What is the solution for x(e^{y/x}-1)y^'=e^{y/x}(y-x),y(1)=0 ?

The solution for x(e^{y/x}-1)y^'=e^{y/x}(y-x),y(1)=0 is -y+xe^{y/x}=1

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x(e^{y/x}-1)y^'=e^{y/x}(y-x),y(1)=0

x (e^{\frac{y}{x}} - 1) y^{^{'}} = e^{\frac{y}{x}} (y - x), y (1) = 0

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

Solution steps

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

Solve exact ODE:

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

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

\frac{d}{d x} ((x^{2} + y^{2}) ln (x + y))

\frac{d}{d x} (\frac{x + 3}{x - 5})

\frac{\partial}{\partial y} (\frac{1}{2} ln (x^{2} + y^{2}))

\frac{\partial}{\partial x} (e^{2 x})

\int \frac{8 x ^{3} - 16 x ^{2} + 7}{x ^{2} - 2 x} d x

What is the solution for x(e^{y/x}-1)y^'=e^{y/x}(y-x),y(1)=0 ?
The solution for x(e^{y/x}-1)y^'=e^{y/x}(y-x),y(1)=0 is -y+xe^{y/x}=1