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

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

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(2xy)dx-(1+y)dy=0

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{''}} - 8 y^{^{'}} - 9 y = 21 e^{2 t}

x \to - 1 lim (4 - x)

x \to 5 - lim (\frac{7}{x - 5})

\frac{d}{d x} ((x - 3) (x + 2))

\int \frac{1}{x ^{2} 196 - x ^{2}} d x

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