Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

k

Answer 5

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

Answer 6

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

Answer 7

thats not what my rule says to do so i cant

Answer 8

believe me you can

Answer 9

i know you can but my professor wont allow it

Answer 10

i didnt get the same answer for the integration by parts

Answer 11

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 :(

Answer 12

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

Answer 13

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

Answer 14

i will figure it out that way...thanks

Answer 15

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

Answer 16

yeh exactly what i got....thankss

Answer 17

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

Answer 18

yupp

Answer 19

\[\phi(x)=0\]

so the solution is y+xsin(xy)=c

Answer 20

yup i got that...thanks so much

Answer 21

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

Answer 22

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

Answer 23

where do you go to school

Answer 24

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