Question

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2, with corresponding eigenvectors v1=[ −3 1 ] and v2=[ −1 1 ]. Find the position of the particle at time t, assuming that x(0)=[ −6 1 ].

Answer 1

The required position of the particle at time t is:
`x(t)=\begin{bmatrix}-7.5e^(4t)+1.5e^(2t)\\2.5e^(4t)-1.5e^(2t)\end{bmatrix}`

Explanation:

Consider the provided matrix.

`v_1=\begin{bmatrix}-3\\1 \end{bmatrix}`

`v_2=\begin{bmatrix}-1\\1 \end{bmatrix}`

`\lambda_1=4, \lambda_2=2`

The general solution of the equation
`x'=Ax`

`x(t)=c_1v_1e^(\lambda_1t)+c_2v_2e^(\lambda_2t)`

Substitute the respective values we get:

`x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^(4t)+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^(2t)`

`x(t)=\begin{bmatrix}-3c_1e^(4t)-c_2e^(2t)\\c_1e^(4t)+c_2e^(2t) \end{bmatrix}`

Substitute initial condition
`x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}`

`\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}`

Reduce matrix to reduced row echelon form.

`\begin{bmatrix} 1& 0 & (5)/(2)\\ 0& 1 & (-3)/(2)\end{bmatrix}`

Therefore,
`c_1=2.5,c_2=1.5`

Thus, the general solution of the equation
`x'=Ax`

`x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^(4t)-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^(2t)`

`x(t)=\begin{bmatrix}-7.5e^(4t)+1.5e^(2t)\\2.5e^(4t)-1.5e^(2t)\end{bmatrix}`

The required position of the particle at time t is:
`x(t)=\begin{bmatrix}-7.5e^(4t)+1.5e^(2t)\\2.5e^(4t)-1.5e^(2t)\end{bmatrix}`