A Beautiful Differential Equation

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  • Опубликовано: 6 ноя 2024

Комментарии • 15

  • @FisicTrapella
    @FisicTrapella 5 месяцев назад +3

    I did the same as you... but it is hard to solve without WA and Lambert's W

  • @mcwulf25
    @mcwulf25 4 месяца назад

    As soon as I saw that ln(...) I knew it was going to feature a W.

  • @moeinraies8313
    @moeinraies8313 5 месяцев назад +1

    Hi.I can solve this solution in 2 line.this is very simple and easy.

    • @moeinraies8313
      @moeinraies8313 5 месяцев назад +1

      I can mail it if you want

    • @moeinraies8313
      @moeinraies8313 5 месяцев назад

      my solution is very different and practical for each kind of these issues

    • @SyberMath
      @SyberMath  5 месяцев назад

      sure

  • @freeze2win697
    @freeze2win697 5 месяцев назад +4

    4:53 🤔😱💀👹👺😈👿😠👽🤖

    • @SyberMath
      @SyberMath  5 месяцев назад +1

      😍😊😄😁😜😮

  • @HATTRICK202
    @HATTRICK202 5 месяцев назад +2

    Hello I am from India.
    I wanna know why are programming engineers taught physics and chemistry in 1st year? Mathematics is helpful in further years but these two?

    • @klepikovmd
      @klepikovmd 5 месяцев назад +2

      Probably to understand basics of topics in which you probably will work. Chemical engineering, construction, and science require a lot of programmers

    • @HATTRICK202
      @HATTRICK202 5 месяцев назад +1

      @@klepikovmd
      Okay 👍🏻

  • @NowInAus
    @NowInAus 5 месяцев назад +1

    Lovely but a bit rushed at the end.

  • @davidseed2939
    @davidseed2939 5 месяцев назад +2

    i got y=x²
    y' =2x
    y'' =2
    ζy=x
    y'/y''=x=ζy
    i cant make sense of the given solutions.
    eg y =x² +f(x)
    y'=2x +f'
    y''=2+ f''
    (y'/y '')² = x²+f
    (2x+f')² =( 2+f'')²(x²+f)
    4x² +4f'x +f'² =
    ( 4+4f''+f''²)(x²+f)
    = 4x² +4f + (4f''+f''²)(x²+f)
    so
    4f'x+f'² =4f+f(4f''+f"²) +
    (4f''+f''²)(x²)
    cant see it

    • @bradwilliams7198
      @bradwilliams7198 5 месяцев назад +1

      I also got y=x^2 by looking for a power law solution (assume y = a*x^b and solve for a,b). If you take the expression at 7:25 and let c_1 = 0 (a solution, but not the most general one), then solve for y, you get y = (x+c_2)^2. So y = x^2 is the solution for the case where c_1 and c_2 are both zero.