# Integration factor

Gold Member

## Homework Statement

## 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?

## The Attempt at a Solution

[/B]
d/dx of e^∫pdx is equal to P ....how?

SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member

## Homework Statement

## 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?

## The Attempt at a Solution

[/B]
d/dx of e^∫pdx is equal to P ....how?
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}##

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

Gold Member
ok, Thanks Sammy and James, greetings from Africa