A Homemade Non-Linear Differential Equation

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  • Опубликовано: 17 сен 2024
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Комментарии • 19

  • @user-vm7wh6eb7f
    @user-vm7wh6eb7f 23 часа назад +5

    Will you please provide a PDF of difficult maths problems ?

  • @user-vm7wh6eb7f
    @user-vm7wh6eb7f 23 часа назад +3

    I love you ❤

  • @scottleung9587
    @scottleung9587 22 часа назад +1

    Nice!

  • @user-vm7wh6eb7f
    @user-vm7wh6eb7f 23 часа назад +2

    Please sir❤❤

  • @kuriana100
    @kuriana100 18 часов назад

    I am 42 years old. And i loved maths when I was younger, and your videos make my love for maths to another level.

    • @SyberMath
      @SyberMath  17 часов назад

      I'm glad to hear that! Thanks for watching 😍

    • @kuriana100
      @kuriana100 17 часов назад

      @@SyberMath You are the best. Best wishes from an Indian from Dubai. I have watched all your videos. And this brings a lot of joy to me. Thank you so much.

  • @user-vm7wh6eb7f
    @user-vm7wh6eb7f 23 часа назад +2

    Hello sir

  • @jesusalej1
    @jesusalej1 22 часа назад +1

    If c=0? That value is missed. If c=0, that integral is -1/y=x+k...

  • @jesusalej1
    @jesusalej1 22 часа назад

    That is a known result!

  • @giuseppemalaguti435
    @giuseppemalaguti435 16 часов назад

    (y')'=(y^2)'...y'=y^2+c,..dy/(y^2+c)=dx...(1/√c)arctg(y/√c)=x+C1...y/√c=tg(√c(x+C1))

  • @jesusalej1
    @jesusalej1 22 часа назад

    Have you replaced y in terms of his solution?

  • @vladimirkaplun5774
    @vladimirkaplun5774 20 часов назад

    what if c

  • @icebear771
    @icebear771 22 часа назад

    You didn't consider the cases c=0 and c

  • @phill3986
    @phill3986 21 час назад

    👍✌️☮️😀😀☮️✌️👍

  • @Don-Ensley
    @Don-Ensley 22 часа назад

    problem
    y" = 2yy'
    Note:
    d(y²)/dx = 2yy'
    and by definition
    y" = d(y')/dx
    Setting these equal we get
    d(y')/dx = d(y²)/dx
    Integrate.
    ∫ d(y')/dx dx = ∫ d(y²)/dx dx
    ∫ d(y') = ∫ d(y²)
    y' = y² + C₁
    , where C₁ is a constant of integration.
    dy/dx = y² + C₁
    dy / (y² + C₁) = dx
    Integrate a second time.
    ∫ dy / (y² + C₁) = ∫ dx
    1/C₁ ∫ dy / (1+y²/C₁) = x + C₂
    , where C₂ is a constant of integration.
    Substitute
    z = y/√C₁
    z√C₁ = y
    √C₁ dz = dy
    1/C₁ ∫ √C₁ dz / (1+z²) = x+C₂
    1/√C₁ ∫ dz / (1+z²) = x+C₂
    1/√C₁ tan⁻¹(z) = x+C₂
    1/√C₁ tan⁻¹(y/√C₁) = x+C₂
    tan⁻¹(y/√C₁) = √C₁ (x+C₂)
    y/√C₁ = tan[√C₁ (x+C₂)]
    y = √C₁ tan[√C₁ (x+C₂)]
    answer
    y = √C₁ tan[√C₁ (x+C₂)]