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

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

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

Solve Bernoulli ODE:

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

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

\int_{- 2}^{2} \frac{1}{x ^{2} + 4 x + 13} d x

\int x e - x^{2} d x

f (x) = (8 - x)^{3}

\frac{d}{d x} (8 sin^{2} (x))

\frac{d x}{d t} = 0.6 - \frac{3}{500} x

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