Question

Solve (3x^2*cosxy-x^3*y*sinxy+4x)dx+(8y-x^4*sinxy)dy=0.

Answer 1

\[\Large\rm (3x^2 cosxy-x^3y sinxy+4x)dx+(8y-x^4 sinxy)dy=0\]Is it exact? :o

Answer 2

Integral of (8y-x^4*sinxy)dy=4y^2-y*x^4*sinxy+h(x)=C
0+h'(x)=3x^2*cosxy-x^3*y*sinxy+4x
h(x)=x^3*cosxy-x^4/4*y*sinxy+2x^2
But the answer in the book is x^3*cosxy+4y^2+2x^2=C. Yes, it's an exact equation.

Answer 3

I got x^3*cosxy-y*x^4*sinxy-x^4/4*y*sinxy+4y^2+2x^2=C. Which answer is right?

Answer 4

Ugh it's so hard to read the text :(

Answer 5

\[\Large\rm \int\limits 8y-x^4\sin xy~dy=4y^2+x^3\cos xy+ h(x)\]
Hmm yours looks quite different from this..

Answer 6

How did you get x^3*cosxy?

Answer 7

\[\Large\rm \int\limits -\sin(ay)dy=\frac{1}{a}\cos(ay)\]

Answer 8

We lose a factor of x out front when we integrate in y.

Answer 9

Your sine function didn't change when you integrated :o that was strange hehe

Answer 10

Looks like you applied the power rule to it by mistake :3

Answer 11

Uh oh, did your brain explode? :(
Ideal, noooooooooo!

Answer 12

Then how come my Ti-89 calculator says the integral of -x^4*sinxy dy=-y*x^4*sinxy?

Answer 13

Hmm, I'm not sure :o Calculators can be bonkers sometimes.
Wolfram appears to be giving us the correct information though:
https://www.wolframalpha.com/input/?i=integral+of+-x%5E4sin%28xy%29dy

Answer 14

Stiill don't believe it? :o lol

Answer 15

I do believe it. Right now, I'm working this problem out and see if it really makes sense. Please wait.

Answer 16

Okay, what's the integral of -x^3*y*sinxy?

Answer 17

with respect to which variable?

Answer 18

dx

Answer 19

I think you want to avoid integrating in x,
that's gonna be so much by parts :( painful painful painful.

Answer 20

\[\Large\rm \int\limits\limits 8y-x^4\sin xy~dy=4y^2+x^3\cos xy+ h(x)\]Taking partial with respect to x gives:\[\Large\rm 0+3x^2\cos xy-x^3y \sin xy+h'(x)\]And then we compare that to our thing with the dx on it.

Answer 21

I got it wrong again while doing partial with respect to x.

Answer 22

Taking our partial, respect to x:
\[\Large\rm (4y^2)_x+(x^3 \cos xy)_x+h(x)_x\]

Setting up product rule:\[\Large\rm =0+(x^3)_x \cos xy+ x^3(\cos xy)_x+h'(x)\]

Answer 23

\[\Large\rm =0+(3x^2) \cos xy+ x^3(-y sinxy)+h'(x)\]

Answer 24

Forget your product rule maybe?
Differentiating that cosine can be a little tough also :)

Answer 25

I got it! Thanks a lot for the help on this problem!

Answer 26

yay team \c:/