The special solution of differential equation 2Y (1 + x) y '= 4 + y ^ 2, when x = 3, y = 2

The special solution of differential equation 2Y (1 + x) y '= 4 + y ^ 2, when x = 3, y = 2

2Y(1+x)y'=4+Y^2
[2y/(4+y^2)]dy=[1/(1+x)]dx
The integral of both sides is obtained
Ln (y ^ 2 + 4) = ln (x + 1) + C, C is constant
Substituting x = 3 and y = 2, we get the following results
C=ln2
So: ln (y ^ 2 + 4) = ln (x + 1) + LN2
y^2+4=2(x+1)
y^2=2x-2
Y = + - radical (2x-2)