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Linear Algebra and its Applications
Found in: Page 267
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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 \(x\left( 0 \right) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\).

The required position is:

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

See the step by step solution

Step by Step Solution

System of Differential Equations

The general solution for any system of differential equations with the 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.

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

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

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.

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