Question

@ganeshie8 (O.D.E. Intervals)

Answer 1

I have to find the interval where the solution exists depends on the initial value \(y_0\).
\[y'=\frac{ t^2 }{ y(1+t^3) }\] where the initial condition is \(y(0)=y_0\)

I got \[y = \left( \frac{ 2 }{ 3 }\ln|1+t^3| \right)^{1/2}+y_0\]

Answer 2

Oh wait..

Answer 3

separation of variables?

Answer 4

Yup

Answer 5

\[y \frac{ dy }{ dt }=\frac{ t^2 }{ 1+t^3 }\]
\[y \frac{ dy }{ dt }=t^2+\frac{ 1 }{ t }\]

Answer 6

\[ydy=(t^2+\frac{ 1 }{ t })dt\]

Answer 7

No you cannot do that :P, but that's not what I need help with, I need to find the interval that depends on y_0 xD which I don't know how to do

Answer 8

ohh

Answer 9

\[\frac{ t^2 }{ 1+t^3 } \neq t^2+ \frac{ 1 }{ t }\]

Answer 10

I mean if I look at \[\frac{ dy }{ dt } = \frac{ t^2 }{ y(1+t^3) }\] we have t can't be -1 right

Answer 11

But I have to look at the solution, I guess since we're focusing on the ivp

Answer 12

initial condition

Answer 13

mmm

Answer 14

so are you allowed to solve it though?

Answer 15

Yes

Answer 16

like what you did in the top post?

Answer 17

Yup, that's what I got when I solved it, I guess if you want to check that, would be nice :P

Answer 18

hmm let me think

Answer 19

Yeah I thought you weren't supposed to solve this, more like take the derivative or something I forget.

Answer 20

I had to solve it anyways, not sure how to deal with the interval stuff..

Answer 21

we need to find y(o) in terms of t i rekn

Answer 22

to do the interval stuff

Answer 23

Graph it and see if there are any discontinuities

Answer 24

i don't think its as easy as that..

Answer 25

I'm on mobile and latex is not displayed properly..

Answer 26

let me see how you solved y(t)

Answer 27

\[\frac{ dy }{ dt } = \frac{ t^2 }{ y(1+t^3) } \implies \int\limits y dy = \int\limits \frac{ t^2 }{ (1+t^3) }dt\] doing u sub yadi, yadi, yada haha.

Answer 28

lol i'm out of ideas lel

Answer 29

After it says to talk about the theorem of existence and uniqueness, so I think I may have to do that first, or it at least requires that

Answer 30

you'd know more than me at this stage haha

Answer 31

Haha, well thanks for trying, this interval stuff is pretty difficult, at least for me

Answer 32

my math knowledge stops after 2nd sem eng maths 2nd yr. so i probs won't have much luck with this

Answer 33

I don't think I really understand theorem of existence and uniqueness either, err

Answer 34

haha i hope ganeshie comes and saves the day

Answer 35

Well finding the subset D, yeah me to lol

Answer 36

good luck

Answer 37

Thanks haha

Answer 38

wolfram says the solution is
\[y^2 = y_0^2 + \frac{2}{3}\log (1+t^3) \]

Answer 39

Clearly, the domain is a subset of \((e^{-3/(2y_0^2)}-1, \infty )\)

Answer 40

Also, depending on whether the intiial value, \(y_0\) is "positive or negative", the solution curve is either "top one or bottom one" respectively :

https://www.desmos.com/calculator/ugxiyenayy

Answer 41

Oh I should've kept all the constants then, also how did you get that domain?

Answer 42

\(y^2 = y_0^2 + \frac{2}{3}\log (1+t^3)\)

solve \(y_0^2 + \frac{2}{3}\log (1+t^3)\gt 0\)

Answer 43

Ohhh wow, so if it depends on the initial value you just look at your solution then? I still thought I had to consider everything separately, but that makes a whole lot of sense

Answer 44

notice that i was careful with what i put above, i didn't say that the domain is \((e^{-3/(2y_0^2)}-1, \infty )\)

i only said that \((-\infty , e^{-3/(2y_0^2)}-1]\) must be excluded from the domain

Answer 45

we can say that much by looking at the solution

Answer 46

Ok I see, so that's where it would be unique?

Answer 47

you need to analyze separately for uniqueness

Answer 48

look at the graph,
for each positive value of \(y_0\), there exists an unique curve that passes through \((0, y_0)\)

for each negative value of \(y_0\), there exists an unique curve that passes through \((0, y_0)\)

yes ?

Answer 49

vary \(y_0\) and convince yourself that an unique curve passes through \((0,y_0)\) for each of ur choice

Answer 50

Yes, very interesting

Answer 51

what can we say about the initial value \((0,0)\)

Answer 52

plugin x=0, you get y=0
but that makes the slope undefined, so...

Answer 53

Oh oops it's (0,0) so it does not exist

Answer 54

yes

Answer 55

so can we say, when the initial value \(y_0 = 0\), there is no solution ?

Answer 56

Yes

Answer 57

for all other values of \(y_0\), there exists an "unique" solution ?

Answer 58

I'd think so but, it doesn't look like it, since the negative and positive seem to give similar results right?

Answer 59

Or does that not matter

Answer 60

How ?
(0, -2) gives you some bottom curve as solution
(0, 2) gives you some top curve as solution

Answer 61

Ah squared!

Answer 62

you get different curves depending on whether \(y_0\) is positive or negative

Answer 63

Yes, ok I see

Answer 64

so you didn't need to solve it right? existance and uniqueness fails if y' is undefined at at (x0,y0)

Answer 65

I think so...you have to define a subset and take the partial derivative

Answer 66

the existance and uniqness theorem for first order ode goes somthing like y' and d(y')/dy should be continuous around (x0,y0)...

Answer 67

I think I'm just mixing up all of these theorems, there's three of them:
Theorem of existence and unique for linear eq.
General theorem of existence and uniqueness theorem (which I believe is this question)
Theorem of existence and uniqueness, which is not limited to the linear eq. I think 1 and 3 are similar...

Answer 68

Yes that's the general theorem of existence and uniqueness

Answer 69

ohh..ok

Answer 70

this eq isnt linear..so not 1?

Answer 71

Yeah I guess not

Answer 72

I don't really know how to define "unique" in each case though haha, like what exactly does it mean

Answer 73

for the first order eq: there are infinite number of curves that satisfy a diff equation, but for a given point (x0,y0) only one curve passes through it, consequently no two solution curves can be tangent or intersect. thats my understanding :)

Answer 74

That sounds pretty good, and this can only be satisfied with a ivp?

Answer 75

I don't really care about the answers xD, I'm just trying to understand all of this, feel so dumb haha.

Answer 76

am*

Answer 77

yep..thats why its an ivp. essentially an ivp just supplies a point (x0,y0). if existence and uniqueness holds, i.e one and only one solution passes through a given point, then all the information i would need to single out a particular solution from the infinitely many solutions would be a point on the solution curve(i.e x0,y0 )

Answer 78

I see, so can we say all I.V.P.'s are unique?

Answer 79

yes...if existance and uniqueness holds.

Answer 80

but it works a bit differently for higher order ODEs... infinite solutions pass through a given point(even if existance and uniqeness holds), so you would need more info than just a point (x0,y0)....(ignore if you havent reached higer order )

Answer 81

Well we did start second order, but it seems the third theorem I mentioned would fall into the higher order category but I just realized/ looks like, "general theorem..." seems to be for IVPs. Yeah, I also think you would need more initial conditions then correct

Answer 82

By the way thanks everyone, that was very helpful, I'm going to go over these definitions and try to figure out what exactly they are saying haha.

Answer 83

do let me know if you gain any additional insight... :)

Answer 84

Haha, alright, the difficult part will be understanding the subset stuff

Answer 85

Actually, the initial conditions become a subset of of the domain, so that would make the subset unique..hmm haha, and I think specifically that's where the unique solution is, in that subset, ok I think I'm overthinking this lol.

Answer 86

Having a "unique subset" haha I don't think even makes sense

Answer 87

lol...i didnt go into that much detail

Answer 88

Haha, yeah I end up overthinking not sure if that's a good thing or bad

Answer 89

Usually when I see definitions in math I'm clueless, and don't really pay attention, but this is pretty interesting , probably because I actually need it

Answer 90

yep, i ask myself the same thing...overthink too much and you end up in a mathematicians territory...

Answer 91

Haha xD, now I understand why they find math so fascinating, have to pay attention to the definitions, and find the exact meaning of them!

Answer 92

yep :)

Answer 93

to understand existence and uniqueness thoroughly, you need to be an expert in real analysis
let me also tag our analysis gurus @eliassaab @zzr0ck3r

Answer 94

Ok thanks!

Answer 95

Oh @ganeshie8 so the interval we have, is that a subset for the DE pretty much? It's like going backwards from the general and uniqueness theorem kind of right

Answer 96

should be \[I = ([e^{-3/2y_0^2}-1]^{1/3},\infty)\]