Interate 1/[y(1-y)] dy= k dt. Please!?
How can the answer be y = [3^(t-1)]/[1-3^(t-1)]?
If when y = 1/4 when t = 0
When y = 1/2, t = 1?
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How can the answer be y = [3^(t-1)]/[1-3^(t-1)]?
If when y = 1/4 when t = 0
When y = 1/2, t = 1?
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1/[y(1-y)] dy= k dt
∫(1/y + 1/(1 - y)) dy = ∫kdt
ln (y/(1 - y)) = kt + C
If when y = 1/4 when t = 0
ln (1/3) = k(0) + C
C = ln(1/3)
When y = 1/2, t = 1
ln (1) = k + ln(1/3)
k = -ln(1/3) = ln3
ln (y/(1 - y)) = ln(3)t + ln(1/3)
ln (y/(1 - y)) = ln(3)t - ln(3)
ln (y/(1 - y)) = ln(3)(t - 1)
y/(1 - y) = e^(ln(3)(t - 1))
y/(1 - y) = 3^(t - 1)
y = 3^(t - 1) (1 - y)
y = 3^(t - 1) - y3^(t - 1)
y + y3^(t - 1) = 3^(t - 1)
y(1 + 3^(t - 1)) = 3^(t - 1)
y = 3^(t - 1) / (1 + 3^(t - 1))