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cwrw238
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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
 one year ago
 one year ago
cwrw238 Group Title
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
 one year ago
 one year ago

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cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
that should be (x+1) y'  xy = x
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
first divide through by (x+1) right?
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
yes thats what i got
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
right so its (1+x)e^(x) y'  x e^(x) y = x e^(x) ok?
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
i must admit i'm struggling a bit with these
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
right and integrating xe^(x) I got e^(x)(1 + x) + A
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
same here, so far you have. \[\large y(x)(x+1)e^{x}=e^{x}(x+1)+C \]
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.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: \[\Large y(x)=\frac{C}{(x+1)e^{x}}1= \frac{Ce^{x}}{x+1}1 \]
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
yea i dont know why they did that
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
makes no sense to me, I just asked WolframAlpha and our answer seems to be correct.
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
what you did seems correct, just seems like a small failure of plus and minus, in the book apparently.
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
right thanks
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
a typo thanks for your help
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
no problem.
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
oh one thing  when you are working out the Integrating factor you dont add a constant of integration right?
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
yes exactly, you can ignore this one during the process (:
 one year ago
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