Question

Whats the integrating factor for the following FODE - I've worked it out but im not sure if its right.

(x+1)f' - xy = x

Answer 1

that should be

(x+1) y' - xy = x

Answer 2

first divide through by (x+1) right?

Answer 3

So far I end up at \[ \large \mu(x) = (x+1)e^{-x} \] In case you got the same, otherwise I would have to take a look again

Answer 4

yes thats what i got

Answer 5

alright, in this case we just need to confirm now if it works out, for the LHS it should be possible to be written in the form of a product rule.

Answer 6

right so its

(1+x)e^(-x) y' - x e^(-x) y = x e^(-x)

ok?

Answer 7

i must admit i'm struggling a bit with these

Answer 8

Exactly, and you can verify for yourself that this is equal as writing

\[ \Large \frac{d(y(x+1)e^{-x})}{dx}=xe^{-x} \]
It's a bit edgy to multiply it out, but it worked for me.

Answer 9

right and integrating xe^(-x|) I got -e^(-x)(1 + x) + A

Answer 10

same here, so far you have.
\[\large y(x)(x+1)e^{-x}=-e^{-x}(x+1)+C \]

Answer 11

right
so i next divided rhough by e ^(-x)

to get

y(x+1) = -1((1+x) + Ce^x

y(x + 1) = Ce^x -x - 1

and thats my result but the book gives y(x+1) = Ce^x -x + 1

Answer 12

hmm I wonder why they divide like that, usually you want to solve a differential equation for the explicit form, therefore solving for y(x), this gives me:

\[\Large y(x)=\frac{C}{(x+1)e^{-x}}-1= \frac{Ce^{x}}{x+1}-1 \]

Answer 13

yea i dont know why they did that

Answer 14

makes no sense to me, I just asked WolframAlpha and our answer seems to be correct.

Answer 15

what you did seems correct, just seems like a small failure of plus and minus, in the book apparently.

Answer 16

right
thanks

Answer 17

a typo
thanks for your help

Answer 18

no problem.

Answer 19

oh one thing -
when you are working out the Integrating factor you dont add a constant of integration right?

Answer 20

yes exactly, you can ignore this one during the process (-: