Find the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ given that $y = 1$, when $x = 0$.

$y = \frac{1}{2x^2 + 1}$

The correct answer is Option (1) → $y = \frac{1}{2x^2 + 1}$ ##

If $y \neq 0$, the given differential equation can be written as

$\frac{dy}{y^2} = -4x \, dx \quad \dots (1)$

Integrating both sides of equation (1), we get

$\int \frac{dy}{y^2} = -4 \int x \, dx$

$\text{or } -\frac{1}{y} = -2x^2 + C$

$\text{or } y = \frac{1}{2x^2 - C} \quad \dots (2)$

Substituting $y = 1$ and $x = 0$ in equation (2), we get, $C = -1$.

Now substituting the value of $C$ in equation (2), we get the particular solution of the given differential equation as $y = \frac{1}{2x^2 + 1}$.