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## anonymous 5 years ago solve the DE using exact method: [sin(xy) + xycos(xy)]dx + [1 + x^2cos(xy)]dy=0

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1. anonymous

i know its exact I am getting stuck on doing the integral of Mdx

2. anonymous

Integration by parts on the xycos(xy) doesnt work?

3. anonymous

it works but then i keep ending up having to do it several times over and over so something is wrong

4. anonymous

k

5. anonymous

k so for the integration by parts i got xsin(xy) + cos(xy)/y

6. anonymous

take the integral of [1 + x^2cos(xy)} instead

7. anonymous

thats not what my rule says to do so i cant

8. anonymous

believe me you can

9. anonymous

i know you can but my professor wont allow it

10. anonymous

i didnt get the same answer for the integration by parts

11. anonymous

it would be the integral of x/ysin(xy)dx so i took out of the 1/y since its considered a constant so your left with xsin(xy)dx which requires IP again :(

12. anonymous

take the integral of [1 + x^2cos(xy) with respect to y then take the ppartial with respect to t - do you see what I am saying -- you do it the same but opposite

13. anonymous

then i would be using the h(y) as h(x) since i am using the complete opposite formula now

14. anonymous

i will figure it out that way...thanks

15. anonymous

you will get $y+ xsin(xy)$ this is the interal of [1 + x^2cos(xy) with respect to y then take the partial with respect to x

16. anonymous

yeh exactly what i got....thankss

17. anonymous

dont forget your constant $\phi(x)$

18. anonymous

yupp

19. anonymous

$\phi(x)=0$ so the solution is y+xsin(xy)=c

20. anonymous

yup i got that...thanks so much

21. anonymous

that is the good thing about exact form if you cant take the intergral of M try the integral of N

22. anonymous

yeh i am glad i know that now....only one more problem to go with this DE stuff and im done

23. anonymous

where do you go to school

24. anonymous

college of saint elizabeth....small womans college in nj

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