Astrophysics
  • Astrophysics
@ganeshie8 (O.D.E. Intervals)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Astrophysics
  • 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
  • Astrophysics
Oh wait..
anonymous
  • anonymous
separation of variables?

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Astrophysics
  • Astrophysics
Yup
anonymous
  • anonymous
\[y \frac{ dy }{ dt }=\frac{ t^2 }{ 1+t^3 }\] \[y \frac{ dy }{ dt }=t^2+\frac{ 1 }{ t }\]
anonymous
  • anonymous
\[ydy=(t^2+\frac{ 1 }{ t })dt\]
Astrophysics
  • 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
  • anonymous
ohh
Astrophysics
  • Astrophysics
\[\frac{ t^2 }{ 1+t^3 } \neq t^2+ \frac{ 1 }{ t }\]
Astrophysics
  • 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
  • Astrophysics
But I have to look at the solution, I guess since we're focusing on the ivp
Astrophysics
  • Astrophysics
initial condition
anonymous
  • anonymous
mmm
anonymous
  • anonymous
so are you allowed to solve it though?
Astrophysics
  • Astrophysics
Yes
anonymous
  • anonymous
like what you did in the top post?
Astrophysics
  • Astrophysics
Yup, that's what I got when I solved it, I guess if you want to check that, would be nice :P
anonymous
  • anonymous
hmm let me think
Empty
  • Empty
Yeah I thought you weren't supposed to solve this, more like take the derivative or something I forget.
Astrophysics
  • Astrophysics
I had to solve it anyways, not sure how to deal with the interval stuff..
anonymous
  • anonymous
we need to find y(o) in terms of t i rekn
anonymous
  • anonymous
to do the interval stuff
anonymous
  • anonymous
Graph it and see if there are any discontinuities
anonymous
  • anonymous
i don't think its as easy as that..
anonymous
  • anonymous
I'm on mobile and latex is not displayed properly..
anonymous
  • anonymous
let me see how you solved y(t)
Astrophysics
  • 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
  • anonymous
lol i'm out of ideas lel
Astrophysics
  • 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
  • anonymous
you'd know more than me at this stage haha
Astrophysics
  • Astrophysics
Haha, well thanks for trying, this interval stuff is pretty difficult, at least for me
anonymous
  • anonymous
my math knowledge stops after 2nd sem eng maths 2nd yr. so i probs won't have much luck with this
Astrophysics
  • Astrophysics
I don't think I really understand theorem of existence and uniqueness either, err
anonymous
  • anonymous
haha i hope ganeshie comes and saves the day
Astrophysics
  • Astrophysics
Well finding the subset D, yeah me to lol
anonymous
  • anonymous
good luck
Astrophysics
  • Astrophysics
Thanks haha
ganeshie8
  • ganeshie8
wolfram says the solution is \[y^2 = y_0^2 + \frac{2}{3}\log (1+t^3) \]
ganeshie8
  • ganeshie8
Clearly, the domain is a subset of \((e^{-3/(2y_0^2)}-1, \infty )\)
ganeshie8
  • 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
  • Astrophysics
Oh I should've kept all the constants then, also how did you get that domain?
ganeshie8
  • 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
  • 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
  • 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
  • ganeshie8
we can say that much by looking at the solution
Astrophysics
  • Astrophysics
Ok I see, so that's where it would be unique?
ganeshie8
  • ganeshie8
you need to analyze separately for uniqueness
ganeshie8
  • 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
  • ganeshie8
vary \(y_0\) and convince yourself that an unique curve passes through \((0,y_0)\) for each of ur choice
Astrophysics
  • Astrophysics
Yes, very interesting
ganeshie8
  • ganeshie8
what can we say about the initial value \((0,0)\)
ganeshie8
  • ganeshie8
plugin x=0, you get y=0 but that makes the slope undefined, so...
Astrophysics
  • Astrophysics
Oh oops it's (0,0) so it does not exist
Astrophysics
  • Astrophysics
yes
ganeshie8
  • ganeshie8
so can we say, when the initial value \(y_0 = 0\), there is no solution ?
Astrophysics
  • Astrophysics
Yes
ganeshie8
  • ganeshie8
for all other values of \(y_0\), there exists an "unique" solution ?
Astrophysics
  • Astrophysics
I'd think so but, it doesn't look like it, since the negative and positive seem to give similar results right?
Astrophysics
  • Astrophysics
Or does that not matter
ganeshie8
  • ganeshie8
How ? (0, -2) gives you some bottom curve as solution (0, 2) gives you some top curve as solution
Astrophysics
  • Astrophysics
Ah squared!
ganeshie8
  • ganeshie8
you get different curves depending on whether \(y_0\) is positive or negative
Astrophysics
  • Astrophysics
Yes, ok I see
baru
  • baru
so you didn't need to solve it right? existance and uniqueness fails if y' is undefined at at (x0,y0)
Astrophysics
  • Astrophysics
I think so...you have to define a subset and take the partial derivative
baru
  • 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
  • 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
  • Astrophysics
Yes that's the general theorem of existence and uniqueness
baru
  • baru
ohh..ok
baru
  • baru
this eq isnt linear..so not 1?
Astrophysics
  • Astrophysics
Yeah I guess not
Astrophysics
  • Astrophysics
I don't really know how to define "unique" in each case though haha, like what exactly does it mean
baru
  • 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
  • Astrophysics
That sounds pretty good, and this can only be satisfied with a ivp?
Astrophysics
  • Astrophysics
I don't really care about the answers xD, I'm just trying to understand all of this, feel so dumb haha.
Astrophysics
  • Astrophysics
am*
baru
  • 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
  • Astrophysics
I see, so can we say all I.V.P.'s are unique?
baru
  • baru
yes...if existance and uniqueness holds.
baru
  • 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
  • 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
  • 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
  • baru
do let me know if you gain any additional insight... :)
Astrophysics
  • Astrophysics
Haha, alright, the difficult part will be understanding the subset stuff
Astrophysics
  • 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
  • Astrophysics
Having a "unique subset" haha I don't think even makes sense
baru
  • baru
lol...i didnt go into that much detail
Astrophysics
  • Astrophysics
Haha, yeah I end up overthinking not sure if that's a good thing or bad
Astrophysics
  • 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
  • baru
yep, i ask myself the same thing...overthink too much and you end up in a mathematicians territory...
Astrophysics
  • 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
  • baru
yep :)
ganeshie8
  • 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
  • Astrophysics
Ok thanks!
Astrophysics
  • 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
  • Astrophysics
should be \[I = ([e^{-3/2y_0^2}-1]^{1/3},\infty)\]

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