5. Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.\\
\begin{eqnarray*}
&& (1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1 \\
&& y_{1}(t)=e^{t}, y_{2}(t)=t.
\end{eqnarray*}
$$(1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1;y_{1}(t)=e^{t}, y_{2}(t)=t$$
Hence,
$\left\{
\begin{array}{l}
y_{1}(t)=e^{t} \\
y_{1}'(t)=e^{t} \\
y_{1}''(t)=e^{t}
\end{array}
\right.$
and
$\left\{
\begin{array}{l}
y_{2}(t)=t\\
y_{2}'(t)=1\\
y_{2}''(t)=0
\end{array}
\right.$
Substitute back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$
\ \\
Verified that $y_{1}(t)$ and $y_{2}(t)$ both satisfy the corresponding homogeneous equation.\\
And the complementary solution $y_{c}(t)=c_{1}e^{t}+c_{2}$\\
Now divide both sides of the original equation by $1-t$:\\
$$y''+\dfrac{t}{1-t}-\dfrac{1}{1-t}=-2(t-1)e^{-t}$$
\ \\
Then\\
$$p(t)=\dfrac{t}{1-t},q(t)=-\dfrac{1}{1-t},g(t)=-2(t-1)e^{-t}$$
$$W[y_{1},y_{2}](t)=
\left|
\begin{array}{cc}
y_{1}(t) & y_{2}(t) \\
y_{1}'(t) & y_{2}'(t)
\end{array}
\right|
=(1-t)e^{t}$$
Since the particular solution has the form:\\
$$Y(t)=u_{1}(t)y_{1}(t)+u_{2}(t)y_{2}(t)$$
and\\
\begin{eqnarray}
&&u_{1}(t)=-\int \dfrac{y_{2}(t)g(t)}{W[y_{1},y_{2}](t)}dt\\
&&\quad\quad\ =-\int \dfrac{t\cdot (-2(t-1)e^{-t})}{(1-t)e^{t}}dt\\
&&\quad\quad\ =-2\int te^{-2t}dt\\
&&\quad\quad\ =(t+\dfrac{1}{2})e^{-2t}
\end{eqnarray}
\begin{eqnarray}
&&u_{2}(t)=\int \dfrac{y_{1}(t)g(t)}{W[y_{1},y_{2}](t)}dt\\
&&\quad\quad\ =\int \dfrac{e^{t}\cdot (-2(t-1)e^{-t})}{(1-t)e^{t}}dt\\
&&\quad\quad\ =2\int e^{-t}dt \\
&&\quad\quad\ =-2e^{-t}
\end{eqnarray}
Therefore,\\
$$Y(t)=(t+\dfrac{1}{2})e^{-2t}\cdot e^{t}+(-2e^{-t})\cdot t=(\dfrac{1}{2}-t)e^{-t}$$
Hence, the general solution:\\
\begin{eqnarray}
&&y(t)=y_{c}(t)+Y(t)\\
&&\quad\quad=c_{1}e^{t}+c_{2}t+(\dfrac{1}{2}-t)e^{-t}
\end{eqnarray}
Therefore, the particular solution of the given nonhomogeneous equation is\\
$$Y(t)=(\dfrac{1}{2}-t)e^{-t}$$
\end{document}
