Question

differential equation...
pls help!

Answer 1

\[\rm [ysin(xy)+xy^2\cos(xy)]dx+[xsin(xy)+x^2ycos(xy)]dy=0\]

Answer 2

i got it as exact
but please do verify
im struggling with the integration part

Answer 3

just by looking at it, we can say its exact,

Mdx + Ndy

if you interchange x and y, you interchange M and N too.

now whats the integral you got?

Answer 4

Are you struggling to do the integral, @rvc

Answer 5

\[\rm \int\limits_{}^{}Mdx=c ~will~be~the~ans~correct?\]

Answer 6

Sort of.

Answer 7

if you take the integral of M dx keep in mind you are missing a function of y.

Answer 8

i need to find the integral
somewhat i forgot how to find the integral xD

Answer 9

Integrate by parts, and take y to be a constant.

Answer 10

yep thats what im doing

Answer 11

recall:
\(\int u dv=uv-\int v du\)

Answer 12

WAIT y is a constant!!!!!!!!!!!!!!!
i totally forgot about that!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 13

*facepalm* :P

Answer 14

\[\int\limits_{}^{}uvdx=u \int\limits v dx - \int\limits{\int\limits vdx \cdot u'}dx\]

Answer 15

ya :)
recall that your solution is in form \(\Psi=C\)
and \(\Psi\) is basically a potential function here, and an exact equation "kind of" gives you the gradient of the function with M=\(\Psi_x\) and N=\(\Psi_y\) and because those are partial derivatives, y is treated as a constant when you take the derivative, so y will be a constant when you integrate

Answer 16

lol
m gone mad today
how could i forget that? xD
what will be the integral of xcos(xy)?

Answer 17

take u=x, and v=cos(xy) and let's find out :)

Answer 18

yep i did the same

Answer 19

sin(xy)- \(\rm \int \frac{sin(xy)}{y}dx \)

Answer 20

ohk i got it
again i was considering y to be a variable lol

Answer 21

what'd you get?

Answer 22

im sorry
the page was loading

Answer 23

no problem, let's keep going!

Answer 24

\[-->\int\limits_{}^{}ysin(xy)dx+\int\limits xy^2\cos(xy)dx\\ --> -\cos(xy)+y^2\int\limits [x\cdot \cos(xy)]dx\\ --> -\cos(xy)+[x(\sin(xy)/y+\int\limits(\sin(xy)/y))]\]

Answer 25

where did your y^2 term go?

Answer 26

oops
forgot to type :(

Answer 27

typos are there


final ans : xysin(xy)

Answer 28

okay now what

Answer 29

done !
tysm <3

Answer 30

np :)

Answer 31

this was a simple problem n i forgot to take y as a constant lol

Answer 32

tysm everyone

Answer 33

\[[\frac{ y^2 }{ (y-x)^2 }-\frac{ 1 }{ x }] dx+(\frac{ 1 }{ y }-\frac{ x^2 }{(x-y)^2 })dy=0\]

Answer 34

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