What is the solution for y^'*e^{x+y}-x=0 ?

The solution for y^'*e^{x+y}-x=0 is y=ln(-xe^{-x}-e^{-x}+c_{1})

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y^'*e^{x+y}-x=0

y^{^{'}} \cdot e^{x + y} - x = 0

y = ln (- x e^{- x} - e^{- x} + c_{1})

Solution steps

y^{'} (x) \cdot e^{x + y} - x = 0

Solve separable ODE:

y = ln (- x e^{- x} - e^{- x} + c_{1})

y = ln (- x e^{- x} - e^{- x} + c_{1})

y^{^{'}} = y + \frac{y}{x ln ( x )}, y (e) = 1

y^{^{'}} = e^{2 x} + y \cdot e^{2 x}

(x - 1)^{2} y^{^{'}} + 8 (x - 1) y^{^{'}} + 12 y = 0

\frac{d y}{d x} = \frac{4 x}{y ^{2}}

y^{^{'}} = \frac{2 y ^{4} + 2}{4 x y ^{3}}

What is the solution for y^'*e^{x+y}-x=0 ?
The solution for y^'*e^{x+y}-x=0 is y=ln(-xe^{-x}-e^{-x}+c_{1})