## cwrw238 one year ago 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

1. cwrw238

that should be (x+1) y' - xy = x

2. cwrw238

first divide through by (x+1) right?

3. Spacelimbus

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

4. cwrw238

yes thats what i got

5. Spacelimbus

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.

6. cwrw238

right so its (1+x)e^(-x) y' - x e^(-x) y = x e^(-x) ok?

7. cwrw238

i must admit i'm struggling a bit with these

8. Spacelimbus

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.

9. cwrw238

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

10. Spacelimbus

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

11. cwrw238

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

12. Spacelimbus

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

13. cwrw238

yea i dont know why they did that

14. Spacelimbus

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

15. Spacelimbus

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

16. cwrw238

right thanks

17. cwrw238

a typo thanks for your help

18. Spacelimbus

no problem.

19. cwrw238

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

20. Spacelimbus

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