Undergrad Integration of 1 variable in 2 different ways.

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The discussion revolves around the integration of the equation (V-M)(dP/dx) + 3P(dV/dx) = 0, where M, P, and V are constants. Two integration methods are presented, leading to different results: one yielding M=4V and the other resulting in a logarithmic expression. Participants highlight that if V and P are constants, the derivatives dV/dx and dP/dx equal zero, making the equation trivial. The consensus suggests that if V and P are functions of x, the second integration method is more appropriate, and the importance of including the integration constant is emphasized. Ultimately, the discussion underscores the need for clarity on whether variables are constants or functions.
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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