What is the solution for y^'=((3x^6e^{y/x}+x^4y^2))/(x^5y) ?

The solution for y^'=((3x^6e^{y/x}+x^4y^2))/(x^5y) is y=-\W((3ln(x)+c_{1})/e)x-x

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y^'=((3x^6e^{y/x}+x^4y^2))/(x^5y)

y^{^{'}} = \frac{( 3 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2} )}{x ^{5} y}

y = - W (\frac{3 ln ( x ) + c _{1}}{e}) x - x

Solution steps

y^{'} (x) = \frac{( 3 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2} )}{x ^{5} y}

Solve using substitution:

y = - W (\frac{3 ln ( x ) + c _{1}}{e}) x - x

y = - W (\frac{3 ln ( x ) + c _{1}}{e}) x - x

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

\int_{0}^{π} (sin^{2} (x)) d x

x \to 0 lim (\frac{cos ( 10 x ) - 1}{x ^{2}})

\int (4 x + 1)^{2} e^{2 x} d x

What is the solution for y^'=((3x^6e^{y/x}+x^4y^2))/(x^5y) ?
The solution for y^'=((3x^6e^{y/x}+x^4y^2))/(x^5y) is y=-\W((3ln(x)+c_{1})/e)x-x