Question number $7.$

Give the general solution to

$x^2{y}^{\text{'}}+5xy=x^{-3}cos(4x)$

$y=\frac{1}{4}{x}^{5}sin(4x)+C{x}^5$

$y=-\frac{1}{4}{x}^{-5}sin(4x)+C{x}^{-5}$

$y=\frac{1}{4}{x}^{-5}sin(4x)+C{x}^5$

$y=-\frac{1}{4}{x}^{5}sin(5x)+C{x}^5$

$y=\frac{1}{4}{x}^{-5}sin(4x)+C{x}^{-5}$

None of the above.