Integration of 1 variable in 2 different ways.

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Discussion Overview

The discussion revolves around the integration of a differential equation involving variables P, V, and M, where the participants explore different methods of integration and the implications of treating these variables as constants or functions of x. The scope includes mathematical reasoning and technical explanation.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents an integration approach leading to the result M=4V, questioning the correctness of their method.
  • Another participant points out that if V and P are constants, then their derivatives with respect to x would be zero, leading to a trivial equation.
  • Some participants emphasize the importance of including the integration constant in their calculations.
  • There is a suggestion that if P and V are functions of x, the integration method used in the first approach may not be valid.
  • Multiple participants express confusion regarding the integration methods and their outcomes, indicating a lack of clarity on the correct approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on which integration method is correct. There are competing views on whether P and V should be treated as constants or functions, and the implications of this choice remain unresolved.

Contextual Notes

Participants highlight the need for clarity regarding the assumptions about the variables involved, particularly whether they are constants or functions of x. The discussion also reflects on the necessity of including integration constants, which some participants initially overlooked.

sreerajt
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I have to do a integration which goes like this:
(V-M)(dP/dx)+3P(dV/dx)=0, (where M,P and V are constants).
If you integrate with dx, you will get:
∫[(V-M)dP]+∫[3PdV]=0.
which ultimately results in the answer M=4V.
Now, i can put the first equation in this form also:
(dP/P)=-3[dV/(V-M)]. Integration will give you,
lnP=-3ln(V-M) , where 'ln' is the logarithm to base e. This give lnP+ln[(V-M)3]=0.
This gives entirely different answer. So which is the correct way??
Thanks in advance...
 
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In both you forget the constant integration ##c## ...
 
sreerajt said:
I have to do a integration which goes like this:
(V-M)(dP/dx)+3P(dV/dx)=0, (where M,P and V are constants).
Something doesn't add up here.
If ##V, P## are constants, then ##\frac{dV}{dx}=0, \frac{dP}{dx}=0##. Your equation then becomes ##0+0=0##.
sreerajt said:
If you integrate with dx, you will get:
∫[(V-M)dP]+∫[3PdV]=0.
which ultimately results in the answer M=4V.
Now, i can put the first equation in this form also:
(dP/P)=-3[dV/(V-M)]. Integration will give you,
lnP=-3ln(V-M) , where 'ln' is the logarithm to base e. This give lnP+ln[(V-M)3]=0.
This gives entirely different answer. So which is the correct way??
Thanks in advance...
 
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Second as @Samy_A said not all can be constant ...
 
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OMG :nb). Ya, that's correct.
But, if P and V is a function of x, then ? In fact i did that calculation keeping this in mind that P and V is a function of x.
 
sreerajt said:
OMG :nb). Ya, that's correct.
But, if P and V is a function of x, then ? In fact i did that calculation keeping this in mind that P and V is a function of x.
If V and P are functions, how can you integrate ∫[(V-M)dP]+∫[3PdV] the way you did?
Only your second method makes sense, and as @Ssnow mentioned, don't forget the integration constant.
 
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Samy_A said:
If V and P are functions, how can you integrate ∫[(V-M)dP]+∫[3PdV] the way you did?
Only your second method makes sense, and as @Ssnow mentioned, don't forget the integration constant.
I made lot many mistakes...
thanks for pointing out...
thank you...
 
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