Solve the following differential equation: $\frac{dy}{dx} = x^3 \csc y$, given that $y(0) = 0$.

$\cos y = 1 - \frac{x^4}{4}$

The correct answer is Option (1) → $\cos y = 1 - \frac{x^4}{4}$ ##

$\frac{dy}{dx} = x^3 \csc y; \quad y(0) = 0$

$\int \frac{dy}{\csc y} = \int x^3 dx$

$\int \sin y \, dy = \int x^3 dx$

$-\cos y = \frac{x^4}{4} + c$

$-1 = c \quad (∵y=0, \text{ when } x=0)$

$\cos y = 1 - \frac{x^4}{4}$