(a) Find two independent solutions of \[ \left\{\begin{array}{l} \dot{x}=3 x+R y \\ \dot{y}=x+4 y \end{array}\right. \] (Hint: The eigenvectors of $\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$ are $\left[\begin{array}{l}1 \\ 1\end{array}\right]$ and $\left[\begin{array}{r}-2 \\ 1\end{array}\right]$, with corresponding eigenvalues 5 and 2.) (b) Find the general solution.

Step 1 ：Given the system of equations: \[\left\{\begin{array}{l} \dot{x}=3 x+R y \\ \dot{y}=x+4 y \end{array}\right.\]

Step 2 ：We are given that the eigenvectors of the matrix \[\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]\] are \[\left[\begin{array}{l}1 \\ 1\end{array}\right]\] and \[\left[\begin{array}{r}-2 \\ 1\end{array}\right]\] with corresponding eigenvalues 5 and 2.

Step 3 ：Using the formula for the general solution of a system of linear differential equations, we have: \[x(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2\] where \(\lambda_1, \lambda_2\) are the eigenvalues and \(v_1, v_2\) are the corresponding eigenvectors.

Step 4 ：Substituting the given eigenvalues and eigenvectors into the formula, we get: \[x(t) = c_1 e^{5t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{2t} \begin{bmatrix} -2 \\ 1 \end{bmatrix}\]

Step 5 ：\(\boxed{x(t) = c_1 e^{5t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{2t} \begin{bmatrix} -2 \\ 1 \end{bmatrix}}\) is the general solution of the system.