What is the solution for (y+1)y^'=y(x-1)^2 ?

The solution for (y+1)y^'=y(x-1)^2 is y=\W(1/(e^{(-x^3+3x^2-3x-c_{1))/3}})

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(y+1)y^'=y(x-1)^2

(y + 1) y^{^{'}} = y (x - 1)^{2}

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x ^{2} - 3 x - c _{1}}{3}}})

Solution steps

(y + 1) y^{'} (x) = y (x - 1)^{2}

Solve separable ODE:

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x ^{2} - 3 x - c _{1}}{3}}})

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x ^{2} - 3 x - c _{1}}{3}}})

y^{^{'}} = 4 x + x y

y^{^{'}} (x) = \frac{15 + 3 x}{3 y ^{2} - 12 y}, y (0) = 2

\frac{d x}{2 x} = d t

\frac{2 d y}{y} = \frac{2 cos ( 2 x )}{sin ( 2 x )} d x

\frac{d v}{d t} = 9.81 - 10 v, v (0) = 0

What is the solution for (y+1)y^'=y(x-1)^2 ?
The solution for (y+1)y^'=y(x-1)^2 is y=\W(1/(e^{(-x^3+3x^2-3x-c_{1))/3}})