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

The answer to solvefor y,xy+e^y=e^x-x is y=-(\W(1/(e^{\frac{x-e^x){x}}x})x-e^x+x)/x

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

solve for

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

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

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

e^{u} u = \frac{1}{e ^{\frac{x - e ^{x}}{x}} x} : u = W (\frac{1}{e ^{\frac{x - e ^{x}}{x}} x})

u = W (\frac{1}{e ^{\frac{x - e ^{x}}{x}} x})

Substitute back

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

solve for

y

Solve

- y + \frac{e ^{x} - x}{x} = W (\frac{1}{e ^{\frac{x - e ^{x}}{x}} x}) : y = - \frac{W ( \frac{1}{e ^{\frac{x - e ^{x}}{x}} x} ) x - e ^{x} + x}{x}

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

- 6^{x - 6} - 9 = - 14

3^{16} = 9^{2 x}

1. 1^{x} = 2

\frac{e ^{x + 5}}{e ^{5}} = 3

9^{x} - 3^{x + 1} + 2 = 0

What is the answer to solvefor y,xy+e^y=e^x-x ?
The answer to solvefor y,xy+e^y=e^x-x is y=-(\W(1/(e^{\frac{x-e^x){x}}x})x-e^x+x)/x