Question

Find an integrating factor and solve the given equation.


\((3x^2y+2xy+y^3)+(x^2+y^2)y\prime=0\)

Answer 1

Well in Order to use an integrating factor this differential equation must be in the form:

\[\frac{ dy }{ dt } + P(t)y=g(t)\]

Answer 2

Problem is I don't need to see how to isolate the y' into that form.

Answer 3

Nope, we are dealing with "exact" differential equations. Different sort of problem.... unfortunately.

Answer 4

Nvm then. Sorry. Can't help.

Answer 5

Thanks anyway!

Answer 6

I have a couple ideas, but no-one is here to help.... I have a basic idea, but I am just so unsure of myself.
*goes and cries in corner*

Answer 7

Sorry, I am not certain of Exact DE myself. Looking at some of my resources right now to remember how this one worked since seemed really weird in comparison to integrating factors / separable DE. :P

Answer 8

Ok... I have an idea. I believe the integrating factor is in the form,

\(\dfrac{M_y(x,y)-N_x(x,y)}{n(x,y)}\mu\)

\(M=3x^2y+2xy+y^3\)
\(\dfrac{\partial M}{dy}(3x^2y+2xy+y^3)=3x^2+2x+3y^2\)

\(N=x^2+y^2\)
\(\dfrac{\partial N}{dx}(x^2+y^2)=2x\)

So that means....

\(\dfrac{3x^2+2x+3y^2-2x}{x^2+y^2}\mu=\dfrac{3x^2+3y^2}{x^2+y^2}\mu\)

Answer 9

o.o Interesting.
I didn't know that integrating factors existed for exact DE as well. That's something I never learned about. :)

But I have a feeling with an answer like that you might be on the right track.

Answer 10

I made a small error...
\(\dfrac{d\mu}{dx}=\dfrac{(3x^2+3y^2)\mu}{x^2+y^2}\)

And I need to find \(\mu(x)\).

Answer 11

\(\dfrac{d\mu}{dx}=3\mu\)
I have arrived here.....

Answer 12

Would that mean that \(\mu(x)=0\)?

Answer 13

that was unfortunate. got refreshed for some reason and lost reply.

anyways is that specifically solving the DE or was there another factor involved.

Since there are more solutions to that DE than mu = 0, generally exponential solutions, but I wasn't sure if there was an assumption to mu that made it mu = 0.

Answer 14

I am getting help from another site and they are saying that it would be an exponential of some sort, but I am unsure of how to even..... ugh.

Answer 15

Have you learned about separable DE?

Answer 16

I think so, but I am drawing a big blank right now.

Answer 17

basically, we would want to bring our mu's to one side and put the x's on the other

\( \displaystyle \frac{d\mu}{dx} = 3 \mu \)

\( \displaystyle \frac{1}{\mu} \frac{d\mu}{dx} = 3 \) (Assumes \(\mu \neq 0\)

I've seen it written like \( \displaystyle \frac{1}{\mu} d \mu = 3 \; dx \) and integrated, although we could integrate both sides with respect to x and the dmu /dx x would sort of cancel out as well.

Answer 18

uh, not sure if the latex just stopped working for me only or its really not displaying properly at all.

Answer 19

I can see the \(\LaTeX\), but the site has been buggy for me today.

Answer 20

yea, same. anyways, does the process seem familiar what i wrote (in some invisible ink)?

Answer 21

Yeah, I remember now. I suppose I just need some caffeine of some kind.

Answer 22

hmm looks like this is not an Exact diff equ

Answer 23

It isn't, we are getting an integrating factor to make it an exact equation. Which has been calculated to be \(\mu(x)=e^{3x}\)

Answer 24

ahh i see just finished reading all the posts

Answer 25

Now I have,

\(\mu(x)(3x^2y+2xy+y^3)+\mu(x)(x^2+y^2)y\prime=0\)

Right? And that should make it exact.

Answer 26

yes.

Answer 27

I don't know why, but when you take \[\frac{M_y-N_x}{N}= 3\]you have the \(\mu = e^3x\)just plug it in and check, you have the problem exact, now, just solve as usual. I think so

Answer 28

*\(\mu = e^{3x}\)

Answer 29

\((3e^{3x}x^2y+2e^{3x}xy+e^{3x}y^3)+(e^{3x}x^2+e^{3x}y^2)y\prime=0\)
?

Answer 30

Holy crap this is gonna suck to solve.

Answer 31

\(C=(3x^2y+y^3)e^{3x}\)
Does this look correct?

Answer 32

yes thats what i get

Answer 33

Very cool, that took much longer than I had hoped lol. Thanks @AccessDenied @Loser66 and @dumbcow
Appreciate the help!
( ^_^）o自自o（^_^ ）

Answer 34

(3x^2y + y^3)e^(3x)
dp/dx = 6xy e^(3x) + 9x^2 y e^(3x) + 3y^3 e^(3x)= 3(2xy e^(3x) + 3x^2 y e^(3x) + y^3 e^(3x))
dp/dy = 3x^2 e^(3x) + 3y^2 e^(3x) = 3(x^2 e^(3x) + y^2 e^(x) )

dp/dx dx + dp/dy dy = 0
3(2xy e^(3x) + 3x^2 y e^(3x) + y^3 e^(3x)) + 3(x^2 e^(3x) + y^2 e^(3x)) dy/dx = 0
2xy e^(3x) + 3x^2 y e^(3x) + y^3 e^(3x) + (x^2 e^(3x) + y^2 e^(3x))dy/dx = 0
looks to be what we started with so yea, good job!

Glad to be of some help! :)

Answer 35

Are you Chinese or Japanese or Korean?

Answer 36

Me? No... why?