What is the solution for y^'=(8x^3+1)/(3y^2) ?

The solution for y^'=(8x^3+1)/(3y^2) is y=\sqrt[3]{2x^4+x+c_{1}}

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y^'=(8x^3+1)/(3y^2)

y^{^{'}} = \frac{8 x ^{3} + 1}{3 y ^{2}}

Solution steps

y^{'} (x) = \frac{8 x ^{3} + 1}{3 y ^{2}}

Solve separable ODE:

\frac{d y}{d x} = - \frac{0.9}{cos ( y )}, y (0) = \frac{π}{3}

y^{^{'}} = t (1 + \frac{y}{4}), y (1) = 2.4

\frac{d y}{d x} = \frac{5 x}{e ^{y}}

\frac{d v}{d t} = - 2 v

\frac{x ^{2} - 9}{x} \frac{d y}{d x} + y = 0

What is the solution for y^'=(8x^3+1)/(3y^2) ?
The solution for y^'=(8x^3+1)/(3y^2) is y=\sqrt[3]{2x^4+x+c_{1}}