Results 1 to 6 of 6

Math Help - General Solution of x'(t)=Ax(t)

  1. #1
    Member diddledabble's Avatar
    Joined
    Jul 2009
    Posts
    80

    Post General Solution of x'(t)=Ax(t)

    General Solution of x'(t)=Ax(t)

    <br />
A=\begin{array}{cc}1&3\\12&1\end{array}<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    You need to solve these DEs

    x_1'(t) = x_1(t)+3x_2(t)

     x_2'(t) = 12x_1(t)+x_2(t)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Treat A as a constant (which it is).

    We know that the general solution of this type of equation is x(t)=e^{At}

    And when A is a matrix, e^{At}=\sum_{n\geq 0} \frac{A^n t^n}{n!}

    See here : http://www.mathhelpforum.com/math-he...al-matrix.html for an example of computing this exponential.

    If you don't understand, I can explain.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by diddledabble View Post
    General Solution of x'(t)=Ax(t)

    <br />
A=\begin{array}{cc}1&3\\12&1\end{array}<br />
    You can also convert this into a pair of uncoupled second order homogeneous constant coefficient ODE's.

    To do this write them out in full as two first order ODE's differnetiate one of them again and substitute the derivative from the other to eliminate one of the variables ...

    CB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member diddledabble's Avatar
    Joined
    Jul 2009
    Posts
    80

    Still don't get it

    I still am not sure what to do with the matrix.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member Haven's Avatar
    Joined
    Jul 2009
    Posts
    197
    Thanks
    8

    EigenValue Method

    Okay here's an easy way to solve it:

     x'(t) = Ax(t)

    We assume the form of the solution that  x(t) = v(t)*e^{rt}

    It turns out the r, corresponds to the eigenvalue of the matrix A and  v(t) are the eigenvectors

    A=\left[\begin{array}{cc}1&3\\12&1\end{array}\right]

    det(A - rI) = \left|\begin{array}{cc}1-r&3\\12&1-r\end{array}\right| = r^2 -2r -35 = (r+5)(r-7) = 0 for an aribitrary constant  c_1

    so the roots of this polynomial are:  r = -5, 7

    Now we plug our values of r back in
     A - (-5)I = \left[\begin{array}{cc}6&3\\12&6\end{array}\right]\\\rightarrow\\\left[\begin{array}{cc}2&1\\0&0\end{array}\right]

    So we see that -2v_1 = v_2 , which means our eigenvector is v_1(t) = \left[\begin{array}{cc}1\\-2\end{array}\right] and our first solution is x_1(t) = \left[\begin{array}{cc}1\\-2\end{array}\right]\\e^{-5t}

    Similarly for  r =7
    we compute  A - (7)I = \left[\begin{array}{cc}-6&3\\12&-6\end{array}\right]\\\rightarrow\\\left[\begin{array}{cc}-2&1\\0&0\end{array}\right]

    eigenvector: v_1(t) = \left[\begin{array}{cc}1\\2\end{array}\right]

    second solution x_2(t) = \left[\begin{array}{cc}1\\2\end{array}\right]\\e^{7t}

    Our final solution is a linear combination to the two solutions x(t) = {c_1}x_1(t) + {c_2}x_2(t) = c_1\left[\begin{array}{cc}1\\-2\end{array}\right]\\e^{-5t}\\+\\c_2\left[\begin{array}{cc}1\\2\end{array}\right]\\e^{7t}

    Where  c_1, c_2 are arbitrary constants
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. General Solution of a differential solution
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 11th 2010, 02:49 AM
  2. GENERAL SOLUTION f(x,t) please help
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: April 11th 2010, 04:36 PM
  3. Help w/ General Solution
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: February 19th 2010, 12:58 AM
  4. Finding the general solution from a given particular solution.
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: October 7th 2009, 01:44 AM
  5. find the general solution when 1 solution is given
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: March 4th 2009, 09:09 PM

Search Tags


/mathhelpforum @mathhelpforum