@ganeshie8 (O.D.E. Intervals)

- Astrophysics

@ganeshie8 (O.D.E. Intervals)

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- Astrophysics

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

- Astrophysics

Oh wait..

- anonymous

separation of variables?

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## More answers

- Astrophysics

Yup

- anonymous

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

- anonymous

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

- Astrophysics

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

- anonymous

ohh

- Astrophysics

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

- Astrophysics

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

- Astrophysics

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

- Astrophysics

initial condition

- anonymous

mmm

- anonymous

so are you allowed to solve it though?

- Astrophysics

Yes

- anonymous

like what you did in the top post?

- Astrophysics

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

- anonymous

hmm let me think

- Empty

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

- Astrophysics

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

- anonymous

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

- anonymous

to do the interval stuff

- anonymous

Graph it and see if there are any discontinuities

- anonymous

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

- anonymous

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

- anonymous

let me see how you solved y(t)

- Astrophysics

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

- anonymous

lol i'm out of ideas lel

- Astrophysics

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

- anonymous

you'd know more than me at this stage haha

- Astrophysics

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

- anonymous

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

- Astrophysics

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

- anonymous

haha i hope ganeshie comes and saves the day

- Astrophysics

Well finding the subset D, yeah me to lol

- anonymous

good luck

- Astrophysics

Thanks haha

- ganeshie8

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

- ganeshie8

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

- ganeshie8

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

- Astrophysics

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

- ganeshie8

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

- Astrophysics

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

- ganeshie8

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

- ganeshie8

we can say that much by looking at the solution

- Astrophysics

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

- ganeshie8

you need to analyze separately for uniqueness

- ganeshie8

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 ?

- ganeshie8

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

- Astrophysics

Yes, very interesting

- ganeshie8

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

- ganeshie8

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

- Astrophysics

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

- Astrophysics

yes

- ganeshie8

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

- Astrophysics

Yes

- ganeshie8

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

- Astrophysics

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

- Astrophysics

Or does that not matter

- ganeshie8

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

- Astrophysics

Ah squared!

- ganeshie8

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

- Astrophysics

Yes, ok I see

- baru

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

- Astrophysics

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

- baru

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

- Astrophysics

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...

- Astrophysics

Yes that's the general theorem of existence and uniqueness

- baru

ohh..ok

- baru

this eq isnt linear..so not 1?

- Astrophysics

Yeah I guess not

- Astrophysics

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

- baru

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 :)

- Astrophysics

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

- Astrophysics

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

- Astrophysics

am*

- baru

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 )

- Astrophysics

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

- baru

yes...if existance and uniqueness holds.

- baru

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 )

- Astrophysics

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

- Astrophysics

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.

- baru

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

- Astrophysics

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

- Astrophysics

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.

- Astrophysics

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

- baru

lol...i didnt go into that much detail

- Astrophysics

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

- Astrophysics

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

- baru

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

- Astrophysics

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!

- baru

yep :)

- ganeshie8

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

- Astrophysics

Ok thanks!

- Astrophysics

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

- Astrophysics

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

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