First we find the solution for the homogeneous system
\begin{equation}
y^{(3)}-6y^{(2)}+11y^{(1)}-6y=0
\end{equation}
The corresponding characteristic equation is
\begin{equation}
r^3-6r^2+11r-6=0
\end{equation}
Three roots are
\begin{equation}
r_1=1
\end{equation}
\begin{equation}
r_2=2
\end{equation}
\begin{equation}
r_3=3
\end{equation}
Then the solution for the homogeneous system is
\begin{equation}
y_c(t)=c_1e^{x}+c_2e^{2x}+c_3e^{3x}
\end{equation}
where
\begin{equation}
y_1(t)=e^{x}
\end{equation}
\begin{equation}
y_2(t)=e^{2x}
\end{equation}
\begin{equation}
y_3(t)=e^{3x}
\end{equation}
Then we follow to find the required solution to the nonhomogeneous equation. We use Variation of Parameters. We have
\begin{equation}
W[y_1,y_2,y_3]=2e^{6x}
\end{equation}
\begin{equation}
W_1[y_1,y_2y_3]=e^{5x}
\end{equation}
\begin{equation}
W_2[y_1,y_2y_3]=-2e^{4x}
\end{equation}
\begin{equation}
W_3[y_1,y_2，y_3]=e^{3x}
\end{equation}
and
\begin{equation}
g(x)=2\frac{e^{3x}}{e^{x}+1}
\end{equation}
And then we have the following integration
\begin{equation}
\int \frac{W_1\cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{e^{2x}}{e^{x}+1}
\end{equation}
\begin{equation}
\int\frac{e^{2x}}{e^{x}+1}=e^{x}-\ln(e^{x}+1)+c_4
\end{equation}
\begin{equation}
\int \frac{W_2\cdot g(x)dx}{W[y_1,y_2,y_3]}=-2\int\frac{e^{x}}{e^{x}+1}
\end{equation}
\begin{equation}
-2\int\frac{e^{x}}{e^{x}+1}=-2\ln(e^{x}+1)+c_5
\end{equation}
\begin{equation}
\int \frac{W_3 \cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{1}{e^{x}+1}
\end{equation}
\begin{equation}
\int\frac{1}{e^{x}+1}=x-\ln(e^{x}+1)+c_6
\end{equation}
And finally, the required general solution $y(t)$
\begin{equation}
y(t)=\sum_{i=1}^3 y_i(t)\cdot\int\frac {W_i\cdot g(x)dx}{W[y_1,y_2,y_3]}
\end{equation}
