What is the solution for (-3(2x-u)+6x+2)dx-(u-1)(2dx-du)=0 ?

The solution for (-3(2x-u)+6x+2)dx-(u-1)(2dx-du)=0 is u=-5\W(1/(5e^{(-x+c_{1))/5}})-4

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(-3(2x-u)+6x+2)dx-(u-1)(2dx-du)=0

(- 3 (2 x - u) + 6 x + 2) d x - (u - 1) (2 d x - d u) = 0

u = - 5 W (\frac{1}{5 e ^{\frac{- x + c _{1}}{5}}}) - 4

Solution steps

(- 3 (2 x - u) + 6 x + 2) d x - (u - 1) (2 d x - d u) = 0

Solve separable ODE:

u = - 5 W (\frac{1}{5 e ^{\frac{- x + c _{1}}{5}}}) - 4

u = - 5 W (\frac{1}{5 e ^{\frac{- x + c _{1}}{5}}}) - 4

(e^{x} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{3} e^{- x} d y = 0

\frac{d y}{d t} + 2 y = 4

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

y^{^{'}} = y t^{2} - 1.1 y, y (0) = 1

y^{^{'}} = \frac{3 t ^{2}}{4 + 2 y}

What is the solution for (-3(2x-u)+6x+2)dx-(u-1)(2dx-du)=0 ?
The solution for (-3(2x-u)+6x+2)dx-(u-1)(2dx-du)=0 is u=-5\W(1/(5e^{(-x+c_{1))/5}})-4