Find solution to the initial value problem?

y ' = 3y(1 - (y/4)) , y(0) = 2

please show steps. thanks

dy/dx = 3y - 3y^2/4

dy = (3y - 3y^2/4) dx

dx = dy/(3y - 3y^2/4)

dx = 4 dy / (12y - 3y^2)

1/4 dx = dy / (3 * (4y - 3y^2))

3/4 dx = dy / (4y - y^2)

-3/4 dx = dy/(y^2 - 4y)

-3/4 dx = dy/[y*(y - 4)]

use partial fractions

A/y + B/(y - 4) = 1/[(y)(y - 4)]

A*(y - 4) + B*(y) = 1

y = 0

-4A = 1

A = -1/4

y = 4

4B = 1

B = 1/4

-3/4 dx = 1/(4*(y - 4)) - 1/(4y) dy

multiply everything out by four

-3 dx = 1/(y - 4) - 1/y dy

3 dx = 1/y - 1/(y - 4) dy

3 dx = 1/y + 1/(4 - y) dy

3x + C = ln(y) - ln(4 - y)

3x + C = ln[y/(4 - y)]

e^(3x + C) = y/(4 - y)

(4 - y) * e^(3x + C) = y

4 * e^(3x + C) - y * e^(3x + C) = y

4 * e^(3x + C) = y + y * e^(3x + C)

4 * e^(3x + C) = y * (e^(3x + C) + 1)

y = 4 * e^(3x + C) / [1 + e^(3x + C)]

y(x = 0) = 4 * e^(3x + C) / [1 + e^(3x + C)] = 2

4 * e^(3*0 + C) / [1 + e^(3*0 + C)] = 2

4 * e^C / (1 + e^C) = 2

4 * e^C = 2 + 2 * e^C

2 * e^C = 2

e^C = 1

C = 0

Final answer is

4 * e^(3x) / [1 + e^(3x)]