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

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$x (t) = c_{1} e^{5 t} [\begin{matrix} 1 \\ 1 \end{matrix}] + c_{2} e^{2 t} [\begin{matrix} - 2 \\ 1 \end{matrix}]$ is the general solution of the system.

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Step 1 ：Given the system of equations: ${\begin{cases} \dot{x} = 3 x + R y \\ \dot{y} = x + 4 y \end{cases}$

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

Step 3 ：Using the formula for the general solution of a system of linear differential equations, we have: $x (t) = c_{1} e^{λ_{1} t} v_{1} + c_{2} e^{λ_{2} t} v_{2}$ where $λ_{1}, λ_{2}$ 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: $x (t) = c_{1} e^{5 t} [\begin{matrix} 1 \\ 1 \end{matrix}] + c_{2} e^{2 t} [\begin{matrix} - 2 \\ 1 \end{matrix}]$

Step 5 ： $x (t) = c_{1} e^{5 t} [\begin{matrix} 1 \\ 1 \end{matrix}] + c_{2} e^{2 t} [\begin{matrix} - 2 \\ 1 \end{matrix}]$ is the general solution of the system.

(a) Find two independent solutions of{x˙=3x+Ryy˙=x+4y(Hint: The eigenvectors of [3214] are [11] and [−21], with corresponding eigenvalues 5 and 2.)(b) Find the general solution.

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