Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized: y′′−2y′+y=sin(t)

Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized:
$y^{″}-2y^{\prime }+y=\sin (t)$
The way that I solved this doesn't involve Euler's formula, and I was wondering how I might use the formula here.

My approach:
The formula can be written as $y(t)=y_h(t)+y_p(t)$ where $y_h(t)$ is the "homogeneous version" of the ODE and $y_p(t)$ is the particular solution that we'll obtain via the basic rule of the method of undetermined coefficients.

$y_h(t)$:
Putting $r(t)=\sin (t)=0$ in the original equation, the ODE we need to solve is:
$y^{″}-2y^{\prime }+y=0$
where we can set the general solution as $y=e^{\lambda t}$ and obtain the characteristic equation:
${\lambda }^2-2\lambda +1=0$
which has a real double root, hence giving us the solution:
$y_h(t)=(c_1+c_{2}t)e^t$

$y_p(t)$:
Judging by the fact that $r(t)$ is shape $k\sin (\omega t)$ and we know that $\omega =1$ we can set the general solution to be of form:
$\begin{array}{rl}{y}_p(t)& =\phantom{-}K\cos (t)+M\sin (t)\\ y_p^{\prime }(t)& =-K\sin (t)+M\cos (t)\\ y_p^{″}(t)& =-K\cos (t)-M\sin (t)\end{array}$
substituting these equations into the original equation and then simplifying gives us:
$y_p(t)=\frac{1}{2}\cos (t)$
And in conclusion, we can write that the solution to the given ODE is:
$\begin{array}{rl}y(t)& =y_h(t)+y_p(t)\\ & =(c_1+c_{2}t)e^t+\frac{1}{2}\cos (t)\end{array}$
How would we be able to derive this conclusion via Euler's formula? Thanks in advance.