Answer
$$ y= -x^4\cos x +\frac{x^4}{3}\cos^3x+Cx^4$$
Work Step by Step
Given $$4 x^{3} y+x^{4} y^{\prime}=\sin ^{3} x$$
Rewrite the equation
$$ y'+ \frac{4}{x} y= \frac{\sin ^{3} x}{x^4} $$
Since the equation is linear and $p(x)= 4/x$ and $q(x)= \frac{\sin ^{3} x}{x^4}$, then
\begin{align*}
\mu &=e^{\int p(x)}dx\\
&= e^{\int (4/x)dx}\\
&=e^{4\ln x}\\
&=e^{\ln x^4}\\
&= x^4
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(x)y&=\int \mu(x)q(x)dx\\
x^4 y&= \int x^4 \frac{\sin ^{3} x}{x^4}dx\ \ \ \\
&= \int \sin ^{3} xdx\\
&= \int (1-\cos^2 x)\sin xdx\\
&= -\cos x +\frac{1}{3}\cos^3x+C
\end{align*}
Then
$$ y= -x^4\cos x +\frac{x^4}{3}\cos^3x+Cx^4$$
