$$(1-t)y’’+ty’-y=2(t-1)^2e^{-t}, 0< t<1; y_1(t)=e^t, y_2(t)=t$$
Hence,$$\cases{y_1(t)=e^t\\y_1’(t)=e^t\\y_1’’(t)=e^t} \text{and} \cases{y_2(t)=t\\y_2’(t)=1\\y_2’’(t)=0}$$
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 $y_c(t)=c_1e^t+c_2$ $\\$
Now divide both sides of the original equation by $1-t$:
$$y’’+{t\over 1-t}-{1\over 1-t}=-2(t-1)e^{-t}$$
Then $$p(t)={t\over 1-t}, q(t)=-{1\over 1-t}, g(t)=-2(t-1)e^{-t}$$
$$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t)&y_2(t)\\y_1’(t)&y_2’(t)\end{array}=(1-t)e^t$$
Since the particular solution has the form: $$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$$
and $$\begin{align}u_1(t)&=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}dt\\&=-\int{t\cdot (-2(t-1)e^{-t})\over (1-t)e^t}dt\\&=-2\int{te^{-2t}}dt\\&=(t+{1\over2})e^{-2t}\end{align}$$
$$\begin{align}u_2(t)&=\int{y_1(t)g(t)\over W[y_1,y_2](t)}dt\\&=\int{e^t\cdot (-2(t-1)e^{-t})\over (1-t)e^t}dt\\&=2\int{e^{-t}}\\&=-2e^{-t}\end{align}$$
Therefore, $$Y(t)=(t+{1\over2})e^{-2t}\cdot e^t+ (-2e^{-t})\cdot t=({1\over2}-t)e^{-t}$$
Hence, the general solution:
$$\begin{align}y(t)&=y_c(t)+Y(t)\\&=c_1e^t+c_2t+({1\over2}-t)e^{-t}\end{align}$$
Therefore, the particular solution of the given nonhomogeneous equation is $$Y(t)=({1\over2}-t)e^{-t}$$
