Question

Refreshing Differential Equations:

How do i solve
\( \Large e^{-x} sin y-e^{-x} cos y ~ y'-x+yy'=0 \)

Answer 1

\(\Large e^{-x} sin y-e^{-x} cos y ~ y'-x+yy'=0\)

Answer 2

huh???

Answer 3

is this Bernoullis? cuz I haven't learned it yet.

Answer 4

there is a sin y in there...so i don't think its bernoullis

Answer 5

\[-e^{-x}cosy y' +yy' +e^{-x}siny-x = 0\]

Answer 6

\[y'(-e^{-x}cosy+y)+e^{-x}siny-x=0\]

Answer 7

integrating factor???

Answer 8

oh wait no...

Answer 9

separable?!

Answer 10

exact? oh gawd not substitution D:

Answer 11

not separable.
can't think of any substitution as of now...

Answer 12

hmmm for a first order ode there are 5 methods, integrating factor, separable, substitution, exact, and ...... ugh I need to search for one more. I am getting a brain freeze here.

Answer 13

i will check whether its exact or not..

Answer 14

homogeneous x+!_@

Answer 15

wait I got a supplement book... lists everything

Answer 16

nevermind.. it had what I listed

Answer 17

woah, yes its exact

Answer 18

Looks exact.

Answer 19

exact equation

Answer 20

so,
now finding Integrating factor

Answer 21

because integrating requires the y' +p(x)y=q(x) and there wasn't any y so forget that

Answer 22

don't you have to take the antiderivative in terms of x and take the derivative in terms of y

Answer 23

then whatever it was + h'(y) = N and something must cancel... and then take the antiderivative.

Answer 24

wouldn't i just do
\(\int M dx
\\\int N dy\)
?
and then combine those ?

Answer 25

what?

Answer 26

\[
u = \exp \left( \int \frac{{M_y-N_x} }{N} dx \right)
\]

Answer 27

oh wait ... are we trying to find the integrating factor first and then multiply it all the way through so when we take the partial derivatives it will be exact and then we do whatever I wrote earlier.

Answer 28

That is one option.

Answer 29

\[M(x,y)dx + N(x,y)dy = 0\]

Answer 30

Mdx + Ndy = 0
thats the type of DE, i have now.
and it is exact

so don't i just integrate M and N, and then combine ?
why do i even need to find Integrating factor! its already exact lol

Answer 31

Take yo M right... TAKE ANTIDERIVATIVE IN TERMS OF X AND TAKE THE DERIVATIVE IN TERMS OF Y! WEEEEEEEEEEEEEEEEEEEE

Answer 32

Okay, if it is exact, you still need to find \(f\)

Answer 33

What did you get for \(f\)?

Answer 34

where is f ? :O
are you asking for final answer ?

Answer 35

I feel like these methods sort of hide the fact that you're really just reverse-engineering the derivative.\[\Large \frac{d \Psi}{dx}=\Psi_x \frac{dx}{dx}+\Psi_y \frac{dy}{dx}\] It makes it easier to see what you have to integrate with respect to what.

Answer 36

:/

Answer 37

\[
\Large
(y-e^{-x} \cos y )~ dy + ( e^{-x} \sin y -x)~dx=0
\]

Answer 38

@hartnn When I say 'looks exact' I mean 'looks like making it exact could work'. Integrating factor makes it exact.

Answer 39

yeah I got this without integrating factor... doesn't look exact