What is the solution for xy^'=y+e^{y/x} ?

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

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xy^'=y+e^{y/x}

x y^{^{'}} = y + e^{\frac{y}{x}}

y = - x ln (\frac{1}{x} - c_{1})

Solution steps

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

Solve using substitution:

y = - x ln (\frac{1}{x} - c_{1})

y = - x ln (\frac{1}{x} - c_{1})

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