I desperately need to solve a coupled system of linear second order differential equations of the form: $$x''+ax'+bx-cy'-dy=0$$ $$y''+ay'+by+cx'+dx=0$$ where both "x" and "y" are functions of time and {a,b,c,d} are constants, I need a method other than that of Laplace transform, any suggestions?

Edit: Is it possible to solve this system by means of "matrix exponential"? Is there any textbook on differential equations that covers solving this sort of systems?


The most general approach to these problems is to write your system as a four dimensional first order ODE system: \begin{align} x' &= \xi\\ \xi' &= - b x - a \xi +d y + c \eta\\ y ' &= \eta\\ \eta' &= -d x -c \xi -b y -a \eta \end{align} which can be written in matrix form \begin{equation} \mathbf{x}' = A\, \mathbf{x} \end{equation} with \begin{equation} \mathbf{x} = \begin{pmatrix} x \\ \xi \\ y \\ \eta \end{pmatrix}\qquad \text{and} \qquad A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ -b & -a & d & c \\ 0 & 0 & 0 & 1 \\ -d & -c & -b & -a \end{pmatrix}. \end{equation} Surprisingly, the eigenvalues of this matrix are not particularly hard to find, as are the eigenvectors. The general solution to the above system is then of the form \begin{equation} \mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{x}_1 + c_2 e^{\lambda_2 t}\mathbf{x}_2 + c_3 e^{\lambda_3 t}\mathbf{x}_3 + c_4 e^{\lambda_4 t}\mathbf{x}_4, \end{equation} where $\lambda_{1,2,3,4}$ are the eigenvalues of the matrix $A$ and $\mathbf{x}_{1,2,3,4}$ are the associated eigenvectors. The constants $c_{1,2,3,4}$ are determined by the initial conditions.

As for your question for a literature reference: I can't imagine a textbook on differential equations not treating the above approach, so pick your favourite.

  • $\begingroup$ Many thanks man, I think this is exactly what I needed to know. $\endgroup$ Nov 26 '15 at 15:09
  • 1
    $\begingroup$ @FritsVeerman Aaand you also could suggest a matrix exponent as a solution to $x' = Ax$ :) $\endgroup$
    – Evgeny
    Nov 27 '15 at 15:07
  • $\begingroup$ @Evgeny: Definitely. Whether that would be particularly helpful in this case, has to be seen :-) $\endgroup$ Nov 27 '15 at 15:41

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