What is the solution for (x+y)dx+(3x+3y-4)dy=0 ?

The solution for (x+y)dx+(3x+3y-4)dy=0 is y=-x+(2\W(1/(4e^{-x-c_{1)}}))/3+2

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(x+y)dx+(3x+3y-4)dy=0

(x + y) d x + (3 x + 3 y - 4) d y = 0

y = - x + \frac{2 W ( \frac{1}{4 e ^{- x - c_{1}}} )}{3} + 2

Solution steps

(x + y) d x + (3 x + 3 y - 4) d y = 0

Solve using substitution:

y = - x + \frac{2 W ( \frac{1}{4 e ^{- x - c_{1}}} )}{3} + 2

y = - x + \frac{2 W ( \frac{1}{4 e ^{- x - c_{1}}} )}{3} + 2

\frac{d}{d x} (e^{- 4 x^{3} - 2 x + 4})

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

e^{x} - x^{7}

x \to 0 - lim (9 + \frac{2}{x})

2 sin (10 x), tan (10 x), [- 0.1, 0.1]

What is the solution for (x+y)dx+(3x+3y-4)dy=0 ?
The solution for (x+y)dx+(3x+3y-4)dy=0 is y=-x+(2\W(1/(4e^{-x-c_{1)}}))/3+2