What is the solution for (x+y+4)dx+(2x+2y+1)dy=0 ?

The solution for (x+y+4)dx+(2x+2y+1)dy=0 is y=-x+(7\W(2/(7e^{(-x-c_{1))/7}}))/(2)+3

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(x+y+4)dx+(2x+2y+1)dy=0

(x + y + 4) d x + (2 x + 2 y + 1) d y = 0

y = - x + \frac{7 W ( \frac{2}{7 e ^{\frac{- x - c _{1}}{7}}} )}{2} + 3

Solution steps

(x + y + 4) d x + (2 x + 2 y + 1) d y = 0

Solve using substitution:

y = - x + \frac{7 W ( \frac{2}{7 e ^{\frac{- x - c _{1}}{7}}} )}{2} + 3

y = - x + \frac{7 W ( \frac{2}{7 e ^{\frac{- x - c _{1}}{7}}} )}{2} + 3

\frac{d y}{d x} (25 - x^{2}) + 5 x y = 0

\frac{d}{d x} (\frac{( 1 - x ) ^{2}}{x ^{2}})

y^{^{'}} = y x e^{x^{2}} - y

x \to 0 - lim (x^{2} ln (x))

y^{^{'}} - x^{2} = 2 x^{2} y

What is the solution for (x+y+4)dx+(2x+2y+1)dy=0 ?
The solution for (x+y+4)dx+(2x+2y+1)dy=0 is y=-x+(7\W(2/(7e^{(-x-c_{1))/7}}))/(2)+3