How to solve the following diferential equation with initial value: y' = 2x / (1 + 2y) where y(2) = 0

I know that I need to use separations of variables. After solve, how to apply the initial value?

Solve to find the general solution, then find the value of the constant:
y' = 2x / (1 + 2y)
dy / dx = 2x / (1 + 2y)
(1 + 2y) dy = 2x dx
∫ (1 + 2y) dy = ∫ 2x dx
y + y² = x² + C
When x = 2, y = 0
0 + 0² = 2² + C
0 = 4 + C
C = -4
y + y² = x² - 4
y² + y = x² - 4
4y² + 4y = 4x² - 16
(2y + 1)² - 1 = 4x² - 16
(2y + 1)² = 4x² - 15
2y + 1 = ±√(4x² - 15)
2y = -1 ± √(4x² - 15)
y = [-1 ± √(4x² - 15)] / 2

dy/dx = (2x)/(1 + 2y), y(2) = 0

dy/(1 + 2y) = 2x dx

Integrating both sides:

1/2*ln|1 + 2y| = x² + C

1 + 2y = C*e^(2x²)

y = C*e^(2x²) - 1/2

0 = C*e^(8) - 1/2

C = 1/(2e^8)

y = 1/2*e^(2x² - 8) - 1/2