Results 1 to 2 of 2

Thread: Application to Differential Equations

  1. #1
    Junior Member
    Feb 2010

    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

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Apr 2005
    First, since web browsers don't respect tabs, lets clean that up using LaTex:
    A has eigenvalues 4 with corresponding eigenvector $\displaystyle \begin{bmatrix}-3 \\ 1\end{bmatrix}$ and 2 with corresponding eigenvector $\displaystyle \begin{bmatrix}-1 \\ 1\end{bmatrix}$. That means that, if we used those two eigenvectors as basis vectors, we can write the equation as $\displaystyle X'= \begin{bmatrix}x' \\ y'\end{bmatrix}= \begin{bmatrix}4 & 0 \\ 0 & 2\end{bmatrix}$ which is equivalent to the two equations x'= 4x and y'= 2y which have solutions $\displaystyle x(t)= Ce^{4t}$ and $\displaystyle y(t)= De^{2t}$.

    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: $\displaystyle X(0)= \begin{bmatrix}-6 \\ 1\end{bmatrix}= a\begin{bmatrix}-3 \\ 1\end{bmatrix}+ b\begin{bmatrix}- 1& 1\end{bmatrix}$ 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, $\displaystyle X(t)= \begin{bmatrix}Ce^{4t} \\ De^{2t}\end{bmatrix}$ must satisfy $\displaystyle X(0)= \begin{bmatrix} C \\ D\end{bmatrix}= \begin{bmatrix}7/4 \\ -3/4\end{bmatrix}$ so C= 7/4, D= -3/4, and $\displaystyle X(t)= \begin{bmatrix}\frac{7}{4}e^{4t} \\ -\frac{3}{4}e^{2t}\end{bmatrix}$.

    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,
    $\displaystyle X(t)= \frac{7}{4}e^{4t}\begin{bmatrix} -3 \\ 1\end{bmatrix}+ \frac{3}{4}e^{2t}\begin{bmatrix}-1 \\ 1\end{bmatrix}$$\displaystyle = \begin{bmatrix}-\frac{21}{4}e^{4t}- \frac{3}{4}e^{2t}\\\frac{7}{4}e^{4t}+ \frac{3}{4}e^{2t}\end{bmatrix}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differential Equations Application Problems
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: Sep 11th 2010, 03:57 AM
  2. Application of Differential Equations
    Posted in the Differential Equations Forum
    Replies: 13
    Last Post: Jul 18th 2010, 03:16 AM
  3. Application of 1st order differential equations 2
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 19th 2009, 03:52 AM
  4. Replies: 2
    Last Post: May 18th 2009, 03:49 AM
  5. Application of Differential Equations.
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Nov 17th 2008, 09:54 PM

Search Tags

/mathhelpforum @mathhelpforum