Answer
See answer below
Work Step by Step
$\frac{dy}{dx}=y^3 \sin x$
$\frac{1}{y^3}\frac{dy}{dx}=\sin x$
Integrating,
$\int \frac{1}{y^3} dy=\int\sin x \;dx$
$-\frac{1}{2y^2}=-\cos (x) + c_1$
Solve for $y$:
$-2y^2=\frac{1}{-\cos (x) + c_1}$
$y^2=\frac{-1}{2(c_1-\cos (x))}$
$y=\pm \frac{1}{\sqrt 2\sqrt c_1-\cos (x)}$
Using initial condition $y(0)=0$
$0=\pm \frac{1}{\sqrt 2(c_1-\cos (0))}$
When we solve for $c_1$, we would end up with an undefined value. Therefore, the solution must be a constant. We choose 0.
Thus, the solution is $y(x)=0$
