Integration of 1 variable in 2 different ways.

  • #1
39
1
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...
 

Answers and Replies

  • #2
Ssnow
Gold Member
563
173
In both you forget the constant integration ##c## ...
 
  • #3
Samy_A
Science Advisor
Homework Helper
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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).
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##.
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...
 
  • #4
Ssnow
Gold Member
563
173
Second as @Samy_A said not all can be constant ...
 
  • #5
39
1
OMG :nb). Ya, that's correct.
But, if P and V is a function of x, then ??? Infact i did that calculation keeping this in mind that P and V is a function of x.
 
  • #6
Samy_A
Science Advisor
Homework Helper
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OMG :nb). Ya, that's correct.
But, if P and V is a function of x, then ??? Infact 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.
 
  • #7
39
1
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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