$$
\begin{align}
u_1(t)&=-\int\dfrac{y_2(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=-\int\dfrac{t[-2(t-1)e^{-t}]}{(1-t)e^t}dt\\
\notag \\
&=-2\int te^{-2t}dt\\
\notag \\
&=(t+\dfrac{1}{2})e^{-2t}\\
\end{align}
$$
$$
\begin{align}
u_2(t)&=\int\dfrac{y_1(t)g(t)}{ W[y_1(t),y_2(t)]}dt\\
\notag \\
&=\int\dfrac{e^t[-2(t-1)e^{-t}]}{(1-t)e^t}dt\\
\notag \\
&=2\int e^{-t}dt\\
\notag \\
&=-2e^{-t}\\
\end{align}
$$
Since,
$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$
Therefore,
$$
\begin{align}
Y(t)&= (t+\dfrac{1}{2})e^{-2t} \cdot e^t+(-2e^{-t}) \cdot t \notag\\
\notag \\
&=(\dfrac{1}{2}-t)e^{-t} \notag
\end{align}
$$
Therefore, the particular solution of the given nonhomogeneous equation is
$Y(t)= (\dfrac{1}{2}-t)e^{-t}$