You just have to integrate the left with respect to dN(t) and the right with respect to dt (lambda is just a constant).

So the integral of 1/N(t) =ln(t) + C and the integral of -lambda dt = -(lambda)t + c

Hopefully that helps if you can follow my pretty bad notation.


I may be wrong but I do not think you can just integrate one side of an equation with respect to one variable and the other side with repect to another and still say that the two integrals are equal.
Reply 6
Avatar for Magu1re
Magu1re
OP
3
Original post by Farhan.Hanif93
I think you guys have misunderstood his question. He's asking why it's acceptable to separate the variables, rather than how to evaluate the integrals.


I actually half wanted to know how to perform the integration. With there being steps missing between the two lines it appeared to me as though an integral sign had been stuck in front of both sides of the equation and they had used the ds on either side as the "with respect to" bits.

However this is not the case. Integrating both sides with respect to t and then cancelling is the actual method used (someone told me this was the case :smile: ). I had not realised that this had been done.

So I can understand this step so long as I accept the "cancelling" of dt's etc.
Reply 7
Avatar for Manitude
Manitude
20
Original post by Magu1re
I may be wrong but I do not think you can just integrate one side of an equation with respect to one variable and the other side with repect to another and still say that the two integrals are equal.


You can...I've done it many many times before.
I can't offer a proof for it, someone else may be able to, but I think you should just accept that sep of variables is a valid technique for solving DEs until then.
Reply 8
Avatar for WJM
WJM
Integration of f'(x)/ f(x) dx = Ln|f(x)| + k
(edited 13 years ago)
Original post by Magu1re
I actually half wanted to know how to perform the integration. With there being steps missing between the two lines it appeared to me as though an integral sign had been stuck in front of both sides of the equation and they had used the ds on either side as the "with respect to" bits.

They have literally put integral signs on both sides after 'multiplying' up by dt and dividing by N(t). That's the concept behind the separation on variables.

However this is not the case. Integrating both sides with respect to t and then cancelling is the actual method used (someone told me this was the case :smile: ). I had not realised that this had been done.

Cancelling what? Nothing will cancel if you integrated both sides w.r.t. t.
I may have misunderstood what you've said, but I am 95% sure that they integrated one side w.r.t. N(t) and the other w.r.t. t after bringing functions of the same variables to the same sides.

So I can understand this step so long as I accept the "cancelling" of dt's etc.


By cancelling dt's, do you mean multiplying both sides by dt right at the start. This so-called cancellation would take place before the integrals are considered.
Original post by Magu1re
I may be wrong but I do not think you can just integrate one side of an equation with respect to one variable and the other side with repect to another and still say that the two integrals are equal.

You can.

I think the proof involving a decent bit of complicated, university-level maths so I probably can't justify it for you.
Reply 11
Avatar for anshul95
anshul95
11
Original post by Farhan.Hanif93
You can.

I think the proof involving a decent bit of complicated, university-level maths so I probably can't justify it for you.

i think I remember someone posting for some help on trying to find the conditions for a separable differential equation - it was definitely university maths.
Reply 12
Avatar for Magu1re
Magu1re
OP
3
Original post by Farhan.Hanif93
They have literally put integral signs on both sides after 'multiplying' up by dt and dividing by N(t). That's the concept behind the separation on variables.


Cancelling what? Nothing will cancel if you integrated both sides w.r.t. t.
I may have misunderstood what you've said, but I am 95% sure that they integrated one side w.r.t. N(t) and the other w.r.t. t after bringing functions of the same variables to the same sides.



By cancelling dt's, do you mean multiplying both sides by dt right at the start. This so-called cancellation would take place before the integrals are considered.


I mean:

Write the integral of both sides with respect to t. Then, because these are equal, you can divide through by dt and it will disappear from the LHS and cancel a dt on the RHS. Now both sides are still equal and so the LHS integrated with respect to n in the same as the integral of the RHS with respect to t.

This is integrating with respect to different variables in the end but originally the "integration" was with respect to the same variable. Does that make more sense? I apologise for the confusion lol.
Original post by Magu1re
I mean:

Write the integral of both sides with respect to t. Then, because these are equal, you can divide through by dt and it will disappear from the LHS and cancel a dt on the RHS. Now both sides are still equal and so the LHS integrated with respect to n in the same as the integral of the RHS with respect to t.

This is integrating with respect to different variables in the end but originally the "integration" was with respect to the same variable. Does that make more sense? I apologise for the confusion lol.

Fair enough. What I'm saying is that you wouldn't be able to divide through by the dt after the integral signs have been put in place (I think; someone more knowledgeable, please feel free to correct me).
Reply 14
Avatar for Magu1re
Magu1re
OP
3
Original post by Farhan.Hanif93
Fair enough. What I'm saying is that you wouldn't be able to divide through by the dt after the integral signs have been put in place (I think; someone more knowledgeable, please feel free to correct me).


Good point. Dividing through by dt beforehand sorts this issue out anyhow (I know this is still dodgy but hey :smile: ).
Original post by Magu1re
Good point. Dividing through by dt beforehand sorts this issue out anyhow (I know this is still dodgy but hey :smile: ).

Yeah, this is all very dodgy territory. It's more a case of notation than anything else.
Reply 16
Avatar for limetang
limetang
19
Original post by Magu1re
I may be wrong but I do not think you can just integrate one side of an equation with respect to one variable and the other side with repect to another and still say that the two integrals are equal.


You can do it... but I don't know why you can.

I'm desperately hoping someone can offer a proof as to why this works. Although I'm thinking the proof may be a bit beyond most of us to be honest.
(edited 13 years ago)
Reply 17
Avatar for Magu1re
Magu1re
OP
3
Original post by limetang
You can do it... but I don't know why you can.

I'm desperately hoping someone can offer a proof as to why this works. Although I'm thinking the proof may be a bit beyond most of us to be honest.


I think the issue is working with dts etc. which we cannot really comprehend. dx/dt for example, has meaning to us as an entity, but when you split it up or start to cancel dts then things become a lot, lot more dodgy. :smile:
Suppose dy/dx = f(x)/g(y)

Define I=∫0y(x)g(t)dt−∫0xf(x) dtI = \int_0^{y(x)} g(t) dt - \int_0^x f(x)\,dt

Then dIdx=dydxg(y)−f(x)=0\dfrac{dI}{dx} = \dfrac{dy}{dx} g(y) - f(x) = 0 (using fundamental theorem of calculus and the chain rule).

So I is a constant, and so ∫g(y) dy=C+∫f(x) dx\int g(y) \, dy = C + \int f(x) \,dx for some arbitrary constant C.
(edited 13 years ago)
Reply 19
Avatar for anshul95
anshul95
11
Original post by DFranklin
Suppose dy/dx = f(x)/g(y)

Define I=∫0y(t)g(t)dt−∫0xf(x) dtI = \int_0^{y(t)} g(t) dt - \int_0^x f(x)\,dt

Then dIdx=dydxg(y)−f(x)=0\dfrac{dI}{dx} = \dfrac{dy}{dx} g(y) - f(x) = 0 (using fundamental theorem of calculus and the chain rule).

So I is a constant, and so ∫g(y) dy=C+∫f(x) dx\int g(y) \, dy = C + \int f(x) \,dx for some arbitrary constant C.

cool. I would never have thought of that. +ve rep for sure.

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