Question

Find the general solution of the differential equation \[{dy\over dx} = {{3x^2 + y}\over{2y^2 + x}}\]

Answer 1

this is going to be one of those things where we divide top and bottom by x and sub in v=y/x I think...

Answer 2

okay, I'll give it a go

Answer 3

I'm not so good at these, but here is a good set of notes on this stuff
http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx

Answer 4

i wonder if we can exact it

Answer 5

yeah I'm getting stuck with my sub
it looks exact

Answer 6

(2y^2+x) dy - (3x^2+y)dx=0

Answer 7

I think that it would be exact if it wasn't for the minus sign

Answer 8

I got stuck with that substitution too

Answer 9

yeah, when i do exact i keep kitting a spurious 2 in there

Answer 10

\[v=y/x\]\[v+xv'={3x+v\over2xv^2+1}\]\[2xv^3+2x^2v^2v'+v+xv'=3x+v\]I don't suppose this can be separated or linearized...

Answer 11

Thats where I got stuck with that substitution, I can't see what to do with that

Answer 12

wolfram isn't helping either
http://www.wolframalpha.com/input/?i=solve+dy%2Fdx%3D%283x%5E2%2By%29%2F%282y%5E2%2Bx%29

Answer 13

all I can do is request backup
@lalaly @myininaya DE help?

Answer 14

yeah, the wolf gives up too; we can try the euler method :)

Answer 15

2y(x)^2 + x = 0

y(x)^2 = -x/2

y(x) = sqrt(-x/2) ; looks like its good across the reals at least

Answer 16

draw a slope field?

Answer 17

I don't think that will cut it, we haven't even covered that in this class :)

Answer 18

slope fields is like the first thing covered in class

Answer 19

I remember it being mentioned in calculus 2 but not everyone has seen it in this DE class

Answer 20

you sure you typed it up correctly?

Answer 21

Yes, I've checked and rechecked. I've made that mistake before

Answer 22

\[\frac{dy}{dx}=\sum_{n=0}^{inf}3(-2)^nx^{1-n}y^n\]

not that it helps ...

Answer 23

and i mighta messed up the long division

Answer 24

- (3x^2+y)dx + (2y^2+x) dy=0
it is exact ODE

Answer 25

1º: integrate - (3x^2+y)dx on x
2º diferentiate the result by y
3º make equal to (2y^2+x) to find the constant
and good job

Answer 26

how do you get over that -1 not= 1?

Answer 27

.....:(

Answer 28

See I am not sure whether this is the right method or not, but please check
\[\int\limits_{}^{}(2y^2+x)dy=\int\limits_{}^{}(3x^2+y)dx\]
\[2\int\limits_{}^{}y^2dy+x \int\limits_{}^{}dy= 3\int\limits_{}^{}x^2dx +y \int\limits_{}^{}dx\]
\[2{{y^3}\over{3}} + xy = 3{{x^3}\over{3}} +xy+C\]
\[2{{y^3}\over{3}} = 3{{x^3}\over{3}}+c \]

Answer 29

tried it

Answer 30

Is it right that I can assume x to be constant while integrating with dy ??

Answer 31

i think you need to account, not for a C but for a f(x) for the dy and an g(y) for the dx

Answer 32

you can, but the resulting integrations dont necessarily have an aribitrary C, the have an arbitrary function of x or y that zeros out instead

Answer 33

But it's not exact, you can't just integrate both sides and find f(x) and g(y)

Answer 34

but since the Fxy and Fyx arent exactly the same, it throws in a monkey wrench

Answer 35

@amistre64 , thanks for confirmation :) I will try this sum in a different method

Answer 36

where did this problem come from?

Answer 37

I've been wondering if there is some integrating factor that would make this exact but haven't had any luck finding one

Answer 38

if you ignore the "-" can we get it to work??

Answer 39

This is the take-home portion of our midterm. We are encouraged to work on them in groups

Answer 40

@amistre64 I'm pretty sure we can't ignore the -

Answer 41

This thing would be so trivial if it wasn't for that darn -

Answer 42

\[C_0=-x^3-xy+xy+\frac{2}{3}y^3+C_1\]

seems to work

Answer 43

where did that come from o-0 ?

Answer 44

lol, playing around with some ideas; its prolly wrong tho

Answer 45

i dont see a way to get to it ....

Answer 46

you got any initial conditions?

Answer 47

no, no initial conditions

Answer 48

you can get back at your teacher by asking them to solve it :)

Answer 49

I could definitely use some kind of hint! Hopefully I can get one from her :)

Answer 50

@satellite73 , Can you help please??

Answer 51

maybe this helps:
opening parentesis:
\[-3x ^{2}dx -ydx +2y ^{2}dy +xdy=0\]
grouping terms trying to get exact differentials:
\[[-ydx+xdy] -[3x ^{2}dx-2y ^{2}dy] =0\]
the first term is equal to :
\[x ^{2} d(y/x)\]
i bet 2º one is something similar...

Answer 52

@myko , The second one cannot be written as you have written in first one or anything similar.

Answer 53

i think it can

Answer 54

See the position of dx and dy in first one and second one.

Answer 55

it should be the differential of oposit: d(x/y), but maybe squared or somthing like that

Answer 56

i saw this type of equations befor. You solve them this way....

Answer 57

Not possible Even of form d(xy) , d(x/y) . Nothing can be possible. Had it been
3x^2 dy−2y^2dx
then the story would have been different but it is
3x^2dx−2y^2dy
:(

Answer 58

just muliply the diffrential by x^2/y^2 and you got it

Answer 59

??

Answer 60

like i did with first one...

Answer 61

I just got an email back from the instructor and there was a typo, it should have been:
\[{dy \over dx} = {{3x^2 + y \over 2y^2 - x}}\]
It is exact!

Answer 62

lol so u still need help with it or u got it?