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

- anonymous

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

- jamiebookeater

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

k

- anonymous

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

- anonymous

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

- anonymous

thats not what my rule says to do so i cant

- anonymous

believe me you can

- anonymous

i know you can but my professor wont allow it

- anonymous

i didnt get the same answer for the integration by parts

- 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 :(

- 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

- anonymous

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

- anonymous

i will figure it out that way...thanks

- 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

- anonymous

yeh exactly what i got....thankss

- anonymous

dont forget your constant \[\phi(x)\]

- anonymous

yupp

- anonymous

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

- anonymous

yup i got that...thanks so much

- anonymous

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

- anonymous

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

- anonymous

where do you go to school

- anonymous

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

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