What is the solution for y^'=-(-(x^2)/(3y)) ?

The solution for y^'=-(-(x^2)/(3y)) is y=(sqrt(c_{1)+2x^3})/3 ,y=-(sqrt(c_{1)+2x^3})/3

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y^'=-(-(x^2)/(3y))

y^{^{'}} = - (- \frac{x ^{2}}{3 y})

y = \frac{c _{1} + 2 x ^{3}}{3}, y = - \frac{c _{1} + 2 x ^{3}}{3}

Solution steps

y^{'} (x) = - (- \frac{x ^{2}}{3 y})

Solve separable ODE:

y = \frac{c _{1} + 2 x ^{3}}{3}, y = - \frac{c _{1} + 2 x ^{3}}{3}

y = \frac{c _{1} + 2 x ^{3}}{3}, y = - \frac{c _{1} + 2 x ^{3}}{3}

\frac{d y}{d x} = 2 x^{3} y

(x^{2} + 1) \frac{d y}{d x} + 3 x y = 3 x

\frac{d y}{d x} y = 2 x^{2} - 13 x + 6

\frac{d y}{d x} = \frac{( x + x y ^{2} )}{4 y}

\frac{d y}{d x} = \frac{6 x ^{2} + 4}{8 y ^{3} + 1}, y (0) = 1

What is the solution for y^'=-(-(x^2)/(3y)) ?
The solution for y^'=-(-(x^2)/(3y)) is y=(sqrt(c_{1)+2x^3})/3 ,y=-(sqrt(c_{1)+2x^3})/3