Answer
$y = -\frac{cos~x}{x^4}+\frac{cos^3~x}{3x^4}+\frac{C}{x^4}$
Work Step by Step
A first-order linear differential equation is one that can be put into the form:
$\frac{dy}{dx} + P(x)~y = Q(x)$
We can consider the given equation:
$4x^3y+x^4y' = sin^3~x$
$y'+\frac{4}{x}y = \frac{sin^3~x}{x^4}$
Let $P(x) = \frac{4}{x}$
Let $Q(x) = \frac{sin^3~x}{x^4}$
We can find the integrating factor:
$I(x) = e^{\int \frac{4}{x} ~dx} = e^{4ln~x} = e^{ln~x^4} = x^4$
We can multiply by the integrating factor:
$x^4~y'+4x^3~y = sin^3~x$
$\frac{d}{dx}(x^4~y) = sin^3~x$
We can integrate both sides of the equation:
$x^4~y = \int sin^3~x~dx$
$x^4~y = \int (1-cos^2~x)~sin~x~dx$
Let $u = cos~x$
$\frac{du}{dx} = -sin~x$
$dx = -\frac{du}{sin~x}$
$x^4~y = \int -(1-u^2)~sin~x~\frac{du}{sin~x}$
$x^4~y = \int -(1-u^2)~du$
$x^4~y = -u+\frac{1}{3}u^3+C$
$x^4~y = -cos~x+\frac{1}{3}cos^3~x+C$
$y = -\frac{cos~x}{x^4}+\frac{cos^3~x}{3x^4}+\frac{C}{x^4}$
