Let \(A\) be a \(2 \times 2\) matrix with eigenvalues \( - {\bf{3}}\) and \( - {\bf{1}}\), and corresponding eigenvectors \({v_1} = \left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right)\) and \({v_2} = \left( {\begin{aligned}{ {20}{c}}1\\1\end{aligned}} \right)\). Let \(x\left( t \right)\) be the position of a particle at time \(t\). Solve the initial value problem \(x' = Ax\), \(x\left( 0 \right) = \left( {\begin{aligned}{ {20}{c}}{\bf{2}}\\{\bf{3}}\end{aligned}} \right)\).

The general solutionfor any system of differential equations withthe eigenvalues\({\lambda _1}\)and\({\lambda _2}\)with the respective eigenvectors\({v_1}\)and\({v_2}\)is given by:

\(x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\)

Here, \({c_1}\) and \({c_2}\) are the constants from the initial condition.

According to the question;

Consider the eigenvalues of\(A\)be\({\lambda _1} = - 3\)and\({\lambda _2} = - 1\)with respective eigenvectors:

\({v_1} = \left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right)\)and\({v_2} = \left( {\begin{aligned}{ {20}{c}}1\\1\end{aligned}} \right)\)

Then the general solution of the equation\(x' = A{\rm{x}}\)is:

\(\begin{aligned}{c}x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right){e^{ - 3t}} + {c_2}\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right){e^{ - t}}\\ = \left( {\begin{aligned}{ {20}{c}}{ - {c_1}{e^{ - 3t}} + {c_2}{e^{ - t}}}\\{{c_1}{e^{ - 3t}} + {c_2}{e^{ - t}}}\end{aligned}} \right)\end{aligned}\)

Substituting the initial condition\(x(0) = \left( {\begin{aligned}{ {20}{l}}2\\3\end{aligned}} \right)\), we get:

\(\left( {\begin{aligned}{ {20}{c}}{ - {c_1} + {c_2}}\\{{c_1} + {c_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}2\\3\end{aligned}} \right)\)

Solving this system, we have:\({c_1} = 0.5,\;\;\;{c_2} = 2.5\), and the solution is:

\(\begin{aligned}{c}x(t) = 0.5\left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right){e^{ - 3t}} + 2.5\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right){e^{ - t}}\\ = \left( {\begin{aligned}{ {20}{c}}{ - 0.5{e^{ - 3t}} + 2.5{e^{ - t}}}\\{0.5{e^{ - 3t}} + 2.5{e^{ - t}}}\end{aligned}} \right)\end{aligned}\)

Hence, this is the required solution.