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
    2,367
    Thanks
    42
    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
    15,697
    Thanks
    1469
    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