[tex]xy'=y+7x^2\sin x[/tex]
[tex]xy'-y=7x^2\sin x[/tex]
[tex]\dfrac1xy'-\dfrac1{x^2}y=7\sin x[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=7\sin x[/tex]
[tex]\dfrac1xy=\displaystyle\int7\sin x\,\mathrm dx[/tex]
[tex]\dfrac1xy=-7\cos x+C[/tex]
[tex]y=-7x\cos x+Cx[/tex]
With the initial value [tex]y(4\pi)=0[/tex], we have
[tex]0=-28\pi+4\piC\implies C=7[/tex]
so that the particular solution to the ODE is
[tex]y=-7x\cos x+7x[/tex]
