Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation.
\begin{gather*}
(1-t)y'' + ty' - y =2(t-1)^2e^{-t}, \qquad 0 <t<1;\\
y_1(t) = e^t, \qquad y_2(t) = t
\end{gather*}
Solution:
first check if $y_1$,$y_2$ are solutions to the homogeneous equation
\begin{equation*}
\begin{cases}
y_1 = e^t\\
y_1'' = e^t\\
y_1'' = e^t\\
\end{cases}
\\(1-t)e^t + te^t- e^t = 0
\end{equation*}
\begin{equation*}
\begin{cases}
y_2 = t\\
y_2'' = 1\\
y_2''= 0\\
\end{cases}
\\t - t = 0
\end{equation*}
Thus, $y_1, y_2$ are solutions to the homogeneous equation.
Then, find the Wronskian
\begin{equation*}
W(y_1,y_2)(t) =
\begin{vmatrix}
e^t&t\\
e^t&1\\
\end{vmatrix}
= e^t - te^t
\end{equation*}
\begin{equation*}
g(x) = \frac{2(t-1)^2e^{-t}}{1-t} = 2(1-t)e^{-t}\\
c_1 = -\int \frac{2t(1-t)e^{-t}}{e^t - te^t}dt = \int-2te^{-2t}dt = (t+\frac{1}{2})e^{-2t}\\
c_2 = \int\frac{2e^t(1-t)e^{-t}}{e^t - te^t}dt = \int2e^{-t}dt = -2e^{-t}
\end{equation*}
Thus, the particular solution is
$$y_p = ((t+\frac{1}{2})e^{-2t})e^t + (-2e^{-t})t = (\frac{1}{2}-t)e^{-t}$$
