Results 1 to 4 of 4

Math Help - Systems of differential equations...

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    34

    Systems of differential equations...

    Determine the General solution to the differential Equation system X'=AX
    A=
    0 -4
    4 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,324
    Thanks
    8
    Quote Originally Posted by Raidan View Post
    Determine the General solution to the differential Equation system X'=AX
    A=
    0 -4
    4 0
    For this problem x' = -4y, y' = 4x, it's actually easier to eliminate a variable, say y, giving

     <br />
x'' + 16 x = 0<br />

    giving the solution x = c_1 \sin 4 x + c_2 \cos 4x then use y = - \frac{x'}{4} to find y.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    For a simple problem like this, danny arrigo's suggestion is best, but you can do as a matrix problem.

    Find the eigenvalues of A= \begin{bmatrix}0 & -4 \\ 4 & 0\end{bmatrix}. It should be easy to see that they are 4i and -4i. Since those are distinct, the matrix can be diagonalize (over the complex numbers).

    With eigenvalue 4i, we have \begin{bmatrix}0 & -4 \\ 4 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} 4ix & 4iy\end{bmatrix} which gives the two equations -4y= 4ix and 4x= 4iy. Those both reduce to y=ix so one eigenvector is <1, i>.

    With eigenvalue -4i, we have \begin{bmatrix}0 & -4 \\ 4 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} -4ix & -4iy\end{bmatrix} which gives the two equations -4y= -4ix and 4x= -4iy. Those both reduce to y=-ix so another, independent, eigenvector is <1, i>.

    If we use those as columns in matrix P, then P= \begin{bmatrix}1 & 1 \\ i & -i\end{bmatrix} and P^{-1}= \begin{bmatrix}\frac{1}{2} & -\frac{1}{2}i \\ \frac{1}{2} & \frac{1}{2}i\end{bmatrix}

    Then it is easy to see that P^{-1}AP= \begin{bmatrix}\frac{1}{2} & -\frac{1}{2}i \\ \frac{1}{2} & \frac{1}{2}i\end{bmatrix}\begin{bmatrix}0 & -4 \\ 4 & 0\end{bmatrix}\begin{bmatrix}1 & 1 \\ i & -i\end{bmatrix}= \begin{bmatrix}4i & 0 \\0 & -4i\end{bmatrix}.

    If we multiply, on the left, by the constant matrix P^{-1}, we have P^{-1}X'= (P^{-1}X)'= P^{-1}AX.

    Let Y= P^{-1}X so that X= PY and substitue PY for X in the equation: Y'= P^{-1}A(PY)= (P^{-1}AP) Y so our differential equation becomes Y'= \begin{bmatrix}2i & 0 \\ 0 & -2i\end{bmatrix}Y.

    The general solution to that, just like any y'= ay, is Y= Ce^{\begin{bmatrix}2it & 0 \\0 & -2it\end{bmatrix}} and because that is a diagonal matrix it is easy to show that we get Y= C\begin{bmatrix}e^{2it} & 0 \\ 0 & e^{-2it}\end{bmatrix}

    = \begin{bmatrix}C_1 & C_2\end{bmatrix}\begin{bmatrix}e^{2it} & 0 \\ 0 & e^{-2it}\end{bmatrix}

    = \begin{bmatrix}C_1e^{2it}\\ C_2e^{-2it}\end{bmatrix}
    Now we use X= P^{-1}Y to write X(t)= \begin{bmatrix}\frac{1}{2} & -\frac{1}{2}i \\ \frac{1}{2} & \frac{1}{2}i\end{bmatrix}\begin{bmatrix}C_1e^{2it} \\ C_2e^{-2it}\end{bmatrix} = \begin{bmatrix}\frac{C_1}{2}e^{2it}- \frac{C_2}{2}e^{-2it} \\ \frac{C_1}{2}e^{2it}+ \frac{C2}{2}e^{-2it}\end{bmatrix}.

    And, then use the fact that e^{2it}= cos(2t)+ i sin(2t), e^{-2it}= cos(t2)- i sin(2t) to change into sine and cosine, getting danny aririgo's answer.

    Whew! Well, I said that, for a simple problem like this danny arrigo's solution was best!
    Last edited by HallsofIvy; May 10th 2009 at 06:52 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2009
    Posts
    34
    Ahh.. thanx guys..I got it now..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Systems of linear Differential Equations
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 24th 2011, 02:05 AM
  2. Systems of differential equations and equilibrium points...
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: May 30th 2011, 02:15 PM
  3. Solving Systems of Vector Differential Equations
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: August 8th 2009, 05:02 PM
  4. Linear systems of differential equations
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 29th 2007, 11:45 AM
  5. Replies: 0
    Last Post: April 14th 2007, 03:41 PM

Search Tags


/mathhelpforum @mathhelpforum