Suggested languages for you:

Americas

Europe

Q1E

Expert-verified
Found in: Page 267

### 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

# 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

## 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.