[ \boxed{y(x) = \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}). solve the differential equation. dy dx = 6x2y2
[ \frac{1}{y} = -2x^3 - C ]
Since (C) is an arbitrary constant, we can rename it: [ \boxed{y(x) = \frac{1}{C - 2x^3}} ] where
Set (C = C_2 - C_1):
[ \int 6x^2 , dx = 6 \cdot \frac{x^3}{3} + C_2 = 2x^3 + C_2 ] ✅ 📌 Example with initial condition: If (y(0)
Left side: