What is the solution for 2x(y+1)-y^'y=0 ?

The solution for 2x(y+1)-y^'y=0 is y=-\W(-1/(e^{-c_{1)+x^2}})-1

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2x(y+1)-y^'y=0

2 x (y + 1) - y^{^{'}} y = 0

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

Solution steps

2 x (y + 1) - y^{'} (x) y = 0

Solve Bernoulli ODE:

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

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

\frac{d y}{d x} + y^{2} cos (x) = 0

y^{^{'}} = \frac{x ^{2} - 1}{2 x + x y}

y d x + (2 + x) d y = 0, y (- 1) = 4

y^{^{'}} = x y + x - y - 1

y^{^{'}} + \frac{2 x}{1 + x ^{2}} y = \frac{- x}{1 + x ^{2}}

What is the solution for 2x(y+1)-y^'y=0 ?
The solution for 2x(y+1)-y^'y=0 is y=-\W(-1/(e^{-c_{1)+x^2}})-1