1. Jul 1, 2016 #1

   chwala

   

   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?

    

   chwala, Jul 1, 2016
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3. Jul 1, 2016 #2

   SammyS

   

   Staff Emeritus

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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}##

    

   SammyS, Jul 1, 2016
4. Jul 5, 2016 #3

   James R

   

   Science Advisor

   Homework Helper

   Gold Member

   To see that a bit more clearly, start by putting ##u=\int p~dx##. Then
   $$\frac{d}{dx}(ye^u) = \frac{dy}{dx}e^u+y(\frac{d}{du}e^u)\frac{du}{dx}$$
   $$=y'e^u + ye^u.p$$
   $$=e^u(y'+py)=(py+y')e^{\int p~dx}$$

    

   James R, Jul 5, 2016
5. Jul 5, 2016 #4

   chwala

   

   ok, Thanks Sammy and James, greetings from Africa

    

   chwala, Jul 5, 2016

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