Explicit form?

Is it like this?

dy/dx = 1 -4y^2

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

dy/[1^2 -(2y)^2] = dx

[(1/2)(2dy)]/[1^2 -(2y)^2] = dx --------(i)

Using the Table, where u^2 is less than a^2,

INT.[du / (a^2 -u^2)] = [1/(2a)]*ln{(a+u)/(a-u)} -------(ii)

The left side of (i) falls into (ii) because

>>>maximum (2y)^2 is (2 * 1/2)^2 or 1^2. But (y = -1/2) and (y = 1/2) are not part of the interval of y, so maximum (2y)^2 is a little less than 1^2.

>>>a^2 is 1^2, so u^2 is less than a^2.

Integrate both sides of (i),

(1/2)[1/(2*1)]*ln{(1 +2y)/(1 -2y)} +C = x

(1/4)ln{(1 +2y)/(1 -2y)} +C = x

Clear the fraction, multiply both sides by 4,

ln{(1 +2y)/(1 -2y)} +4C = 4x -------(iii)

We can find C, or 4C, when y = 1/4 and x=0.

ln{(1 +2(1/4))/(1 -2(1/4))} +4C = 4(0)

ln{(1.5)/(0.5)} +4C = 0

ln(3) +4C = 0

4C = -ln(3) ----***

Substitute that into (iii)

ln{(1 +2y)/(1 -2y)} -ln(3) = 4x

Combine the two ln's,

ln{(1 +2y)/(1 -2y) / 3} = 4x

ln{(1/3)(1 +2y)/(1 -2y)} = 4x

Antilogs,

(1/3)(1 +2y)/(1 -2y) = e^(4x)

Multiply both sides by 3(1 -2y),

1 +2y = 3(1 -2y)e^(4x)

1 +2y = 3e^(4x) -2y(3e^(4x))

1 +2y +2y(3e^(4x)) = 3e^(4x)

1 +2y(1 +3e^(4x)) = 3e^(4x)

2y(1 +3e^(4x)) = 3e^(4x) -1

y = (1/2)[(3e^(4x) -1) / (3e^(4x) +1)] --------answer.

Check, when y=1/4 and x=0,

1/4 =? (1/2)[(3e^0 -1)/(3e^0 +1)]

1/4 =? (1/2)[(3-1)/(3+1)]

1/4 =? (1/2)[2/4]

1/4 =? 1/4

Yes, so, OK.