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

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

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x^2y^'=xy+2x^2e^{y/x}

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

y = - x ln (- 2 ln (x) - c_{1})

Solution steps

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

Solve using substitution:

y = - x ln (- 2 ln (x) - c_{1})

y = - x ln (- 2 ln (x) - c_{1})

f (x) = - \frac{1}{1 - x}

\frac{\partial}{\partial x} (- sin^{2} (x))

\frac{d}{d x} (arcsin (f (x)))

\int tan (4 x) (sec (4 x))^{2} d x

\frac{( s ^{2} - 1 )}{( s ^{2} + 1 ) ^{2}}

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