How can I solve this set of linear coupled system? Consider the matrix $A=\begin{bmatrix} -2k & k \\ k & -2k \end{bmatrix}$
.I have to solve this linear coupled system :
$X'' = A.X$, where $X= \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}$
.Can any one help me or give me links related to this?
 A: I will give you some thoughts to help you to solve the problem:


*

*Since neither intial conditions nor boundary data are specified, the solution of the problem will depend on four constants of integration since the system of differential equations is second order in $t$.

*Assuming $k$ is a constant, the ODEs are linear and with constant coefficients. This leads you many ways to solve this. For example, Laplace transform in $t$, exponential methods, algebraic manipulation, etc.

*Suppose you are interested in solving this problem using exponential methods. You may reduce first the order of the system in order to come up with an expression such that $Y' = M Y$, where $M$ is some matrix. How can you do this? Define $u_1 = x_1, u_2 = x_1'$ and $v_1 = x_2, v_2 = x_2'$ so we have:
\begin{align}
u_2' &= -2k  u_1 +  k  u_2  \\
v_2' &= k u_2 - 2 k u_2  \\
u_1' &= u_2  \\
v_1' &= v_2 
\end{align} which you can re-arrange in matrix form, with $Y = (u_1,u_2,v_1,v_2)$. When you find $M$, then the solution of the problem will be given by:
$$Y(t) = e^{Mt} C,$$ where $C$ is an arbitrary vector.

*What about Laplace transform?. Take Laplace transform on both sides of your system to have $$s X(s) - s X_0 + X'_0 =  A \, X(s)$$ where $X(s) = \mathcal{L}_sX(t)$ and $X_0 = X(0), X'_0 = X'(0)$ (which are unknown constants). The equation above is indeed an algebraic system of equation for the vector $X(s)$, which you can easily solve (if $A-sI$ is invertible, i.e., $s$ is not an eigenvalue of $A$). Take inverse Laplace transform then to find $X(t)$ as a function of the two constants $X_0$ and $X'_0$.
Hope this helps!
