What is the solution for (y^2dy)/(y^3-4)=(xdx)/(x^2+2) ?

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

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(y^2dy)/(y^3-4)=(xdx)/(x^2+2)

\frac{y ^{2} d y}{y ^{3} - 4} = \frac{x d x}{x ^{2} + 2}

Solution steps

\frac{y ^{2} d y}{y ^{3} - 4} = \frac{x d x}{x ^{2} + 2}

Solve separable ODE:

\frac{d r}{d t} = 0.1872 \cdot r, r (0) = 24

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

y^{^{'}} = - \frac{y}{x}, y (1) = 3

\frac{d y}{d x} = \frac{x e ^{2 x - y}}{e ^{x - 2 y}}

\frac{d y}{d t} = (t y^{2})^{2}

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