\[\begin{cases} \dfrac{dx}{dt} = x + 2y + 4z + 3w\\ \dfrac{dy}{dt} = 2z + w\\ \dfrac{dz}{dt} = z + w\\\dfrac{dw}{dt} = 2w\end{cases} \]

The general solution to the above differential system is as follows:

\[\begin{cases} x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t) + c_{3}x_{3}(t) + c_{4}x_{4}(t)\\ y(t) = c_{1}y_{1}(t) + c_{2} y_{2}(t) + c_{3}y_{3}(t) + c_{4}y_{4}t\\ z(t) = c_{1}z_{1}(t) + c_{2}z_{2}(t) + c_{3}z_{3}(t) + c_{4}z_{4}(t)\\
w(t) = c_{1}w_{1}(t) + c_{2}w_{2}(t) + c_{3}w_{3}(t) + c_{4}w_{4}(t)\\
\end{cases} \] where
\[ \begin{vmatrix}{x_{j}(t)} \\{y_{j}(t)} \\ {z_{j}(t)} \\ {w_{j}(t)}\\ \end{vmatrix} =

\begin{vmatrix}{A_{j}} \\{B_{j}} \\ {C_{j}} \\ {D_{j}} \\ \end{vmatrix} * e^{\lambda_{j} t} \]

for \( (1 <= j <= 3) \), and
\[ \begin{vmatrix}{x_{4}(t)} \\{y_{4}(t)} \\ {z_{4}(t)} \\ {w_{4}(t)} \\ \end{vmatrix} =

\begin{vmatrix}{A_{4} + A_{3} * t} \\{B_{4} + B_{3} * t}\\ {C_{4} + C_{3} * t}
\\ {D_{4} + D_{3} * t}
\end{vmatrix} * e^{\lambda_{3}t } \]

If \( B_{1} = 1, \: D_{2} = 1, \: A_{3} = 1, \: A_{4} = 1 \) and \(x(0) = 0\), \(y(0) = 1\), \(z(0) = 2\), and \(w(0) = 0 \), Find \( c_{1} + c_{2} + c_{3} + c_{4}. \)

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