Answer
$y=\dfrac{-\cos x}{x^3}+\dfrac{c}{x^3}=\dfrac{-\cos x+c}{x^3}$
Here, $c$ is an arbitrary constant.
Work Step by Step
Here, we have $xy+3y=\dfrac{\sin x}{x^2}$
or, $x^3y+3y=\sin x$
Also, it can re-written as:
$(x^3y)'=\sin x$
In order to determine the general solution, isolate the x and y terms on one side and integrate both sides.
$\int (x^3y)' dx=\int \sin x dx$ ...(1)
Equation (1) gives:
$x^3 y=-\cos x+c$
Thus, the general solution is as follows:
$y=\dfrac{-\cos x}{x^3}+\dfrac{c}{x^3}=\dfrac{-\cos x+c}{x^3}$
Here, $c$ is an arbitrary constant.
