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Astrophysics
 one year ago
@ganeshie8 (O.D.E. Intervals)
Astrophysics
 one year ago
@ganeshie8 (O.D.E. Intervals)

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I 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 \[y = \left( \frac{ 2 }{ 3 }\ln1+t^3 \right)^{1/2}+y_0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0separation of variables?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y \frac{ dy }{ dt }=\frac{ t^2 }{ 1+t^3 }\] \[y \frac{ dy }{ dt }=t^2+\frac{ 1 }{ t }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ydy=(t^2+\frac{ 1 }{ t })dt\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1No 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ t^2 }{ 1+t^3 } \neq t^2+ \frac{ 1 }{ t }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I mean if I look at \[\frac{ dy }{ dt } = \frac{ t^2 }{ y(1+t^3) }\] we have t can't be 1 right

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1But I have to look at the solution, I guess since we're focusing on the ivp

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1initial condition

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so are you allowed to solve it though?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like what you did in the top post?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yup, that's what I got when I solved it, I guess if you want to check that, would be nice :P

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I thought you weren't supposed to solve this, more like take the derivative or something I forget.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I had to solve it anyways, not sure how to deal with the interval stuff..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we need to find y(o) in terms of t i rekn

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to do the interval stuff

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Graph it and see if there are any discontinuities

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't think its as easy as that..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm on mobile and latex is not displayed properly..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me see how you solved y(t)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dt } = \frac{ t^2 }{ y(1+t^3) } \implies \int\limits y dy = \int\limits \frac{ t^2 }{ (1+t^3) }dt\] doing u sub yadi, yadi, yada haha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol i'm out of ideas lel

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1After 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
 one year ago
Best ResponseYou've already chosen the best response.0you'd know more than me at this stage haha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha, well thanks for trying, this interval stuff is pretty difficult, at least for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my math knowledge stops after 2nd sem eng maths 2nd yr. so i probs won't have much luck with this

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I don't think I really understand theorem of existence and uniqueness either, err

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha i hope ganeshie comes and saves the day

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Well finding the subset D, yeah me to lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0wolfram says the solution is \[y^2 = y_0^2 + \frac{2}{3}\log (1+t^3) \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Clearly, the domain is a subset of \((e^{3/(2y_0^2)}1, \infty )\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Also, 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
 one year ago
Best ResponseYou've already chosen the best response.1Oh I should've kept all the constants then, also how did you get that domain?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(y^2 = y_0^2 + \frac{2}{3}\log (1+t^3)\) solve \(y_0^2 + \frac{2}{3}\log (1+t^3)\gt 0\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ohhh 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
 one year ago
Best ResponseYou've already chosen the best response.0notice 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
 one year ago
Best ResponseYou've already chosen the best response.0we can say that much by looking at the solution

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok I see, so that's where it would be unique?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0you need to analyze separately for uniqueness

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0look 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
 one year ago
Best ResponseYou've already chosen the best response.0vary \(y_0\) and convince yourself that an unique curve passes through \((0,y_0)\) for each of ur choice

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yes, very interesting

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what can we say about the initial value \((0,0)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0plugin x=0, you get y=0 but that makes the slope undefined, so...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh oops it's (0,0) so it does not exist

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so can we say, when the initial value \(y_0 = 0\), there is no solution ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0for all other values of \(y_0\), there exists an "unique" solution ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I'd think so but, it doesn't look like it, since the negative and positive seem to give similar results right?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Or does that not matter

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0How ? (0, 2) gives you some bottom curve as solution (0, 2) gives you some top curve as solution

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0you get different curves depending on whether \(y_0\) is positive or negative

baru
 one year ago
Best ResponseYou've already chosen the best response.0so you didn't need to solve it right? existance and uniqueness fails if y' is undefined at at (x0,y0)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I think so...you have to define a subset and take the partial derivative

baru
 one year ago
Best ResponseYou've already chosen the best response.0the existance and uniqness theorem for first order ode goes somthing like y' and d(y')/dy should be continuous around (x0,y0)...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1Yes that's the general theorem of existence and uniqueness

baru
 one year ago
Best ResponseYou've already chosen the best response.0this eq isnt linear..so not 1?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I don't really know how to define "unique" in each case though haha, like what exactly does it mean

baru
 one year ago
Best ResponseYou've already chosen the best response.0for 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
 one year ago
Best ResponseYou've already chosen the best response.1That sounds pretty good, and this can only be satisfied with a ivp?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I don't really care about the answers xD, I'm just trying to understand all of this, feel so dumb haha.

baru
 one year ago
Best ResponseYou've already chosen the best response.0yep..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
 one year ago
Best ResponseYou've already chosen the best response.1I see, so can we say all I.V.P.'s are unique?

baru
 one year ago
Best ResponseYou've already chosen the best response.0yes...if existance and uniqueness holds.

baru
 one year ago
Best ResponseYou've already chosen the best response.0but 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
 one year ago
Best ResponseYou've already chosen the best response.1Well 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
 one year ago
Best ResponseYou've already chosen the best response.1By 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
 one year ago
Best ResponseYou've already chosen the best response.0do let me know if you gain any additional insight... :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha, alright, the difficult part will be understanding the subset stuff

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Actually, 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
 one year ago
Best ResponseYou've already chosen the best response.1Having a "unique subset" haha I don't think even makes sense

baru
 one year ago
Best ResponseYou've already chosen the best response.0lol...i didnt go into that much detail

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha, yeah I end up overthinking not sure if that's a good thing or bad

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Usually 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
 one year ago
Best ResponseYou've already chosen the best response.0yep, i ask myself the same thing...overthink too much and you end up in a mathematicians territory...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha xD, now I understand why they find math so fascinating, have to pay attention to the definitions, and find the exact meaning of them!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0to understand existence and uniqueness thoroughly, you need to be an expert in real analysis let me also tag our analysis gurus @eliassaab @zzr0ck3r

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh @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
 one year ago
Best ResponseYou've already chosen the best response.1should be \[I = ([e^{3/2y_0^2}1]^{1/3},\infty)\]
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