What is the solution for y^'-e^xe^y=-1 ?

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

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y^'-e^xe^y=-1

y^{^{'}} - e^{x} e^{y} = - 1

y = - ln (- x - c_{1}) - x

Solution steps

y^{'} (x) - e^{x} e^{y} = - 1

Solve using substitution:

y = - ln (- x - c_{1}) - x

y = - ln (- x - c_{1}) - x

\frac{\partial}{\partial x} (e^{1 + x^{2} - y^{2}})

\frac{d}{d x} (6 cos (2 x) - e^{- x})

f (x) = 2 x^{3} + 7 x^{2} - 8 x - 3, at x = - 1

x \to 1 - lim (1 - x - [x] - [1 - x])

x \to 4 - lim (\frac{1}{( x - 4 ) ^{2}})

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