Hey.
I am learning about exponential decay and have come across this article:
http://en.wikipedia.org/wiki/Exponen...ntial_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?

Magu1re
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 02042011 12:38

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Farhan.Hanif93
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 02042011 13:04
(Original post by Magu1re)
Hey.
I am learning about exponential decay and have come across this article:
http://en.wikipedia.org/wiki/Exponen...ntial_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?
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.Last edited by Farhan.Hanif93; 02042011 at 13:07. 
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 02042011 13:08
What do you get if you integrate 1/x ?

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 02042011 13:11
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. 
Farhan.Hanif93
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 02042011 13:13
(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. 
Magu1re
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 02042011 13:55
(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. 
Magu1re
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 02042011 13:59
(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.
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 ). 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. 
Manitude
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 02042011 14:00
(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.
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. 
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 02042011 14:06
Integration of f'(x)/ f(x) dx = Lnf(x) + k
Last edited by WJM; 02042011 at 14:08. 
Farhan.Hanif93
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 02042011 14:06
(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.
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 ). I had not realised that this had been done.
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. 
Farhan.Hanif93
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 02042011 14:11
(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.
I think the proof involving a decent bit of complicated, universitylevel maths so I probably can't justify it for you. 
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 02042011 14:16
(Original post by Farhan.Hanif93)
You can.
I think the proof involving a decent bit of complicated, universitylevel maths so I probably can't justify it for you. 
Magu1re
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 02042011 14:25
(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 socalled cancellation would take place before the integrals are considered.
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. 
Farhan.Hanif93
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 02042011 14:32
(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. 
Magu1re
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 02042011 14:57
(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). 
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 02042011 15:00
(Original post by Magu1re)
Good point. Dividing through by dt beforehand sorts this issue out anyhow (I know this is still dodgy but hey ). 
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 02042011 19:08
(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.
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.Last edited by limetang; 02042011 at 19:09. 
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 03042011 11:29
(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. 
DFranklin
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 03042011 14:08
Suppose dy/dx = f(x)/g(y)
Define
Then (using fundamental theorem of calculus and the chain rule).
So I is a constant, and so for some arbitrary constant C.Last edited by DFranklin; 03042011 at 18:47. 
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 03042011 15:38
(Original post by DFranklin)
Suppose dy/dx = f(x)/g(y)
Define
Then (using fundamental theorem of calculus and the chain rule).
So I is a constant, and so for some arbitrary constant C.
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