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

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

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

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

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

Solution steps

2 x (1 + y) 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^{8} ln (x)

f (x) = e^{3 x}

f (x) = 3^{x l n (x)}

e^{4 t} sin (2 t)

\frac{d}{d y} (\frac{y}{t ^{2} + y ^{2}})

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