Question

HELP! linear algebra & differential question:
looking for exact equation

ex: (y^-3-y^-2sinx)y'+y^-1cosx = 0

my final solution should be y = ?

Answer 1

\[\left(y^{-3}-y^{-2}\sin (x)\right)y'+y^{-1}\cos (x) = 0\]

Answer 2

is this your equation?

Answer 3

yes

Answer 4

and you're looking for EXACT??

Answer 5

yes!

Answer 6

I'm reading this: http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx I'll race you to see who can solve it the fastest!

Answer 7

multiply through by y and it's pretty easy

Answer 8

well obviously if you know it then its easy but i dont so its hard and your reply is very vague...

Answer 9

multiply through by y you'll have something like

f(x,y) *dy/dx + f(x) =0

Answer 10

you use the exact solution method that he linked

Answer 11

So basically by reading this I found out that you have it in a form that's \[\Psi _{x} + \Psi _{y}\frac{ dy }{ dx }=0\] So what you have is the derivative of a function. Now you need to take the derivative of each derivative with respect to the other function to see if they're equal so you know they're an exact equation.

Answer 12

f(x,y) is the partial of some function (the solution) with respect to y so integrate to find F(x,y)
f(x,y) = d ( F(x,y) ) / dy

Answer 13

You'll find that formula I wrote on your own (I did) by using implicit differentiation. It's multivariable calculus but it's not too bad, I can help you out but keep in mind I'm still figuring this out right now lol.

Answer 14

this is what i did im trying to solve what y will equal ..

Answer 15

any help? i just need to solve for y but im not getting what my teacher put on his answer sheet.

Answer 16

I think you're done...

Answer 17

what was your question? you have it all worked out there on the page...

Answer 18

I know how to find y, you multiply your final answer by y on both sides until you can do the quadratic equations.

Answer 19

lol nvm sinx as a coefficient doesn't work, what am I thinking? I need sleep.

Answer 20

what did your teacher get?

Answer 21

I don't know how to solve for y in this case, kinda funky to me.

Answer 22

multiply by y
simplify

Answer 23

Try it out, it doesn't go so easily I'm afraid.

Answer 24

I meant in the beginning, before you do the 'exact soln.' method.... but he already did the longer version... and now he's done, but evidently his teacher got a different answer...

Answer 25

I would like to know what the teacher's answer is, because his work looks correct.

Answer 26

\[\frac{ \sin(x) }{ y } = \frac{ 1 }{ 2y^2 }+C\]

Answer 27

so he forgot the 2 when integrating to find g(y)

Answer 28

basically the aswer should be \[y = [sinx \pm(\sin^2x+c)^1/2]^-1\]

Answer 29

and the negative sign.

Answer 30

wut

Answer 31

Is that allowed adunb8? I guess when I said sin(x) can't be a coefficient I was wrong.

Answer 32

does that even say.

Answer 33

It's the quadratic formula with y in there. Makes sense to me, I stopped myself because it didn't make sense when simplifying it but really we just needed an equation for y. Good job.

Answer 34

iono how he got this!!! he a math genius but wont tell us how he got it!

Answer 35

looks like he just used the quadratic formula after multiplying through by y^2

Answer 36

Yeah, what @Kainui said...

Answer 37

oh i would use quadratic formula to solve this one?

Answer 38

\[C+y^{-1}sinx-\frac{ 1 }{ 2 }y^{-2}=0\] now multiply each side by y^2, totally allowed in algebra. You end up with a=C, b=sinx, and c=-1/2 which are the weirdest coefficients I've ever plugged into the quadratic formula lol.

Answer 39

multiply by y^2 and plug the coefficients on y^2 y etc into the quad. formula

Answer 40

omg how am i suppose to know this for a test =(

Answer 41

@Kainui where did you get the -1/2 from?

Answer 42

@adunb8 fyi, if you have the problem done and you just have a procedural question, you can say that at the beginning and save yourself and everyone else a lot of time.

Answer 43

@adunb8 go back to your integration to find g(y) and check it...

Answer 44

oh.. sorry... well i still needed the refreshment on the exact its still confusing even still..

Answer 45

it was still very helpful for me to look at all the steps.

Answer 46

You're in diffeq and don't know the quadratic formula yet...? eiiieee.... also the -1/2 came from integrating y^-2.

Answer 47

you'll see the coeff. is supposed to be -1/2

Answer 48

i mean i know how to do it but its confusing when he puts it like this and i gotta solve it from scratch..

Answer 49

I literally learned how to solve this problem from that link I showed you after I posted it and found it on google. I've only taken up to calculus II, lol.

Answer 50

That website is awesome lol.

Answer 51

It is a great great site, tons of stuff on there.

Answer 52

Earlier when I said I was racing, I was literally racing you lol. XD Well any more questions or am I off to bed?

Answer 53

no thank you so much for your help =) geez im a buzzkill

Answer 54

s'ok, we love you <3

Answer 55

=)

Answer 56

:)