# Math Help - Application to Differential Equations

1. ## Application to Differential Equations

A particle moving in a planar force field has a position vector x that satisfies x' = Ax. The 2x2 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].

I hate to ask like this..but I really have no clue on where to start this. I think I am trying to find x(t), but I don't know how I would go about doing that. Can anyone explain this to me or give me some hints/starting points explained? Thank you for any help.

2. First, since web browsers don't respect tabs, lets clean that up using LaTex:
A has eigenvalues 4 with corresponding eigenvector $\begin{bmatrix}-3 \\ 1\end{bmatrix}$ and 2 with corresponding eigenvector $\begin{bmatrix}-1 \\ 1\end{bmatrix}$. That means that, if we used those two eigenvectors as basis vectors, we can write the equation as $X'= \begin{bmatrix}x' \\ y'\end{bmatrix}= \begin{bmatrix}4 & 0 \\ 0 & 2\end{bmatrix}$ which is equivalent to the two equations x'= 4x and y'= 2y which have solutions $x(t)= Ce^{4t}$ and $y(t)= De^{2t}$.

Using the eigenvectors as basis had the nice property of separating the equation into two independent equations but now we should also put the intial value in terms of those: $X(0)= \begin{bmatrix}-6 \\ 1\end{bmatrix}= a\begin{bmatrix}-3 \\ 1\end{bmatrix}+ b\begin{bmatrix}- 1& 1\end{bmatrix}$ which gives the two equations -3a+ b= -6 and a+ b= 1. Subtracting the second equation from the first, -4a= -7 so a= 7/4. Then 7/4+ b= 1 so b= 1- 7/4= -3/4.

That is, $X(t)= \begin{bmatrix}Ce^{4t} \\ De^{2t}\end{bmatrix}$ must satisfy $X(0)= \begin{bmatrix} C \\ D\end{bmatrix}= \begin{bmatrix}7/4 \\ -3/4\end{bmatrix}$ so C= 7/4, D= -3/4, and $X(t)= \begin{bmatrix}\frac{7}{4}e^{4t} \\ -\frac{3}{4}e^{2t}\end{bmatrix}$.

But, once again, that is in terms of the eigenvectors as basis- those numbers are the coefficients of the eigenvectors, not the "standard basis". To put them in terms of the standard basis,
$X(t)= \frac{7}{4}e^{4t}\begin{bmatrix} -3 \\ 1\end{bmatrix}+ \frac{3}{4}e^{2t}\begin{bmatrix}-1 \\ 1\end{bmatrix}$ $= \begin{bmatrix}-\frac{21}{4}e^{4t}- \frac{3}{4}e^{2t}\\\frac{7}{4}e^{4t}+ \frac{3}{4}e^{2t}\end{bmatrix}$.