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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?

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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.

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  • $\begingroup$ Many thanks man, I think this is exactly what I needed to know. $\endgroup$ Nov 26 '15 at 15:09
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    $\begingroup$ @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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