What is the answer to solvefor y,xy-e^x+e^y=0 ?

The answer to solvefor y,xy-e^x+e^y=0 is y=-(\W((e^{\frac{e^x)/x})/(x))x-e^x}{x}

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solvefor y,xy-e^x+e^y=0

solve for

y, x y - e^{x} + e^{y} = 0

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

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

- y + \frac{e ^{x}}{x} = u

and

y = - \frac{xu - e ^{x}}{x}

1 = (- (- \frac{xu - e ^{x}}{x}) x + e^{x}) e^{- (- \frac{xu - e ^{x}}{x})}

Simplify

1 = e^{\frac{xu - e ^{x}}{x}} xu

Rewrite

1 = e^{\frac{xu - e ^{x}}{x}} xu

in Lambert form:

e^{u} u = \frac{e ^{\frac{e ^{x}}{x}}}{x}

Solve

e^{u} u = \frac{e ^{\frac{e ^{x}}{x}}}{x} : u = W (\frac{e ^{\frac{e ^{x}}{x}}}{x})

u = W (\frac{e ^{\frac{e ^{x}}{x}}}{x})

Substitute back

u = - y + \frac{e ^{x}}{x},

solve for

y

Solve

- y + \frac{e ^{x}}{x} = W (\frac{e ^{\frac{e ^{x}}{x}}}{x}) : y = - \frac{W ( \frac{e ^{\frac{e ^{x}}{x}}}{x} ) x - e ^{x}}{x}

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

(\frac{1}{27})^{x} = 9

y, x = 4^{y}

t^{n} e^{a t}

9^{- x - 1} \times 9^{3 - 3 x} = 81

2 = e^{2 x}

What is the answer to solvefor y,xy-e^x+e^y=0 ?
The answer to solvefor y,xy-e^x+e^y=0 is y=-(\W((e^{\frac{e^x)/x})/(x))x-e^x}{x}