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

The solution for y(1+xy)dx+(2y-x)dy=0 is y=-x/(2\W(-(x){2e^{\frac{-x^2+c_{1})/4}})}

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

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

y = - \frac{x}{2 W ( - \frac{x}{2 e ^{\frac{- x ^{2} + c _{1}}{4}}} )}

Solution steps

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

Solve using substitution:

y = - \frac{x}{2 W ( - \frac{x}{2 e ^{\frac{- x ^{2} + c _{1}}{4}}} )}

y = - \frac{x}{2 W ( - \frac{x}{2 e ^{\frac{- x ^{2} + c _{1}}{4}}} )}

\int (\frac{2}{x}) d x

f (x) = 6 x^{3} - 6 x^{2} + 4

x \to 0 + lim (2 x^{s i n (3 x)})

\int x e^{5 x + π} d x

x \to - 8 lim (- \frac{3 x}{x + 8})

What is the solution for y(1+xy)dx+(2y-x)dy=0 ?
The solution for y(1+xy)dx+(2y-x)dy=0 is y=-x/(2\W(-(x){2e^{\frac{-x^2+c_{1})/4}})}