What is the solution for y^'= 1/((t-1)(y+1)),y(0)=1 ?

The solution for y^'= 1/((t-1)(y+1)),y(0)=1 is y=(-2+sqrt(8ln(-t+1)+16))/2

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y^'= 1/((t-1)(y+1)),y(0)=1

y^{^{'}} = \frac{1}{( t - 1 ) ( y + 1 )}, y (0) = 1

y = \frac{- 2 + 8 ln ( - t + 1 ) + 16}{2}

Solution steps

y^{'} (t) = \frac{1}{( t - 1 ) ( y + 1 )}

Solve linear ODE:

y = \frac{- 2 + 8 ln ( - t + 1 ) + 16}{2}

y = \frac{- 2 + 8 ln ( - t + 1 ) + 16}{2}

\int \frac{1}{s ^{2} - 2 s + 5} d s

x \to - 4 lim (\frac{4 - ∣ x ∣}{4 + x})

\frac{d}{d x} (csec x)

\frac{\partial}{\partial x} (5 - x^{2} + 3 y^{2})

3 x^{2}, at x = 2

What is the solution for y^'= 1/((t-1)(y+1)),y(0)=1 ?
The solution for y^'= 1/((t-1)(y+1)),y(0)=1 is y=(-2+sqrt(8ln(-t+1)+16))/2