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

The solution for y^'= 1/((t-1)(y+1)) is y=-1+sqrt(2(c_{1)+ln(t-1))+1},y=-sqrt(2(c_{1)+ln(t-1))+1}-1

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

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

y = - 1 + 2 (c_{1} + ln (t - 1)) + 1, y = - 2 (c_{1} + ln (t - 1)) + 1 - 1

Solution steps

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

Solve separable ODE:

y = - 1 + 2 (c_{1} + ln (t - 1)) + 1, y = - 2 (c_{1} + ln (t - 1)) + 1 - 1

y = - 1 + 2 (c_{1} + ln (t - 1)) + 1, y = - 2 (c_{1} + ln (t - 1)) + 1 - 1

f (x) = x^{3} (x - 6)^{2}

f (x) = x - 1 + x + 1

y = (\frac{( x + 2 )}{( x - 2 )})^{2}, at x = 3

\int_{0}^{π} 14 sin^{4} (x) d x

\int \frac{x - 1}{( x + 1 ) ^{3}} d x

What is the solution for y^'= 1/((t-1)(y+1)) ?
The solution for y^'= 1/((t-1)(y+1)) is y=-1+sqrt(2(c_{1)+ln(t-1))+1},y=-sqrt(2(c_{1)+ln(t-1))+1}-1