a) First, we should solve for y"-y=0.
Let r^{2}-1=0,r_{1}=1,r_{2}=-1.
thus, y_{c}(t)=C_{1}e^{t}+C_{2}e^{-t}.
Second, we should calculate W.
W=\begin{bmatrix}e^{t} & e^{-t}\\ e^{t} & -e^{-t} \end{bmatrix}=-2,W_{1}=\begin{bmatrix}0 & e^{-t}\\ 1 & -e^{-t} \end{bmatrix}=-e^{-t},W_{2}=\begin{bmatrix}e^{t} & 0\\ e^{t} & 1 \end{bmatrix}=e^{t}.
therefore, y_{p}(t)=e^{t}\int\frac{W_{1}(t)g(t)}{W(t)}dt-e^{-t}\int\frac{W_{2}(t)g(t)}{W(t)}dt.
y_{p}(t)=6e^{t}\int\frac{e^{-t}}{e^{t}+1}dt+6e^{-t}\int\frac{e^{t}}{e^{t}+1}dt.
y_{p}(t)=6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).
Thus, y(t)=C_{1}e^{t}+C_{2}e^{-t}+6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).
b) y'(t)=C_{1}e^{t}-C_{2}e^{-t}+6e^{t}ln\left|e^{t}+1\right|+6e^{t}\frac{e^{t}}{e^{t}+1}-6te^{t}-6e^{t}-6e^{-t}ln\left|e^{t}+1\right|+6e^{-t}\frac{e^{t}}{e^{t}+1}.
plug in, we can get C_{1}=6-6ln(2),C_{2}=6ln(2).
Therefore, y(t)=(6-6ln(2))e^{t}+6ln(2)e^{-t}+6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).
OK, except LaTeX sucks:
2) "operators" should be escaped: \cos, \sin, \tan, \ln
\boxed{ y= 6\Bigl(-e^{-t}+\ln (1+e^{-t})+1-\ln(2) \Bigr)e^{t} + 6\Bigl(\ln (e^t+1)-\ln(2)\Bigr)e^{-t}. }