Starting with: $$ -\frac{1}{y} = 2x^3 + C $$
The given differential equation is $\frac{dy}{dx} = 6x^2y^2$. This is a first-order nonlinear differential equation.
Left side:
For a particular solution, we would need an initial condition, such as $y(x_0) = y_0$. Without a specific initial condition, we cannot determine a unique particular solution. solve the differential equation. dy dx = 6x2y2
$$ -\frac{1}{y} = 2x^3 + C $$
[ y = \frac{1}{C - 2x^3} ]
$$ \int \frac{1}{y^2} , dy = \int 6x^2 , dx $$ Starting with: $$ -\frac{1}{y} = 2x^3 + C
The general solution to the differential equation y=−12x3+Cy equals negative the fraction with numerator 1 and denominator 2 x cubed plus cap C end-fraction
∫y-2dy=∫6x2dxintegral of y to the negative 2 power d y equals integral of 6 x squared d x
Therefore, the complete solution set is: $$ y(x) = -\frac{1}{2x^3 + C} \quad \text{and} \quad y=0 $$ Without a specific initial condition, we cannot determine
Right side:
$$ \frac{dy}{dx} = 6x^2y^2 $$
Now, we combine the results. We only need one constant of integration, usually denoted as $C$.