Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
$$
(1-t)y''+ty'-y=2(t-1)^2e^{-t}, 0<t<1;y1(t)=e^t \quad y2(t)=t
$$
$$y1(t)=e^t \quad y'1(t)=e^t \quad y''1(t)=e^t$$
$$y2(t)=t \quad y'2(t)=1 \quad y''2(t)=0$$
Substitude back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$
Verified that y1(t) and y2(t) both satisfy the corresponding homogeneous equation.
$$y_c(t)=C_1e^t+C_2t$$
Divide both side by 1-t, then
$$p(t)=\frac{t}{1-t} \quad q(t)=-\frac{1}{1-t} \quad g(t)=-2(t-1)e^{-t}$$
$$W(t)=(1-t)e^t$$
$$u1(t)=-\int{\frac{y2(t)g(t)}{W(t)}}dt=-2\int{te^{2t}}=(t+\frac{1}{2})e^{-2t}$$
$$u2(t)=\int\frac{y1(t)g(t)}{W(t)}dt=2\int{e^{-t}}=-2e^{-t}$$
Therefore, the particular solution is:
$$Y(t)=u1(t)y1(t)+u2(t)y2(t)=(t+\frac{1}{2})e^{-2t}*e^t+(-2e^{-t})*t=(\frac{1}{2}-t)e^{-t}$$
Hence, the general solution:
$$y(t)=y_c(t)+Y(t)=C_1e^t+C_2t+(\frac{1}{2}-t)e^{-t}$$
The particular solution is:
$$Y(t)=(\frac{1}{2}-t)e^{-t}$$
