@ganeshie8 (O.D.E. Intervals)

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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 \[y = \left( \frac{ 2 }{ 3 }\ln|1+t^3| \right)^{1/2}+y_0\]
Oh wait..
separation of variables?

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Other answers:

Yup
\[y \frac{ dy }{ dt }=\frac{ t^2 }{ 1+t^3 }\] \[y \frac{ dy }{ dt }=t^2+\frac{ 1 }{ t }\]
\[ydy=(t^2+\frac{ 1 }{ t })dt\]
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
ohh
\[\frac{ t^2 }{ 1+t^3 } \neq t^2+ \frac{ 1 }{ t }\]
I mean if I look at \[\frac{ dy }{ dt } = \frac{ t^2 }{ y(1+t^3) }\] we have t can't be -1 right
But I have to look at the solution, I guess since we're focusing on the ivp
initial condition
mmm
so are you allowed to solve it though?
Yes
like what you did in the top post?
Yup, that's what I got when I solved it, I guess if you want to check that, would be nice :P
hmm let me think
Yeah I thought you weren't supposed to solve this, more like take the derivative or something I forget.
I had to solve it anyways, not sure how to deal with the interval stuff..
we need to find y(o) in terms of t i rekn
to do the interval stuff
Graph it and see if there are any discontinuities
i don't think its as easy as that..
I'm on mobile and latex is not displayed properly..
let me see how you solved y(t)
\[\frac{ dy }{ dt } = \frac{ t^2 }{ y(1+t^3) } \implies \int\limits y dy = \int\limits \frac{ t^2 }{ (1+t^3) }dt\] doing u sub yadi, yadi, yada haha.
lol i'm out of ideas lel
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
you'd know more than me at this stage haha
Haha, well thanks for trying, this interval stuff is pretty difficult, at least for me
my math knowledge stops after 2nd sem eng maths 2nd yr. so i probs won't have much luck with this
I don't think I really understand theorem of existence and uniqueness either, err
haha i hope ganeshie comes and saves the day
Well finding the subset D, yeah me to lol
good luck
Thanks haha
wolfram says the solution is \[y^2 = y_0^2 + \frac{2}{3}\log (1+t^3) \]
Clearly, the domain is a subset of \((e^{-3/(2y_0^2)}-1, \infty )\)
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
Oh I should've kept all the constants then, also how did you get that domain?
\(y^2 = y_0^2 + \frac{2}{3}\log (1+t^3)\) solve \(y_0^2 + \frac{2}{3}\log (1+t^3)\gt 0\)
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
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
we can say that much by looking at the solution
Ok I see, so that's where it would be unique?
you need to analyze separately for uniqueness
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 ?
vary \(y_0\) and convince yourself that an unique curve passes through \((0,y_0)\) for each of ur choice
Yes, very interesting
what can we say about the initial value \((0,0)\)
plugin x=0, you get y=0 but that makes the slope undefined, so...
Oh oops it's (0,0) so it does not exist
yes
so can we say, when the initial value \(y_0 = 0\), there is no solution ?
Yes
for all other values of \(y_0\), there exists an "unique" solution ?
I'd think so but, it doesn't look like it, since the negative and positive seem to give similar results right?
Or does that not matter
How ? (0, -2) gives you some bottom curve as solution (0, 2) gives you some top curve as solution
Ah squared!
you get different curves depending on whether \(y_0\) is positive or negative
Yes, ok I see
so you didn't need to solve it right? existance and uniqueness fails if y' is undefined at at (x0,y0)
I think so...you have to define a subset and take the partial derivative
the existance and uniqness theorem for first order ode goes somthing like y' and d(y')/dy should be continuous around (x0,y0)...
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...
Yes that's the general theorem of existence and uniqueness
ohh..ok
this eq isnt linear..so not 1?
Yeah I guess not
I don't really know how to define "unique" in each case though haha, like what exactly does it mean
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 :)
That sounds pretty good, and this can only be satisfied with a ivp?
I don't really care about the answers xD, I'm just trying to understand all of this, feel so dumb haha.
am*
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 )
I see, so can we say all I.V.P.'s are unique?
yes...if existance and uniqueness holds.
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 )
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
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.
do let me know if you gain any additional insight... :)
Haha, alright, the difficult part will be understanding the subset stuff
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.
Having a "unique subset" haha I don't think even makes sense
lol...i didnt go into that much detail
Haha, yeah I end up overthinking not sure if that's a good thing or bad
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
yep, i ask myself the same thing...overthink too much and you end up in a mathematicians territory...
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!
yep :)
to understand existence and uniqueness thoroughly, you need to be an expert in real analysis let me also tag our analysis gurus @eliassaab @zzr0ck3r
Ok thanks!
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
should be \[I = ([e^{-3/2y_0^2}-1]^{1/3},\infty)\]

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