What is the solution for (xdy)/(dx)=y+xe^{-y/x} ?

The solution for (xdy)/(dx)=y+xe^{-y/x} is y=ln(ln(x)+c_{1})x

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(xdy)/(dx)=y+xe^{-y/x}

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

y = ln (ln (x) + c_{1}) x

Solution steps

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

Solve using substitution:

y = ln (ln (x) + c_{1}) x

y = ln (ln (x) + c_{1}) x

\frac{d}{d x} ((ln (x) + 3 x)^{2})

\frac{\partial}{\partial x} (ln (x + 9 y + 2 z))

\frac{d}{d x} (2 tan (x + \frac{π}{12}))

\frac{d}{d x} ((x + 1)^{\frac{2}{3}} (3 x^{2} - 2)^{3})

\int (sec (2 x) tan (2 x)) d x

What is the solution for (xdy)/(dx)=y+xe^{-y/x} ?
The solution for (xdy)/(dx)=y+xe^{-y/x} is y=ln(ln(x)+c_{1})x