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

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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

Differential Equations
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that should be (x+1) y' - xy = x
first divide through by (x+1) right?
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

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yes thats what i got
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.
right so its (1+x)e^(-x) y' - x e^(-x) y = x e^(-x) ok?
i must admit i'm struggling a bit with these
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.
right and integrating xe^(-x|) I got -e^(-x)(1 + x) + A
same here, so far you have. \[\large y(x)(x+1)e^{-x}=-e^{-x}(x+1)+C \]
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
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: \[\Large y(x)=\frac{C}{(x+1)e^{-x}}-1= \frac{Ce^{x}}{x+1}-1 \]
yea i dont know why they did that
makes no sense to me, I just asked WolframAlpha and our answer seems to be correct.
what you did seems correct, just seems like a small failure of plus and minus, in the book apparently.
right thanks
a typo thanks for your help
no problem.
oh one thing - when you are working out the Integrating factor you dont add a constant of integration right?
yes exactly, you can ignore this one during the process (-:

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