A differential equation problem Sketch the general solution to  dy /dx = y2 - 3y. ?

Sketch the general solution to

 dy /dx = y2 - 3y. 

4 Answers

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  • 4 weeks ago

    dy/dx = y^2 - 3y

     dy = y(y - 3)dx

    ....dy

    ∫----------- =∫ dx

    ..y(y - 3)

    ......Ady............Bdy

     ------------ + ---------- = dx

    ........y.............y - 3

    A(y - 3) + By = 1

    when y = 3

    A(3 - 3) + 3B = 1

      3B = 1

      B = 1/3

    solving for A when y = 0

    A(0 - 3) + B(0) = 1

      - 3A = 1

       A = - 1/3

    ...- 1/3dy.........1/3dy

     ∫-------------- + ∫--------- = ∫dx

    .......y.................y - 3

    - 1/3ln(y) + 1/3ln(y - 3) = x + C

    1/3[ln(y - 3) - ln(y)] = x + C

    1/3[ln[(y - 3)/y]]= x + C

    ln[(y - 3)/y] = 3(x + C)

       (y - 3)/y = e^(3x+ C)

      y - 3 = ye^(3x + C)

     y - ye^(3x + C) = 3

    y[1 - e^(3x + C)] = 3

    .............3

    y = -------------------- Answer//

    .......1 - e^(C + 3x)

  • Ian H
    Lv 7
    4 weeks ago

    1/(y – 3) – 1/y = 3/(y^2 – 3y) = 3dx/dy

    ln(y – 3) – ln(y) = ln[(y – 3)/y] = 3x + k

    (y – 3)/y = e^k * e^3x = Ce^3x

    y = 3/[1 - Ce^3x]

    Here are some examples of that family of curves

    https://www.wolframalpha.com/input/?i=y+%3D+3%2F%5...

  • 4 weeks ago

    dy/dx = y² - 3y

    y' = y² - 3y

    general coordinate pairs for example for the domain [-2, 2]

    (-2,0), y' = 0

    (-2,1), y' = -2

    (-2,2), y' = -2

    (-2,3), y' = 0

    (-2,4), y' = 4

    (-2,5), y' = 15

    (-1,0), y' = 0

    (-1,1), y' = -2

    (-1,2), y' = -2

    (-1,3), y' = 0

    (-1,4), y' = 4

    (-1,5), y' = 15

    (0,0), y' = 0

    (0,1), y' = -2

    (0,2), y' = -2

    (0,3), y' = 0

    (0,4), y' = 4

    (0,5), y' = 15

    (1,0), y' = 0

    (1,1), y' = -2

    (1,2), y' = -2

    (1,3), y' = 0

    (1,4), y' = 4

    (1,5), y' = 15

    (2,0), y' = 0

    (2,1), y' = -2

    (2,2), y' = -2

    (2,3), y' = 0

    (2,4), y' = 4

    (2,5), y' = 15

    https://flic.kr/p/2jkMdwe

  • dy / (y^2 - 3y) = dx

    dy / (y * (y - 3)) = dx

    a/y + b/(y - 3) = 1/(y^2 - 3y)

    a * (y - 3) + by = 0y + 1

    ay + by - 3a = 0y + 1

    ay + by = 0y

    a + b = 0

    -3a = 1

    a = -1/3

    -1/3 + b = 0

    b = 1/3

    (-1/3) * dy / y  +  (1/3) * dy / (y - 3)  =  dx

    (1/3) * (dy / (y - 3) - dy / y) = dx

    dy / (y - 3) - dy / y = 3 * dx

    ln|y - 3| - ln|y| = 3x + C

    ln|(y - 3) / y| = 3x + C

    (y - 3) / y = e^(3x + C)

    Let e^(3x + C) = k for now

    (y - 3) / y = k

    y - 3 = ky

    y - ky = 3

    y * (1 - k) = 3

    y = 3 / (1 - k)

    y = 3 / (1 - e^(3x + C))

    e^C is just some constant C, so we can rewrite this as:

    y = 3 / (1 - e^C * e^(3x))

    y = 3 / (1 - C * e^(3x))

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