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Integration factor

  1. Jul 1, 2016 #1
    1. The problem statement, all variables and given/known data
    ## d/dx (ye^∫pdx = Py+ y'e^∫pdx##
    now i know how they got ## y'e^∫pdx## .
    How do you differentiate ##ye^∫pdx## to get the first part i.e## Py ## presumably by product rule?

    2. Relevant equations


    3. The attempt at a solution
    d/dx of e^∫pdx is equal to P ....how?
     
  2. jcsd
  3. Jul 1, 2016 #2

    SammyS

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    Corrected (LaTeX) version of your first equation? ## \ d/dx (ye^{∫pdx}) = py+ y'e^{∫pdx}##
    Well they are not equal.
    However, a set of parentheses can fix that.
    ## \ d/dx (ye^{∫pdx}) = (py+ y')e^{∫pdx}##
     
  4. Jul 5, 2016 #3

    James R

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    To see that a bit more clearly, start by putting ##u=\int p~dx##. Then
    $$\frac{d}{dx}(ye^u) = \frac{dy}{dx}e^u+y(\frac{d}{du}e^u)\frac{du}{dx}$$
    $$=y'e^u + ye^u.p$$
    $$=e^u(y'+py)=(py+y')e^{\int p~dx}$$
     
  5. Jul 5, 2016 #4
    ok, Thanks Sammy and James, greetings from Africa
     
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