What is the answer to solvefor x,xe^y+ye^x=2 ?

The answer to solvefor x,xe^y+ye^x=2 is x=-(\W(y/(e^{-2e^{-y)+y}})e^y-2)/(e^y)

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solvefor x,xe^y+ye^x=2

solve for

x, x e^{y} + y e^{x} = 2

x = - \frac{W ( \frac{y}{e ^{- 2 e^{- y} + y}} ) e ^{y} - 2}{e ^{y}}

Solution steps

x e^{y} + y e^{x} = 2

Prepare

x e^{y} + y e^{x} = 2

for Lambert form:

y = (2 - x e^{y}) e^{- x}

Rewrite the equation with

- x + \frac{2}{e ^{y}} = u

and

x = - \frac{e ^{y} u - 2}{e ^{y}}

y = (2 - (- \frac{e ^{y} u - 2}{e ^{y}}) e^{y}) e^{- (- \frac{e ^{y} u - 2}{e ^{y}})}

Simplify

y = e^{\frac{e ^{y} u - 2}{e ^{y}}} e^{y} u

Rewrite

y = e^{\frac{e ^{y} u - 2}{e ^{y}}} e^{y} u

in Lambert form:

e^{u} u = \frac{y}{e ^{- 2 e^{- y} + y}}

Solve

e^{u} u = \frac{y}{e ^{- 2 e^{- y} + y}} : u = W (\frac{y}{e ^{- 2 e^{- y} + y}})

u = W (\frac{y}{e ^{- 2 e^{- y} + y}})

Substitute back

u = - x + \frac{2}{e ^{y}},

solve for

x

Solve

- x + \frac{2}{e ^{y}} = W (\frac{y}{e ^{- 2 e^{- y} + y}}) : x = - \frac{W ( \frac{y}{e ^{- 2 e^{- y} + y}} ) e ^{y} - 2}{e ^{y}}

x = - \frac{W ( \frac{y}{e ^{- 2 e^{- y} + y}} ) e ^{y} - 2}{e ^{y}}

(\frac{1}{2})^{x} = 0.0625

5^{5 x + 2} = 45

5^{x + 1} = 1 0^{x}

2^{x + 2} = 8^{- x + 8}

x, 5 (1.08)^{x} = 20

What is the answer to solvefor x,xe^y+ye^x=2 ?
The answer to solvefor x,xe^y+ye^x=2 is x=-(\W(y/(e^{-2e^{-y)+y}})e^y-2)/(e^y)