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

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

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(y+4)*y^'=2xy+4y

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

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

Solution steps

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

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{x}{2})^{n}

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

\frac{d}{d x} (ln (\frac{cos ( x )}{x}))

\frac{d}{d x} ((4 x + 5 x^{2})^{4})

\int \frac{16 - x ^{2}}{x ^{2}} d x

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