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

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

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

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

y = 3 \frac{x ^{3}}{3} + x + c_{1}

Solution steps

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

Solve separable ODE:

y = 3 \frac{x ^{3}}{3} + x + c_{1}

y = 3 \frac{x ^{3}}{3} + x + c_{1}

\int \frac{1}{x ^{2} - 7} d x

- 4 y^{^{''}} + 16 y = 0, y (1) = 2, y^{^{'}} (2) = 1

y = x^{3} (x - 1)^{2}

f (x) = 6 sin^{3} (x)

\int 8 x^{2} + 6 + \frac{9}{x ^{2} + 1} d x

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