What is the solution for (y+4)y^'=y(2x+4) ?

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

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(y+4)y^'=y(2x+4)

(y + 4) y^{^{'}} = y (2 x + 4)

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

Solution steps

(y + 4) y^{'} (x) = y (2 x + 4)

Solve separable ODE:

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

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

n = 0 \sum \infty \frac{1}{2 !}

(6, - 3), (8, 4)

\frac{3 x - 1}{2 x + 1}

x (1.1)

\frac{\partial}{\partial x} (2 x - 3 y)

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