Question

Integration Step Confusion :'(

Answer 1

I'm really confused to as how the circled terms ''disappear'' in the next line.

Can anyone else how the step containing the circled terms goes to the next step?

Answer 2

having trouble reading your writing ...sorry
what is the initial diff equ ? looks like you are using an integrating factor?

Answer 3

Sorry, let me use the equation tool here:

Answer 4

\[e^{-kt} . \frac{ dN }{ dt } - k e^{-kt} . N = 0\]

Is the first step.

Answer 5

It's from an integral application question based on Growth & Decay.

Yes, I'm using the I.F

Answer 6

oh ok i got it now
the reason the next line is:
\[\frac{d}{dt}(e^{-kt}N) = 0\]
due to the product rule working backwords
note
\[(fg)' = f'g + fg'\]

Answer 7

Hmm.....

Answer 8

f = e^(-kt) ......... f' = -ke^(-kt)
g = N ................ g' = dN/dt

Answer 9

Ahhhhhhhhh

e^a = 1/ae^a ?

Answer 10

\[e^{ax} = \frac{ 1 }{ a }e^{ax}\]

Answer 11

It's essentially the reverse ?

Answer 12

wait that would be the anti-derivative....
\[\frac{d}{dx} e^{ax} = a e^{ax}\]

Answer 13

note no integrating has been done yet

Answer 14

\[\int\limits e^{ax} = \frac{1}{a} e^{ax}\]

Answer 15

I see that. Do you have a numerical only example of the reverse product?

Sorry for this :'( It's one step out of a 2 A4 page question thats confusing me

Answer 16

an example:
\[\frac{d}{dx}(xy) = x*\frac{dy}{dx} + 1*y\]

Answer 17

Ah, thats clearer.

Answer 18

Thanks ya Dumbcow! :D

Answer 19

yw....the selection of the I.F. makes this possible every time