What is the solution for xy^'-y=(x^2)/y e^{y/x} ?

The solution for xy^'-y=(x^2)/y e^{y/x} is y=-\W((ln(x)+c_{1})/e)x-x

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xy^'-y=(x^2)/y e^{y/x}

x y^{^{'}} - y = \frac{x ^{2}}{y} e^{\frac{y}{x}}

y = - W (\frac{ln ( x ) + c _{1}}{e}) x - x

Solution steps

x y^{'} (x) - y = \frac{x ^{2}}{y} e^{\frac{y}{x}}

Solve using substitution:

y = - W (\frac{ln ( x ) + c _{1}}{e}) x - x

y = - W (\frac{ln ( x ) + c _{1}}{e}) x - x

y = x^{3} - 2 x + 2, (3, 23)

y^{^{''}} + 6 y^{^{'}} + 25 y = 0

\frac{d}{d x} (1 + 7 sin (x) - tan (x))

y (1 + x^{2}) y^{^{'}} - x (2 + y^{2}) = 0

f (x) = - 8 x^{2} + 10, (0.5, 8)

What is the solution for xy^'-y=(x^2)/y e^{y/x} ?
The solution for xy^'-y=(x^2)/y e^{y/x} is y=-\W((ln(x)+c_{1})/e)x-x