Find the general solution of the given differential equation in variation of parameter.
$$
t y^{\prime}+2 y=\sin (t), t>0
$$
Answer: Find an integrating factor
$$
y^{\prime}+\frac{2}{t} y=\frac{\sin (t)}{t}
$$
$$
\therefore P(t)=\frac{2}{t}
$$
$$
\therefore \mu(t)=e^{\int \frac{2}{t} d t}=e^{2 \ln t}=t^{2}
$$
$$
\text{multiply} \quad\mu(t)\quad \text{on both sides}
$$
$$
\begin{array}{l}{\therefore t^{2} y^{\prime}+2 t y=\sin (t) \cdot t} \\ {\because t^{2} y^{\prime}+2 t y=\left(t^{2} y\right)^{\prime}} \\ {\therefore\left(t^{2} y\right)^{\prime}=\sin (t) \cdot t}\\{\therefore \int\left(t^{2} y\right)^{\prime}=\int \sin (t) \cdot t} \\ {\text { let } u=t \quad d v=\sin (t)} \\ {\therefore d u=d t \quad v=-\cos (t)} \\ {\therefore t^{2} y=t \cdot(-\cos (t))-\int-\cos (t) d t} \\ {\quad y=\frac{-t \cdot \cos (t)+\sin (x)+c}{t^{2}}}\end{array}
$$
