Question

A particle moving in a planar force field has a position vector .\(x\). that satisfies \(x' = Ax\). The \(2 \times 2\) matrix \(A\) has eigenvalues 4 and 2, with corresponding eigenvectors \({v_1} = \left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right)\) and \({v_2} = \left( {\begin{aligned}{{20}{c}}{ - 1}\\1\end{aligned}} \right)\). Find the position of the particle at a time \(t\), assuming that \(x\left( 0 \right) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\).

Short answer

The required position is:

\({\rm{x}}\left( t \right) = \left( {\begin{aligned}{{20}{c}}{ - 7.5{e^{4t}} + 1.5{e^{2t}}}\\{2.5{e^{4t}} - 1.5{e^{2t}}}\end{aligned}} \right)\)

Step-by-step solution

System of Differential Equations

The general solution for any system of differential equations with theeigenvalues \({\lambda _1}\) 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.

Find the position of the particle

According to the question;

Consider the eigenvalues of \(A\) be \({\lambda _1} = 4\) and \({\lambda _2} = 2\) with the respective eigenvectors:

\({v_1} = \left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right)\) and \({v_2} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\1\end{aligned}} \right)\)

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

\(\begin{aligned}{c}x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\\ = {c_1}\left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right){e^{4t}} + {c_2}\left( {\begin{aligned}{{20}{c}}{ - 1}\\1\end{aligned}} \right){e^{2t}}\\ = \left( {\begin{aligned}{*{20}{c}}{ - 3{c_1}{e^{4t}} - {c_2}{e^{2t}}}\\{{c_1}{e^{4t}} + {c_2}{e^{2t}}}\end{aligned}} \right)\end{aligned}\)

Now, substituting the initial condition \(x(0) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\), we get:

\(\left( {\begin{aligned}{{20}{c}}{ - 3{c_1} - {c_2}}\\{{c_1} + {c_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 6}\\1\end{aligned}} \right)\)

Solving this system, we have:

\({c_1} = 2.5,\,\,\,\,{c_2} = - 1.5\)

Thus, the solution can be given as:

\(\begin{aligned}{c}x(t) = 2.5\left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right){e^{4t}} - 1.5\left( {\begin{aligned}{{20}{c}}{ - 1}\\1\end{aligned}} \right){e^{2t}}\\ = \left( {\begin{aligned}{{20}{c}}{ - 7.5{e^{4t}} + 1.5{e^{2t}}}\\{2.5{e^{4t}} - 1.5{e^{2t}}}\end{aligned}} \right)\end{aligned}\)

Hence, this is the required position.