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Oh wait..

separation of variables?

Yup

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

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

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

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)

lol i'm out of ideas lel

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 \((e^{-3/(2y_0^2)}-1, \infty )\)

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

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

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 ?

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

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

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

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

am*

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

yes...if existance and uniqueness holds.

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

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

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

yep :)

Ok thanks!

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