What is the solution for y^'=(x^2-2)/y ?

The solution for y^'=(x^2-2)/y is y=sqrt((2x^3)/3-4x+c_{1)},y=-sqrt((2x^3)/3-4x+c_{1)}

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y^'=(x^2-2)/y

y^{^{'}} = \frac{x ^{2} - 2}{y}

y = \frac{2 x ^{3}}{3} - 4 x + c_{1}, y = - \frac{2 x ^{3}}{3} - 4 x + c_{1}

Solution steps

y^{'} (x) = \frac{x ^{2} - 2}{y}

Solve separable ODE:

y = \frac{2 x ^{3}}{3} - 4 x + c_{1}, y = - \frac{2 x ^{3}}{3} - 4 x + c_{1}

y = \frac{2 x ^{3}}{3} - 4 x + c_{1}, y = - \frac{2 x ^{3}}{3} - 4 x + c_{1}

y^{^{'}} + \frac{y ^{^{'}}}{x} + (1 - \frac{1}{4 x ^{2}}) y = 0

y y^{^{'}} - sin (x) = 0, y (π) = 1

csc (y) d y + sec^{2} (x) d x = 0

y^{^{'}} = \frac{3 x e ^{x}}{y 5 + y ^{2}}

2 y (y^{^{'}}) = 6 t^{2}

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