[ y = \frac{1}{C - 2x^3} ]
where (C) is an arbitrary constant. The solution is valid for (x \neq \sqrt[3]{C/2}) (to avoid division by zero). Also note that (y=0) is a singular equilibrium solution — you can check it satisfies the original ODE: (y=0 \implies dy/dx = 0) and (6x^2y^2 = 6x^2 \cdot 0 = 0). ✅ 📌 Example with initial condition: If (y(0) = 1): (1 = \frac{1}{C - 0} \implies C = 1) So (y(x) = \frac{1}{1 - 2x^3}).
Since (C) is an arbitrary constant, we can rename it: solve the differential equation. dy dx = 6x2y2
[ \frac{1}{y} = -2x^3 - C ]
[ \boxed{y(x) = \frac{1}{C - 2x^3}} ]
Left side:
[ \frac{1}{y^2} , dy = 6x^2 , dx ]
[ \frac{dy}{dx} = 6x^2 y^2 ]