What is the solution for y^'= 1/(e^{-y)-x} ?

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

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

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

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

Solution steps

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

Solve exact ODE:

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

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

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\frac{7}{1 - t}

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\frac{d}{d x} (\frac{ln ( x ^{3} - y + z )}{x y + z})

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