Question

(ODE) Problem regarding the Existence of a Unique Solution theorem, prompt posted below shortly.

Answer 1

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

Answer 2

The first one is something that I literally could copy-paste from the textbook, but I struggle with paraphrasing it in a meaningful way because I don't quite get the way it works.

Answer 3

this is hard
existence and uniqueness sounds like real analysis to me lol

Answer 4

Lol.

I'm trying to find where it talks about it in my book, it's hidden away somewhere.

Answer 5

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

Answer 6

Looking for equation (2) really quick.

Answer 7

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

Answer 8

Lol....god, these guys are too freaking lazy to just rewrite the equation, let me go find equation (1) now.

Answer 9

http://i.imgur.com/E0w3sn9.png

Answer 10

Alright, so that's just an IVP.

Answer 11

So it's the uniqueness/existence of a unique solution to an IVP.

Answer 12

Just gonna try and solve the second part, that should give me a better feel for how the theorem works.

Answer 13

\[y \frac{dy}{dx}=xy+1\]

Answer 14

IVP needs to be in the general form of, \[\frac{dy}{dx}=f(x,y);\]Putting this problem into that form: \[\frac{dy}{dx}=\frac{xy+1}{y}\]

Answer 15

that equation cannot be solved analytically with any of the standard known methods

Answer 16

Alright, now, uh, something something, rectangular region, partial with respect to y, MAGIC

Answer 17

y not

Answer 18

the main thing is thatu cannot have 2 different slopes at a point

Answer 19

solution slope curves cannot intersect

Answer 20

How do I know that this isn't solvable from looking at the problem, or that it does not have a solution via that theorem?

Answer 21

we need to "just show" that an unique solution exists using the existence and uniqueness thm

we don't need to find the actual solution

Answer 22

Yeah, so trying to show that...

Answer 23

Needs to have (5,3) in it, and first partial with respect to y must be continuous, along with the function itself, in that region.

Answer 24

ohk.. im still trying to interpret the statemetn of the thm

Answer 25

Yeah, same here

Answer 26

Well, alright, I'm just going to select a 2x2 region for R and see if it is continuous inside that region. Let's see. \[R=[2,4]\times[4,6]\]

Answer 27

Denominator is nonzero, everything looks sensible, no discontinuities over that given region for either x or y to my understanding, so that part of the theorem I think is fulfilled.

Answer 28

f(x,y) is continuous everywhere except at y=0

Answer 29

Yeah, and that's not in our region, so we're good AFAIK

Answer 30

same with \(\frac{\partial f}{\partial y} = \frac{-1}{y^2}\)

Answer 31

Need to check \[\frac{\partial f}{\partial y}\]

Answer 32

Oh, ok

Answer 33

Alright, lol, welp, hopefully we properly applied the theorem, and it looks like that's that, heh

Answer 34

Wait a minute

Answer 35

Got my point wrong, it was (-5,3), not (5,3)

Answer 36

It still should work, but with a different region R, I believe

Answer 37

yeah pick the region in second quadrant

Answer 38

\[R=[-4,-6]\times[2,4]\]

Answer 39

http://gifsec.com/wp-content/uploads/2014/02/boss-gifs.gif

Answer 40

doesnt work

Answer 41

Wot

Answer 42

you have (0,0) in your region

Answer 43

Lemme, lemme draw the thing I'm imagining

Answer 44

I think we're confused about how I notated the square, the left hand bracket was the x side, from negative four to negative six, and the right hand bracket was the y side/length, from two to four

Answer 45

Got you !!!