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

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

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

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

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

Solution steps

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

Solve using substitution:

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

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

y = x^{3} - 3 x + 2, (4, 54)

\frac{s ^{2} + 12}{s ^{4} - 16}

\frac{d}{d x} (\frac{x ^{2}}{2 - x})

x \to 0 lim (sin (x) ln (x))

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