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

The solution for y^'=(x^2)/y (1+x^3) is y=sqrt((2x^3+x^6+c_{1))/3},y=-sqrt((2x^3+x^6+c_{1))/3}

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y^'=(x^2)/y (1+x^3)

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

y = \frac{2 x ^{3} + x ^{6} + c _{1}}{3}, y = - \frac{2 x ^{3} + x ^{6} + c _{1}}{3}

Solution steps

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

Solve separable ODE:

y = \frac{2 x ^{3} + x ^{6} + c _{1}}{3}, y = - \frac{2 x ^{3} + x ^{6} + c _{1}}{3}

y = \frac{2 x ^{3} + x ^{6} + c _{1}}{3}, y = - \frac{2 x ^{3} + x ^{6} + c _{1}}{3}

\frac{d x}{d t} = x^{2} + \frac{1}{81}

\frac{d x}{d t} = 0.36 - \frac{9 x}{100}

u^{^{'}} = x^{2} H (x)

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

\frac{( d y )}{( d x )} = (2 x + 10) \cdot e^{- y}, y (0) = 0

What is the solution for y^'=(x^2)/y (1+x^3) ?
The solution for y^'=(x^2)/y (1+x^3) is y=sqrt((2x^3+x^6+c_{1))/3},y=-sqrt((2x^3+x^6+c_{1))/3}