What is the solution for y^'=(6x^2-1)/y ,y(0)=-4 ?

The solution for y^'=(6x^2-1)/y ,y(0)=-4 is y=-sqrt(4x^3-2x+16)

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y^'=(6x^2-1)/y ,y(0)=-4

y^{^{'}} = \frac{6 x ^{2} - 1}{y}, y (0) = - 4

y = - 4 x^{3} - 2 x + 16

Solution steps

y^{'} (x) = \frac{6 x ^{2} - 1}{y}

Solve separable ODE:

y = - 4 x^{3} - 2 x + 16

y = - 4 x^{3} - 2 x + 16

\int_{- 2}^{2} (3 x^{3} - 10) d x

y^{^{'}} = (x + y - 4)

\int x cos (n \cdot \frac{π}{3} \cdot x) d x

x \to - \infty lim (x \cdot (x))

y = x^{2}, x = 3, x = - 3

What is the solution for y^'=(6x^2-1)/y ,y(0)=-4 ?
The solution for y^'=(6x^2-1)/y ,y(0)=-4 is y=-sqrt(4x^3-2x+16)