I have the following ordinary differential equation before me: $y''+\sin(x)y'+2y=1$.
I have to find its PI(particular integral).My strategy is to come up with the solution of corresponding homogeneous equation i.e. $y''+\sin(x)y'+2y=0$ and then use the method of variation of parameters to determine PI. But I am stuck at finding the solution of homogeneous equation. Could you please suggest a way out here?
As mentioned in comments, since the coefficient of $y$ and the right-hand side are both constants, a particular solution (integral) is given by the constant function $y=\frac12$, which is what you wanted.
To get solutions of the homogeneous part, however, requires some trickery. Since $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$, suppose $y(x)=z(w)=z(e^{ix})$, then rewrite in terms of $z$ and $w$. We find that $z$ is a holonomic function defined by $$2w^2z''+(-w^2+2w+1)z'-4z=0$$ and while this cannot be simplified to named functions it does give an effective way of computing any solution to the original differential equation as $y(x)=z(e^{ix})+\frac12$ plus initial conditions.