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) 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) 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}}})

\frac{1}{X - 1}, 12

\frac{d}{d x} (2 x^{3} - 2 x)

\int x sin (s) d x

y = (7) (6^{x - 8})

x^{2} \frac{d y}{d x} + 4 x y - e^{x} = 0

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}})