What is the answer to solvefor y,e^{2xy}+2y=2x+2 ?

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

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

solve for

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

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

Solution steps

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

Prepare

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

for Lambert form:

1 = (2 x + 2 - 2 y) e^{- 2 x y}

Rewrite the equation with

- 2 x y + x (2 x + 2) = u

and

y = - \frac{u - x ( 2 x + 2 )}{2 x}

1 = (2 x + 2 - 2 (- \frac{u - x ( 2 x + 2 )}{2 x})) e^{- 2 x (- \frac{u - x ( 2 x + 2 )}{2 x})}

Simplify

1 = e^{u - x (2 x + 2)} (2 x + 2 + \frac{u - x ( 2 x + 2 )}{x})

Rewrite

1 = e^{u - x (2 x + 2)} (2 x + 2 + \frac{u - x ( 2 x + 2 )}{x})

in Lambert form:

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

Solve

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

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

Substitute back

u = - 2 x y + x (2 x + 2),

solve for

y

Solve

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

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

2^{x - 1} = 32

5^{x + 2} = 100

(\frac{1}{2})^{n} = 4^{(n + 1)}

2^{- x} = 4

2^{- x} = 8

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