For the ODE
[tex]ty'+2y=\sin t[/tex]
multiply both sides by t so that the left side can be condensed into the derivative of a product:
[tex]t^2y'+2ty=t\sin t[/tex]
[tex]\implies(t^2y)'=t\sin t[/tex]
Integrate both sides with respect to t :
[tex]t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C[/tex]
Divide both sides by [tex]t^2[/tex] to solve for y :
[tex]y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}[/tex]
Now use the initial condition to solve for C :
[tex]y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}[/tex]
[tex]\implies9=\dfrac4{\pi^2}(1+C)[/tex]
[tex]\implies C=\dfrac{9\pi^2}4-1[/tex]
So the particular solution to the IVP is
[tex]y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}[/tex]
or
[tex]y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}[/tex]
