What is the answer to solvefor x,y^3-2xy=4e^{(2x-y)} ?

The answer to solvefor x,y^3-2xy=4e^{(2x-y)} is x=-(\W(4/(e^{-y^2+y)y})-y^2)/2

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solvefor x,y^3-2xy=4e^{(2x-y)}

solve for

x, y^{3} - 2 x y = 4 e^{(2 x - y)}

x = - \frac{W ( \frac{4}{e ^{- y^{2} + y} y} ) - y ^{2}}{2}

Solution steps

y^{3} - 2 x y = 4 e^{(2 x - y)}

Prepare

y^{3} - 2 x y = 4 e^{(2 x - y)}

for Lambert form:

(y^{3} - 2 x y) e^{- 2 x + y} = 4

Rewrite the equation with

- 2 x + y^{2} = u

and

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

(y^{3} - 2 (- \frac{u - y ^{2}}{2}) y) e^{- 2 (- \frac{u - y ^{2}}{2}) + y} = 4

Simplify

e^{u - y^{2} + y} y u = 4

Rewrite

e^{u - y^{2} + y} y u = 4

in Lambert form:

e^{u} u = \frac{4}{e ^{- y^{2} + y} y}

Solve

e^{u} u = \frac{4}{e ^{- y^{2} + y} y} : u = W (\frac{4}{e ^{- y^{2} + y} y})

u = W (\frac{4}{e ^{- y^{2} + y} y})

Substitute back

u = - 2 x + y^{2},

solve for

x

Solve

- 2 x + y^{2} = W (\frac{4}{e ^{- y^{2} + y} y}) : x = - \frac{W ( \frac{4}{e ^{- y^{2} + y} y} ) - y ^{2}}{2}

x = - \frac{W ( \frac{4}{e ^{- y^{2} + y} y} ) - y ^{2}}{2}

4^{x + 3} = 8^{x}

4^{x - 3} = 3^{2 x}

(7^{x + 3} 7^{- 7})^{2} = 7

x^{9} = e^{x}

e^{- 2 x} = 0.025

What is the answer to solvefor x,y^3-2xy=4e^{(2x-y)} ?
The answer to solvefor x,y^3-2xy=4e^{(2x-y)} is x=-(\W(4/(e^{-y^2+y)y})-y^2)/2