What is the solution for 3x^2e^y+(x^3e^y-1)y^'=0 ?

The solution for 3x^2e^y+(x^3e^y-1)y^'=0 is y=-\W(-(x^3)/(e^{c_{1)}})-c_{1}

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3x^2e^y+(x^3e^y-1)y^'=0

3 x^{2} e^{y} + (x^{3} e^{y} - 1) y^{^{'}} = 0

y = - W (- \frac{x ^{3}}{e ^{c_{1}}}) - c_{1}

Solution steps

3 x^{2} e^{y} + (x^{3} e^{y} - 1) y^{'} (x) = 0

Solve exact ODE:

y = - W (- \frac{x ^{3}}{e ^{c_{1}}}) - c_{1}

y = - W (- \frac{x ^{3}}{e ^{c_{1}}}) - c_{1}

\int_{1}^{8} \frac{3}{x} d x

\int \frac{( x ^{2} )}{( 15 + 6 x - 9 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (e^{x^{\frac{1}{2}}})

f (x) = 2 x^{3} - 3 x^{2} + 3, at x = 1

\frac{1}{2 x ^{6}}

What is the solution for 3x^2e^y+(x^3e^y-1)y^'=0 ?
The solution for 3x^2e^y+(x^3e^y-1)y^'=0 is y=-\W(-(x^3)/(e^{c_{1)}})-c_{1}