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

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

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2x(y+1)dx-ydy=0

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

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

Solution steps

2 x (y + 1) d x - y d 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

x \to 0 lim (arccos (x))

x \to \infty lim (6 x)

\int 5 sin^{3} (x) cos^{2} (x) d x

f (x) = arcsin (6 x + 1)

\int y^{2} z^{3} d x

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