We are asked to solve the following differential equation: $\frac{y'}{\sin x} - y^3 = 0$.

1. Problem Description

We are asked to solve the following differential equation:

$\frac{y'}{\sin x} - y^3 = 0$.

2. Solution Steps

First, we rewrite the differential equation as:

$\frac{y'}{\sin x} = y^3$

$y' = y^3 \sin x$

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

Now, we separate the variables:

$\frac{dy}{y^3} = \sin x \, dx$

Integrate both sides:

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

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

$\frac{y^{-2}}{-2} = -\cos x + C$

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

$\frac{1}{2y^2} = \cos x - C$

$2y^2 = \frac{1}{\cos x - C}$

$y^2 = \frac{1}{2(\cos x - C)}$

$y = \pm \sqrt{\frac{1}{2(\cos x - C)}}$

$y = \pm \frac{1}{\sqrt{2(\cos x - C)}}$

3. Final Answer

$y = \pm \frac{1}{\sqrt{2(\cos x - C)}}$