dy/dx+y/x=y2sin(x).
Let v=1/y, then dv/dx=-(1/y2)(dy/dx), dy/dx=-y2(dv/dx)=-(1/v2)(dv/dx).
The DE becomes:
-(dv/dx)/v2+1/(vx)=sin(x)/v2, -(dv/dx)/v+1/x=sin(x)/v.
Multiply through by -v:
dv/dx-v/x=-sin(x).
Multiply through by 1/x:
(1/x)(dv/dx)-v/x2=-sin(x)/x.
(d/dx)(v/x)=-sin(x)/x.
v/x=-∫(sin(x)/x)dx, 1/(xy)=-∫(sin(x)/x)dx.
There is no standard solution for the indefinite integral, but sin(x)=∑(-1)nx2n+1/(2n+1)! as a series (for integer n≥0).
Therefore, 1/xy=-∫∑(-1)nx2n/(2n+1)!=∑(-1)n+1x2n+1/(2n+1)[(2n+1)!]+C.
y=1/(Cx+∑(-1)n+1x2(n+1)/(2n+1)[(2n+1)!].
I suspect that the original DE should have been dy/dx+y/x=xy2sin(x). If so we get to:
(d/dx)(v/x)=-sin(x). 1/(xy)=cos(x)+C, y=1/(Cx+xcos(x)).