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I Integration of 1 variable in 2 different ways.

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  1. Apr 14, 2016 #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...
     
  2. jcsd
  3. Apr 14, 2016 #2

    Ssnow

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    In both you forget the constant integration ##c## ...
     
  4. Apr 14, 2016 #3

    Samy_A

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    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##.
     
  5. Apr 14, 2016 #4

    Ssnow

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    Second as @Samy_A said not all can be constant ...
     
  6. Apr 14, 2016 #5
    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.
     
  7. Apr 14, 2016 #6

    Samy_A

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    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.
     
  8. Apr 14, 2016 #7
    I made lot many mistakes....
    thanks for pointing out...
    thank you...
     
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