What is the solution for (dy}{dx}=\frac{(x^2+2))/y ?

The solution for (dy}{dx}=\frac{(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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(dy}{dx}=\frac{(x^2+2))/y

\frac{d y}{d x} = \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

\frac{d y}{d 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^{^{'}} = e^{- y} sin (x), y (\frac{π}{2}) = \frac{1}{2}

y^{^{'}} + 2 y = 0, y (0) = 1.5

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

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

\frac{d y}{d x} x y + x = 2

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