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} \cdot (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} \cdot (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^{3} + 4 t^{2} = 65

y^{^{'}} + 4 y^{^{'}} + 3 y = 0

\frac{d y}{d x} sin (y) = 4 x^{4} - 4

y^{^{'}} = 10 x y - 2 x, y (0) = 1

y^{^{'}} = 10 x y - 2 x, y (0) = 3

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}