AAA asked in Science & MathematicsMathematics · 4 weeks ago

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

Sketch the general solution to

dy /dx = y2 - 3y.

Relevance
• 4 weeks ago

dy/dx = y^2 - 3y

dy = y(y - 3)dx

....dy

∫----------- =∫ dx

..y(y - 3)

------------ + ---------- = 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

.......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

• 4 weeks ago

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))