What is the solution for y^'=(2x^4e^{y/x}+x^2y^2)/(x^3y) ?

The solution for y^'=(2x^4e^{y/x}+x^2y^2)/(x^3y) is y=-\W((2ln(x)+c_{1})/e)x-x

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y^'=(2x^4e^{y/x}+x^2y^2)/(x^3y)

y^{^{'}} = \frac{2 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2}}{x ^{3} y}

y = - W (\frac{2 ln ( x ) + c _{1}}{e}) x - x

Solution steps

y^{'} (x) = \frac{2 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2}}{x ^{3} y}

Solve using substitution:

y = - W (\frac{2 ln ( x ) + c _{1}}{e}) x - x

y = - W (\frac{2 ln ( x ) + c _{1}}{e}) x - x

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y^{^{''}} + y^{^{'}} + y = cos (t)

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What is the solution for y^'=(2x^4e^{y/x}+x^2y^2)/(x^3y) ?
The solution for y^'=(2x^4e^{y/x}+x^2y^2)/(x^3y) is y=-\W((2ln(x)+c_{1})/e)x-x