What is the solution for x(e^{y/x}-1)y^'=e^{y/x}(y-x) ?

The solution for x(e^{y/x}-1)y^'=e^{y/x}(y-x) is y=-\W(-e^{(e^{-c_{1}})/x})x+e^{-c_{1}}

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

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

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

Solution steps

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

Solve using substitution:

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

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

What is the solution for x(e^{y/x}-1)y^'=e^{y/x}(y-x) ?
The solution for x(e^{y/x}-1)y^'=e^{y/x}(y-x) is y=-\W(-e^{(e^{-c_{1}})/x})x+e^{-c_{1}}