[tex]x\,\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x\implies\dfrac1x\,\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x[/tex]
Note that in order to do this division, we cannot allow [tex]x=0[/tex]. This means the largest interval on which a solution can exist is either [tex](0,\infty)[/tex] or [tex](-\infty,0)[/tex].
If [tex]y(x)[/tex] is a solution to the ODE, then any term that vanishes as [tex]x\to\infty[/tex] (or [tex]-\infty[/tex], depending on which interval above is used) is a transient term.
Solve the ODE:
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac yx\right]=\sin x\implies\dfrac yx=C-\cos x\implies y=Cx-x\cos x[/tex]
As [tex]x\to\infty[/tex], [tex]\cos x[/tex] will oscillate between -1 and 1, so [tex]x\cos x[/tex] will oscillate between [tex]-\infty[/tex] and [tex]\infty[/tex], so the limit of [tex]y(x)[/tex] as [tex]x\to\infty[/tex] does not exists. There are no transient terms.