MATH 219 Video 3 First order Linear Equations

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  • Опубликовано: 29 окт 2024

Комментарии • 29

  • @gokberkerses107
    @gokberkerses107 4 года назад +13

    00:00 - special class of 1st order ODE w/ explicit solutions
    00:17 - definition of first order linear equations
    01:19 - y appearing just by itself, else is the function of t.
    01:40 - counter-example, non-linear
    02:19 - example with LHS and RHS issues by changing signs of p(t)
    03:30 - solving first order linear ODE’s
    04:48 - what if DE is multiplied with μ(t) ?
    06:45 - condition for DE, “to really have equality” with (μ(t) * y)’
    07:46 - condition turns into a separable equation
    09:25 - μ is found,
    09:40 - comment: redundancy of c, simple form, key analogy, non-zero c
    10:30 - μ is integrating factor
    11:12 - example, all solutions
    12:40 - whether to put +c
    14:59 - essentiality of c after integration, to find all solutions to DE
    15:44 - example w/ a given initial condition, and solution interval is sought
    17:20 - conversion into general form
    19:25 - concerns about |t|^2, (intervals which t is non-zero is to be separately treated)
    22:30 - initial value checks
    23:39 - largest interval of definition, relevant interval is determined w/ initial value check for t
    25:05 - example, which needs a simple change to make it linear
    26:20 - swapping dependent and independent variable
    28:40 - no absolute value since y=0 is problem point
    30:15 - integrate with respect to y
    30:50 - solving quadratic eq. In order to write y in terms of x
    31:28 - checking whether y=0 is a solution, keeping track of all division steps
    32:34 - exponential growth and decay, constant p
    35:37 - example w/ constant q, and an interpretation
    40:00 - a general definition of growing exponential and decaying exponential

  • @utkudenizaltiok860
    @utkudenizaltiok860 4 года назад +10

    Hocam, at 32.00 how can we say we divided it by y so we need to check y = 0 when we flipped the equation. I couldn't see the division by y.
    And at 25.45 can't we say (-inf, 0) U (0,+inf)

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  4 года назад +2

      When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked.

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  4 года назад +3

      The interval of definition should be a connected interval. The other disconnected piece is not meaningful from the standpoint of differential equations, so we ignore it.

  • @ceren6204
    @ceren6204 4 года назад +5

    Great explanation, thank you

  • @parvizhasanzade271
    @parvizhasanzade271 4 года назад +2

    hocam, I did not understand that at 24.29 why do we choose one of the intervals?

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  4 года назад +1

      For differential equations, a connected interval containing the initial point is relevant. Another, disconnected interval may seem ok for the function, but it is physically irrelevant for the differential equation.

  • @mustafabahadryaldiz8979
    @mustafabahadryaldiz8979 Год назад

    25:03 de t=0 da interval'in içinde değil mi? belki Y(t)=1/2 sabit fonksiyon olamaz mı?

  • @mertuzmez8706
    @mertuzmez8706 4 года назад +2

    At 24:40, I didn't understand how did we choose the largest open interval. Can we compare size of these two intervals ,(0,+infty) and (-infty,0)?

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  4 года назад +1

      The term "largest" is used in the sense of set inclusion; if one set is a subset of another, then the set containing the other one is considered "larger". It is not necessarily measured by length, etc. In this question, one of these two infinite intervals does not contain the initial point, so we have to discard it.

  • @edengizmen
    @edengizmen 4 года назад +1

    Hocam, at 23.30 I tested y(1) = 1/2 after leaving y alone and making C absorb (I don't know the term for it) 1/2 and t^2. Is this an okay way of solving it? If not why?

  • @emirb.4910
    @emirb.4910 4 года назад +1

    Hocam bir şey sorucam dakika 20.07de çözdüğün soruda interprating factor fonksiyonunda e^2lnt buluyoruz buarada mutlak olmadığı için; dakika 23.51de çözdüğünüz largest interval da direkt olarak t > 0 dememiz gerekmez mi? fonksiyonun hangi kısımda olduğuna bakmaksınız.

    •  3 года назад +1

      Integrating factoru e^2ln|t| şeklinde yazsaydık da olurdu, netice değişmiyor. 19:22 de bunu açıklıyor aslında.
      Yani initial condition y(-1)=1/4 de verilebilirdi mesela, o durumda da bu integrating factor çalışırdı ve (-infty,0) aralığında tanımlı bir çözüm fonksiyonu bulurduk.

  • @oguzhanydn
    @oguzhanydn 4 года назад +1

    At 32.00 Even if we swap the dependent and independent variables, why do we find out y as a result. Despite of swapping variables we didnt make any change.

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  4 года назад +1

      When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked.

  • @zeynepselvinbaloglu8531
    @zeynepselvinbaloglu8531 3 года назад +1

    I could not understand that at 28.26 why we did not write ln|y| instead of lny

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  3 года назад

      Essentially, it is because both would give us integrating factors, but finding one of them Is enough.

  • @mzan5125
    @mzan5125 4 года назад +2

    Hocam , in 32:00 how did we check y=0 was also a solution?

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  4 года назад +1

      When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked. To actually check it, substitute y=0 to the DE and see if the two sides agree.

    • @narimannovruzov396
      @narimannovruzov396 4 года назад

      @@aliulasozgurkisisel4991 hocam, can you explain it a little bit widely.I did not get it as well. How we actually write that y=0 is also a solution. The rest is satisfactory, only this part remained unclear.

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  4 года назад +1

      @@narimannovruzov396 y=0 should we checked, since at some point we divided by y. We can check it by plugging y=0 in the equation directly and see if the left and right hand sides match.

  • @oussamaawad3606
    @oussamaawad3606 3 года назад

    Hocam what is the justification for inverting the fraction at 26:30? Isn't dy/dx supposed to be just an operator and not an actual fraction?

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  3 года назад +1

      Dear Oussama, the justification is the theorem that the derivative of an inverse function is the multiplicative inverse of the derivative of the original function. The notation is designed so as to capture this fact.

    • @burakturkes3590
      @burakturkes3590 11 месяцев назад

      @@aliulasozgurkisisel4991 Hocam, how do we know the existence of the inverse function of y?

  • @cerencetin9317
    @cerencetin9317 4 года назад +1

    At 20.29 why do we need to thread t0 seperately?

    • @aliulasozgurkisisel4991
      @aliulasozgurkisisel4991  4 года назад

      The antiderivative of 1/t is ln|t|. So it gives different answers for t>0 and t>0. The equation itself has a bad point at t=0: at t=0 p(t) is not continuous. For this reason, the two intervals (0,+infty) and (-infty,0) require separate treatment.

  • @nesedertassk
    @nesedertassk 4 года назад

    Hocam, videolarda anlattığıniz yerlerin (yani kendi yazdığınız yerler) notunu da pdf şeklinde atabilirmisiniz ?