Question

(ODE) another elementary Differential Equation, prompt placed below shortly.

Answer 1

http://i.imgur.com/0T9LWyn.png

Answer 2

Thinking about how to do this, first off.

Answer 3

Is it maybe separable? Gonna see if I can make it a separable eqn. @Concentrationalizing

Answer 4

SHould I actually answer that question or let you figure it out? Haha

Answer 5

it looks homogeneous

Answer 6

I dont want you to waste your time, so go with bernoulli :P

Answer 7

Nevermind, I'm back, site loading issues, keep having to freaking clear my browser history every five minutes while using OpenStudy if I expect it to work.

Answer 8

divide both sides by x^2 * y^2

Answer 9

Thank you! Alright, now I need to remember how to solve a Bernoulli DE...lol.

Answer 10

Alright, going to put it in that general form.

Answer 11

\[\frac{dy}{dx}+\frac{y^2}{x^2}=\frac{y}{x}\]

Answer 12

Now some kind of substitution, m8

Answer 13

Well, you would need the higher power of y on the other side. So
\[\frac{ dy }{ dx }-\frac{ y }{ x }= -\frac{ y^{2} }{ x^{2} }\] Do you remember the substitution to make?

Answer 14

http://i.imgur.com/aR14ol5.png

Answer 15

No, but I found it, lol

Answer 16

Cool, cool.

Answer 17

Alright, one sec, in the process of becoming slowly less incompetent

Answer 18

x^2 * dy/dx + y^2 = xy

subtract xy and y^2 from both sides

x^2 * dy/dx - xy = -y^2

divide both sides by -x^2 * y^2

(dy/dx)/(-y^2) + 1/ ( x*y) = 1/x^2

let v(x) = 1/y

dv/dx = -1/y^2 * dy/dx

so we have

dv/dx + v(x) / x = 1/x^2

this is a linear ode , we can solve by integrating factor

Answer 19

Do you think Bernoulli or Integrating Factor would be better, and why?

Answer 20

this can be worked less painfully without bernouli

Answer 21

notice that the de is homogeneous and simply substitute `y = vx`

Answer 22

you will get a separable equation straight

Answer 23

Alright, cool. I need to learn how to do Bernoulli nonetheless, so I may end up doing it that way.

It's homogeneous once we subtract xy, correct? Previously it's not homogeneous unless you're talking about some homogeneous function stuff that IDR.

Answer 24

yes homogenous is easier approach

Answer 25

homogeneous is a cool thing to learn too

Answer 26

I guess the placement of dy and dx doesnt ruin the idea of homogenous. Guess seeing it the way it was didnt make me thing homogenous :/

Answer 27

Alright, how about we do both? lol. We'll already know the solution at some given point, so reworking it with a different technique won't be bad.

Answer 28

Okay, so is this a homogeneous function thing? Need to look up/remember how to deal with those.

(I know we literally like just did one of those @Concentrationalizing , wasn't this the "of order blahblahblah" thing?)

Answer 29

Yeah. To be proper, you would show that \(f(tx,ty) = t^{\alpha}f(x,y)\) for some \(\alpha\). But looking at the combined total of powers in each term, you can see this would be homogenous degree 2. The total amount of powers in each term is 2.

Answer 30

WHOOOO GUESS WHO JUST HAD TO clEAR HIS HISTORY AGAIN YAY OPENSTUDY

Answer 31

Alright, site is finally back up for me again.

Answer 32

Alright, so homogeneous function of order 2. Now to do the substitution @Concentrationalizing

Answer 33

(PS I am still legitimately intent as solving it as a Bernoulli DE, because I need to familiarize myself with it)

Answer 34

\[y=ux.\]

Answer 35

(I'm currently looking over Ganeshie and Perl's stuff to catch up to what they did, sorry, one minute)

Answer 36

Alright, so just getting this clear, when you're evaluating the order of a DE to see whether it is homogeneous or not, derivatives don't contribute at all, right? e.g. y' won't contribute to the degree/order of the homogeneous function.

Answer 37

But yeah, looks like a homogeneous function of degree 2.

Answer 38

No, it wont. Just check the total degee of each term. All of them are 2 this time so yeah, degree 2.

Answer 39

And I'm going to quickly read over the other problem we just did to make sure I remember how to do this, if it isn't clear yet my general memory span is horrific

Answer 40

In case anybody in the future happens to find this thead through google site search and wants to see a worked Homogeneous function problem I'm mentioning, here you go: http://openstudy.com/study#/updates/54850fbbe4b01e7eabcc0904

Answer 41

Alright, going to try to get it into the form \[M(x,y)dx+N(x,y)dy=0\]now. I think Ganeshie or Perl already did that, but they didn't do it in LaTeX and it was frankly hard to read.

Answer 42

\[x^2 \frac{dy}{dx}+y^2=xy\]

Answer 43

Perl basically went along the lines of bernoulli set-up. So this is completely different.

Answer 44

Well, one of them did, I think Ganeshie did, oh well, w/e

Answer 45

Alright, I want to solve it as Bernoulli first anyways, can we do that?

Answer 46

So, putting it in the appropriate form of a Bernoulli DE:

Answer 47

He's the one who had the idea, but none of us actually did it, lol. But sure, put it into the form ya need.

Answer 48

\[\frac{dy}{dx}+\frac{y^2}{x^2}=\frac{y}{x}\]

Answer 49

Now I'm thinking I need to subtract the term on the RHS over to the LHS and do likewise with the y-term on the LHS over to the RHS so the powers of y are in an appropriate place.

Answer 50

\[\frac{dy}{dx}-\frac{y}{x}=-\frac{y^2}{x^2}\]

Answer 51

Where \[P(x)=-\frac{1}{x}, \ f(x) = -\frac{1}{x^2}, \text{and} \ n=2\]

Answer 52

The substitution made should be:\[u=y^{1-2}=y^{-1}=\frac{1}{y}\]

Answer 53

Good so far.

Answer 54

Alright, so, taking care of the y'/adjusting for it being a function of different variables now:

Answer 55

\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}= \ ? \]

Answer 56

(Having a brain fart, one moment)

Answer 57

Moment granted :D

Answer 58

http://i.imgur.com/0IF6E2y.png

Answer 59

Haha, nice.

Answer 60

Alright, actually legit forgetting how to think, sorry, could you....do the next step, lol?

Answer 61

Okay, so \(u = y^{1-n}\), meaning we have \(u=y^{-1}\). Taking the derivative, we have:
\(du = -y^{-2}y'\). Now we an multiply through by \(-y^{-2}\), giving us:

\[-y^{-2}y' -y^{-2}(\frac{ - y}{ x }) = -y^{-2}(\frac{ -y^{2} }{ x^{2} })\]
SInce \(-y^{-2}y' = du\) and \(y^{-1}\) = u, we can now make the appropriate substitutions:

\[u' + \frac{ u }{ x } = \frac{ 1 }{ x^{2} }\]

Answer 62

Oh man, alright, reading.

Answer 63

Okay, so general procedure is once you've simplified and taken the derivative with respect to y of your substitution is to take everything in that derivative other than y prime itself and multiply through that value across the entire equation?

Answer 64

Exactly. You wouldnt multiply through by the y', just the rest of it.

Answer 65

Okay, one sec.

Answer 66

Alright, I follow everything, how do we move forward from here? Is this essentially just....is using this technique on a Bernoulli DE just essentially transforming it into a form we can solve with already known, other techniques?

Answer 67

Right, we transform it into a first order, linear ODE which is now solvable using an integrating factor.

Answer 68

:C I don't like integrating factors because of obvious reasons, well, alright, and from there, after we solve it with an integrating factor, do we just back-substitute in the values for our transformation in the first place?

Answer 69

Yep. But yeah, as long as you know the process. If we skip the stupid steps, we have
\[\mu = e^{\int\limits_{}^{}\frac{ 1 }{ x }dx} = x\]
\[ux = \int\limits_{}^{}\frac{ 1 }{ x }dx\]
\[ux = \ln|x| + C \implies y^{-1}x = \ln|x| + C\] And blah blah

Answer 70

(brb in about 5 minutes)

Alright, cool

Answer 71

P.S. you still in that Starbucks?

Answer 72

Nah, I went home early in a semi-emergency. Been home for a while now.

Answer 73

Ah. Back.

Answer 74

Alright, solved it as a Bernoulli to the point that we're at another technique, good enough for me.

What I'm going to do now is try to solve it as a homogeneous DE but likewise not finish to solution, but just get it in the general form to be proven as a homogeneous DE, perform the necessary substitutions, and then leave it at that before moving on.

Answer 75

sounds good.

Answer 76

Alright, taking the original setup of the problem:
\[x^2 \frac{dy}{dx}+y^2=xy\]

Answer 77

\[x^2\frac{dy}{dx}=xy-y^2\]

Answer 78

\[\frac{dy}{dx}=\frac{xy-y^2}{x^2}\]

Answer 79

Nevermind, I went further than necessary, didn't I? Backtracking one second.

Answer 80

I would do the substitution first as soon as I knew it was homogenous.

Answer 81

\[x^2 dy=(xy-y^2)dx\]

Answer 82

Well, putting it in the general form of a homogeneous DE sounds appropriate, yes? I just want to make sure I can get it in the right form first. And there is is, right?\[(xy-y^2)dx-x^2dy=0\]Homogeneous function of order II.

Answer 83

Alright, don't tell me how to do the next part, trying to remember how to proceed. Substitution like \[y = ux, \ \ \ x=vy\]

Answer 84

Alright, not saying anything.

Answer 85

\[(xux-u^2x^2)dx-x^2(xdu+udx)=0\]

Answer 86

Is that general setup correct?

Answer 87

Thats fine.

Answer 88

Alright, great, and then...from there, not going to do it yet, but conceptually, you do what again....never did it the first time, going to have to do it eventually.

Answer 89

Now you do a normal separation of variables. Just the algebra to get there might suck a little.

Answer 90

"might suck a little"

That sounds like this entire class, lol

Answer 91

Aww, lol. I dont find too much of it annoying, only the series solutions stuff did I really not like.

Answer 92

Yeah, no, this stuff isn't all that bad, it's just that I didn't learn it right earlier and now hope I'm not going to send $1,000+ down the drain based on this final exam. It's high-stakes testing, and the immediate following morning I have another exam I am woefully unprepared for, and am having to invest all my time into this. Luckily, the other exam stuff is way easier to understand.

Speaking of separation of variables, how the hell do I do that? I think we did a problem on it, give me a second to try to remember it myself.

Answer 93

Let me know if you want me to mess with it :)

Answer 94

(The second exam, like this, is just an issue of time. ODE isn't really hard to understand....all of this stuff just takes time for consistency and performance.)

Answer 95

Alright, separation of variables is where you have a derivative term and two variables expressed in both present in the equation, you isolate all of them to one side and all of them to another and multiply through part of a differential, let's see.

Answer 96

One moment.

Answer 97

(Sorry lol)