What is the solution for (e^xy)y^'=e^{-y}+e^{-2x-y} ?

The solution for (e^xy)y^'=e^{-y}+e^{-2x-y} is y=\W((-3/(e^x)-1/(e^{3x)}+c_{1})/(3e))+1

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(e^xy)y^'=e^{-y}+e^{-2x-y}

(e^{x} y) y^{^{'}} = e^{- y} + e^{- 2 x - y}

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

Solution steps

(e^{x} y) y^{'} (x) = e^{- y} + e^{- 2 x - y}

Solve separable ODE:

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

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

(1 + cos (x)) y^{^{'}} = (1 + e^{- y}) sin (x), y (0) = 0

\frac{d y}{d x} = - \frac{0.6}{cos ( y )}

\frac{d y}{d x} = x y - x y^{2}

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

\frac{d x}{d y} = e^{x + y}

What is the solution for (e^xy)y^'=e^{-y}+e^{-2x-y} ?
The solution for (e^xy)y^'=e^{-y}+e^{-2x-y} is y=\W((-3/(e^x)-1/(e^{3x)}+c_{1})/(3e))+1