Deriving d/dx (ye^∫pdx): Integration Factor Homework

[chwala]

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?

Homework Equations

The Attempt at a Solution

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

 

[SammyS]

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?

Homework Equations

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]

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

 

[chwala]

ok, Thanks Sammy and James, greetings from Africa