Can You Solve An Interesting Differential Equation?

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  • Опубликовано: 6 ноя 2024
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Комментарии • 15

  • @Nobodyman181
    @Nobodyman181 6 месяцев назад +1

    Great❤

  • @scottleung9587
    @scottleung9587 6 месяцев назад +3

    Nice job - I knew I wasn't gonna solve it if you were gonna talk about it for 13 min, but it was fun watching you walk us thru the steps nonetheless.

  • @ivangalechko5174
    @ivangalechko5174 6 месяцев назад +1

    Очень интересные видео

  • @mikecaetano
    @mikecaetano 6 месяцев назад

    Hmm, reminds me a touch of Halley's method for approximating roots.

  • @renedelatorre2138
    @renedelatorre2138 6 месяцев назад +4

    There is a standard way to solve this equation. Notice that the independent variable is not explicitly in the equation. Let y'=p
    y''=(dp/dx)(dy/dy)
    =(dp/dy)(dy/dx)
    y"=pdp/dy
    This will reduce the equation to a first order differential equation without resorting to tricks.

    • @pelledanasten1615
      @pelledanasten1615 6 месяцев назад

      So what would y be? Integral of p + c? I dont see how that is easier

    • @allozovsky
      @allozovsky 6 месяцев назад +1

      Substitute back, so that it will eventually give us the same equation *y' = 2√y + k.* And yeah, that is a standard way to solve such differential equations (without the independent variable explicitly present).

  • @barakathaider6333
    @barakathaider6333 6 месяцев назад

    👍

  • @preoalex8298
    @preoalex8298 6 месяцев назад

    Why aren’t you using differential / leibniz notation. Using lagranges notation is pretty confusing.

  • @christopherellis2663
    @christopherellis2663 6 месяцев назад

    6:08 4=y?

    • @jaspercarran3432
      @jaspercarran3432 6 месяцев назад

      I think you misunderstood the notation, y-prime is different to y^1, although they look very similar

  • @wrc2933
    @wrc2933 6 месяцев назад

    Can I have some help? I got y=(x+c1/2)^2

    • @wrc2933
      @wrc2933 6 месяцев назад

      Never mind I figured it out- I was doing it with respect to y

  • @rob876
    @rob876 6 месяцев назад

    y''/y' = 1/√y
    ∫dy' = ∫dy/√y
    y' = 2√y + c
    ∫dy/(2√y + c) = ∫dx
    let u = 2√y + c
    du = dy/√y
    1/2∫(1 - c/u)du = x + a
    1/2(u - c ln u) = x + a
    1/2(2√y + c - c ln(2√y + c)) = x + a
    (2√y + c)/2 - c/2 ln (2√y + c) = x + a