Resolve the differential equation : xy' + 4 = y^2 with the condition : y(1)=1

I finished with the answer 2x -2/ 1-x = y I think it's wrong though so if anyone could show their steps that'd be great. Thank you.

Solve this differential equation by using an integrating factor:
xy' + 4 = y²
x(dy / dx) + 4 = y²
x(dy / dx) = y² - 4
dy / (y² - 4) = dx / x
∫ 1 / (y² - 4) dy = ∫ 1 / x dx

Decompose the integrand on the left into partial fractions:
1 / (y² - 4) = 1 / (y + 2)(y - 2)
1 / (y² - 4) = A / (y + 2) + B / (y - 2)
1 = A(y - 2) + B(y + 2)
Let y = -2,
-4A = 1
4A = -1
A = -¼
Let y = 2,
4B = 1
B = ¼
1 / (y² - 4) = -1 / 4(y + 2) + 1 / 4(y - 2)
1 / (y² - 4) = 1 / 4(y - 2) - 1 / 4(y + 2)

Use this result to continue solving and find the general solution:
∫ 1 / (y² - 4) dy = ∫ 1 / x dx
∫ [1 / 4(y - 2) - 1 / 4(y + 2)] dy = ∫ 1 / x dx
∫ 1 / (y - 2) dy / 4 - ∫ 1 / (y + 2) dy / 4 = ∫ 1 / x dx
ln|y - 2| / 4 - ln|y + 2| / 4 = ln|x| + C
ln|x| = C + ln|y - 2| / 4 - ln|y + 2| / 4
x = ℮^(C + ln|y - 2| / 4 - ln|y + 2| / 4)
x = ℮^[C + ln|₄√(y - 2)| - ln|₄√(y + 2)|]
x = ℮ᶜ₄√(y - 2) / ₄√(y + 2)
x = C₄√(y - 2) / ₄√(y + 2)
x⁴ = C(y - 2) / (y + 2)
x⁴(y + 2) = C(y - 2)
x⁴y + 2x⁴ = Cy - 2C
Cy - x⁴y = 2x⁴ + 2C
(C - x⁴)y = 2x⁴ + 2C
y = (2x⁴ + 2C) / (C - x⁴)

Use the initial condition to solve for the constant and find the particular solution:
When x = 1, y = 1
(2 + 2C) / (C - 1) = 1
2 + 2C = C - 1
C = -3
y = (2x⁴ - 6) / (-3 - x⁴)
y = (6 - 2x⁴) / (3 + x⁴)
y = (6 - 2x⁴) / (x⁴ + 3)

xy' + 4 = y^2
x [dy/dx] = y^2 - 4
x dy = [y^2 - 4] dx
1/[y^2 - 4] dy = [1/x] dx

∫ 1 / [y^2 - 4] dy = ∫ [1/x] dx
∫ 1 / [(y - 2)(y + 2)] dy = ln|x| + C

1 / [(y - 2)(y + 2)] = A/(y - 2) + B/(y + 2)
1 = A(y + 2) + B(y - 2)

y = -2 ===> B = -1/4
y = 2 ===> A = 1/4

(1/4) ∫ 1/(y - 2) dy - (1/4) ∫ 1 / (y + 2) dy = ln|x| + C
(1/4) ln|y - 2| - (1/4) ln|y + 2| = ln|x| + C