\begin{equation}
\begin{array}{c}{\text { Qusetion: }} \\ {\qquad \begin{array}{c}{(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1} \\ {y_{1}(t)=e^{t}, y_{2}(t)=t}\end{array}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{(1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, 0<t<1 ; y_{1}(t)=e^{t}, y_{2}(t)=t} \\ {\text { Hence, }} \\ {\qquad\left\{\begin{array}{l}{y_{1}(t)=e^{t}} \\ {y_{1}^{\prime}(t)=e^{t}} \\ {y_{1}^{\prime \prime}(t)=e^{t}}\end{array} \text { and }\left\{\begin{array}{l}{y_{2}(t)=t} \\ {y_{2}^{\prime \prime}(t)=1} \\ {y_{2}^{\prime \prime}(t)=0}\end{array}\right.\right.}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Substitute back into the homogeneous equation: }} \\ {\qquad(1-t) y^{\prime \prime}+t y^{\prime}-y=0} \\ {\text { Verfified that } y_{1}(t) \text { and } y_{2}(t) \text { both satisfy the corresponding homogeneous }} \\ {\text { equation. }}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { And the complementary solution } y_{c}(t)=c_{1} e^{t}+c_{2}} \\ {\text { Now divide both sides of the original equation by } 1-t:} \\ {\qquad y^{\prime \prime}+\frac{t}{1-t} y'-\frac{1}{1-t} y=-2(t-1) e^{-t}} \\ {\qquad p(t)=\frac{t}{1-t}, g(t)=-2(t-1) e^{-t}} \\ {\qquad W\left[y_{1}, y_{2}\right](t)=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|=(1-t) e^{t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { since the particular solution has the form: }} \\ {\qquad Y(t)=u_{1}(t) y_{1}(t)+u_{2}(t) y_{2}(t)}\end{array}
\end{equation}
\begin{equation}
\begin{aligned} u_{1}(t) &=-\int \frac{y_{2}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=-\int \frac{t \cdot\left(-2(t-1) e^{-t}\right)}{(1-t) e^{t}} d t \\ &=-2 \int t e^{-2 t} d t \\ &=\left(t+\frac{1}{2}\right) e^{-2 t} \end{aligned}
\end{equation}
\begin{equation}
\begin{aligned} u_{2}(t) &=\int \frac{y_{1}(t) g(t)}{W\left[y_{1}, y_{2}\right](t)} d t \\ &=\int \frac{e^{t^{t} \cdot\left(-2(t-1) e^{-t}\right)}}{(1-t) e^{t}} d t \\ &=2 \int e^{-t} \\ &=-2 e^{-t} \end{aligned}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Therefore, }} \\ {\qquad Y(t)=\left(t+\frac{1}{2}\right) e^{-2 t} \cdot e^{t}+\left(-2 e^{-t}\right) \cdot t=\left(\frac{1}{2}-t\right) e^{-t}}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{\text { Hence, the general solution: }} \\ {\qquad \begin{aligned} y(t) &=y_{c}(t)+Y(t) \\ &=c_{1} e^{t}+c_{2} t+\left(\frac{1}{2}-t\right) e^{-t} \end{aligned}} \\ {\text { Therefore, the particular solution of the given nonhomogeneous equation is }} \\ {\qquad Y(t)=\left(\frac{1}{2}-t\right) e^{-t}}\end{array}
\end{equation}
