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

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

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

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

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

Solution steps

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

Solve using substitution:

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

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

\int_{0}^{2} k (2 - x)

y^{^{'}} + tan (x) y = cos (x)

\int x^{2} + 7 e^{x} + C d x

y^{^{''}} + 2 y^{^{'}} = 1

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{( - 3 e ^{5 x} )}{( x ^{2} + 1 )}

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