Assuming you mean
[tex]5\sin t\dfrac{\mathrm dy}{\mathrm dt}+5y\cos t=\cos t\sin^2t[/tex]
This ODE is linear in [tex]y[/tex], and you can already contract the left hand side as the derivative of a product:
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t[/tex]
Integrating both sides with respect to [tex]t[/tex] yields
[tex]5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt[/tex]
[tex]5y\sin t=\dfrac13\sin^3t+C[/tex]
[tex]y=\dfrac1{15}\sin^2t+C\csc t[/tex]
Given that [tex]y\left(\dfrac\pi2\right)=9[/tex], we have
[tex]9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2[/tex]
[tex]9=\dfrac1{15}+C[/tex]
[tex]C=\dfrac{134}{15}[/tex]
so that the particular solution over the interval is
[tex]y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t[/tex]
