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Astrophysics

  • one year ago

@ganeshie8 (O.D.E. Intervals)

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  1. Astrophysics
    • one year ago
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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 \[y = \left( \frac{ 2 }{ 3 }\ln|1+t^3| \right)^{1/2}+y_0\]

  2. Astrophysics
    • one year ago
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    Oh wait..

  3. anonymous
    • one year ago
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    separation of variables?

  4. Astrophysics
    • one year ago
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    Yup

  5. anonymous
    • one year ago
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    \[y \frac{ dy }{ dt }=\frac{ t^2 }{ 1+t^3 }\] \[y \frac{ dy }{ dt }=t^2+\frac{ 1 }{ t }\]

  6. anonymous
    • one year ago
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    \[ydy=(t^2+\frac{ 1 }{ t })dt\]

  7. Astrophysics
    • one year ago
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    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

  8. anonymous
    • one year ago
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    ohh

  9. Astrophysics
    • one year ago
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    \[\frac{ t^2 }{ 1+t^3 } \neq t^2+ \frac{ 1 }{ t }\]

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

  11. Astrophysics
    • one year ago
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    But I have to look at the solution, I guess since we're focusing on the ivp

  12. Astrophysics
    • one year ago
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    initial condition

  13. anonymous
    • one year ago
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    mmm

  14. anonymous
    • one year ago
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    so are you allowed to solve it though?

  15. Astrophysics
    • one year ago
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    Yes

  16. anonymous
    • one year ago
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    like what you did in the top post?

  17. Astrophysics
    • one year ago
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    Yup, that's what I got when I solved it, I guess if you want to check that, would be nice :P

  18. anonymous
    • one year ago
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    hmm let me think

  19. Empty
    • one year ago
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    Yeah I thought you weren't supposed to solve this, more like take the derivative or something I forget.

  20. Astrophysics
    • one year ago
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    I had to solve it anyways, not sure how to deal with the interval stuff..

  21. anonymous
    • one year ago
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    we need to find y(o) in terms of t i rekn

  22. anonymous
    • one year ago
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    to do the interval stuff

  23. anonymous
    • one year ago
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    Graph it and see if there are any discontinuities

  24. anonymous
    • one year ago
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    i don't think its as easy as that..

  25. anonymous
    • one year ago
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    I'm on mobile and latex is not displayed properly..

  26. anonymous
    • one year ago
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    let me see how you solved y(t)

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

  28. anonymous
    • one year ago
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    lol i'm out of ideas lel

  29. Astrophysics
    • one year ago
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    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

  30. anonymous
    • one year ago
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    you'd know more than me at this stage haha

  31. Astrophysics
    • one year ago
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    Haha, well thanks for trying, this interval stuff is pretty difficult, at least for me

  32. anonymous
    • one year ago
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    my math knowledge stops after 2nd sem eng maths 2nd yr. so i probs won't have much luck with this

  33. Astrophysics
    • one year ago
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    I don't think I really understand theorem of existence and uniqueness either, err

  34. anonymous
    • one year ago
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    haha i hope ganeshie comes and saves the day

  35. Astrophysics
    • one year ago
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    Well finding the subset D, yeah me to lol

  36. anonymous
    • one year ago
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    good luck

  37. Astrophysics
    • one year ago
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    Thanks haha

  38. ganeshie8
    • one year ago
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    wolfram says the solution is \[y^2 = y_0^2 + \frac{2}{3}\log (1+t^3) \]

  39. ganeshie8
    • one year ago
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    Clearly, the domain is a subset of \((e^{-3/(2y_0^2)}-1, \infty )\)

  40. ganeshie8
    • one year ago
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    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

  41. Astrophysics
    • one year ago
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    Oh I should've kept all the constants then, also how did you get that domain?

  42. ganeshie8
    • one year ago
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    \(y^2 = y_0^2 + \frac{2}{3}\log (1+t^3)\) solve \(y_0^2 + \frac{2}{3}\log (1+t^3)\gt 0\)

  43. Astrophysics
    • one year ago
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    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

  44. ganeshie8
    • one year ago
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    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

  45. ganeshie8
    • one year ago
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    we can say that much by looking at the solution

  46. Astrophysics
    • one year ago
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    Ok I see, so that's where it would be unique?

  47. ganeshie8
    • one year ago
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    you need to analyze separately for uniqueness

  48. ganeshie8
    • one year ago
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    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 ?

  49. ganeshie8
    • one year ago
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    vary \(y_0\) and convince yourself that an unique curve passes through \((0,y_0)\) for each of ur choice

  50. Astrophysics
    • one year ago
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    Yes, very interesting

  51. ganeshie8
    • one year ago
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    what can we say about the initial value \((0,0)\)

  52. ganeshie8
    • one year ago
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    plugin x=0, you get y=0 but that makes the slope undefined, so...

  53. Astrophysics
    • one year ago
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    Oh oops it's (0,0) so it does not exist

  54. Astrophysics
    • one year ago
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    yes

  55. ganeshie8
    • one year ago
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    so can we say, when the initial value \(y_0 = 0\), there is no solution ?

  56. Astrophysics
    • one year ago
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    Yes

  57. ganeshie8
    • one year ago
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    for all other values of \(y_0\), there exists an "unique" solution ?

  58. Astrophysics
    • one year ago
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    I'd think so but, it doesn't look like it, since the negative and positive seem to give similar results right?

  59. Astrophysics
    • one year ago
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    Or does that not matter

  60. ganeshie8
    • one year ago
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    How ? (0, -2) gives you some bottom curve as solution (0, 2) gives you some top curve as solution

  61. Astrophysics
    • one year ago
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    Ah squared!

  62. ganeshie8
    • one year ago
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    you get different curves depending on whether \(y_0\) is positive or negative

  63. Astrophysics
    • one year ago
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    Yes, ok I see

  64. baru
    • one year ago
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    so you didn't need to solve it right? existance and uniqueness fails if y' is undefined at at (x0,y0)

  65. Astrophysics
    • one year ago
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    I think so...you have to define a subset and take the partial derivative

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

  67. Astrophysics
    • one year ago
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    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...

  68. Astrophysics
    • one year ago
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    Yes that's the general theorem of existence and uniqueness

  69. baru
    • one year ago
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    ohh..ok

  70. baru
    • one year ago
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    this eq isnt linear..so not 1?

  71. Astrophysics
    • one year ago
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    Yeah I guess not

  72. Astrophysics
    • one year ago
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    I don't really know how to define "unique" in each case though haha, like what exactly does it mean

  73. baru
    • one year ago
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    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 :)

  74. Astrophysics
    • one year ago
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    That sounds pretty good, and this can only be satisfied with a ivp?

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

  76. Astrophysics
    • one year ago
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    am*

  77. baru
    • one year ago
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    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 )

  78. Astrophysics
    • one year ago
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    I see, so can we say all I.V.P.'s are unique?

  79. baru
    • one year ago
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    yes...if existance and uniqueness holds.

  80. baru
    • one year ago
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    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 )

  81. Astrophysics
    • one year ago
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    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

  82. Astrophysics
    • one year ago
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    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.

  83. baru
    • one year ago
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    do let me know if you gain any additional insight... :)

  84. Astrophysics
    • one year ago
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    Haha, alright, the difficult part will be understanding the subset stuff

  85. Astrophysics
    • one year ago
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    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.

  86. Astrophysics
    • one year ago
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    Having a "unique subset" haha I don't think even makes sense

  87. baru
    • one year ago
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    lol...i didnt go into that much detail

  88. Astrophysics
    • one year ago
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    Haha, yeah I end up overthinking not sure if that's a good thing or bad

  89. Astrophysics
    • one year ago
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    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

  90. baru
    • one year ago
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    yep, i ask myself the same thing...overthink too much and you end up in a mathematicians territory...

  91. Astrophysics
    • one year ago
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    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!

  92. baru
    • one year ago
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    yep :)

  93. ganeshie8
    • one year ago
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    to understand existence and uniqueness thoroughly, you need to be an expert in real analysis let me also tag our analysis gurus @eliassaab @zzr0ck3r

  94. Astrophysics
    • one year ago
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    Ok thanks!

  95. Astrophysics
    • one year ago
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    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

  96. Astrophysics
    • one year ago
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    should be \[I = ([e^{-3/2y_0^2}-1]^{1/3},\infty)\]

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