Differential equation corresponding to a linear system of differential equation. Consider  linear system of differential equations $$\frac{dx}{dt}=ax+by$$ $$\frac{dy}{dt}=cx+dy$$ my question is how to find the second order linear differential equation corresponding to above system? I have no idea to obtain corresponding ODE. Please help me. Thanks in advance.
 A: Assuming that the system is $$x'=ax+by\qquad y'=cx+dy$$ Assuming $b\neq 0$, extract $y$ from the first equation $$y=\frac{x'-ax}{b}$$ which makes (assuming that $a$ and $b$ are constants)$$y'=\frac{x''-ax'}{b}$$ So the second equation becomes $$\frac{x''-ax'}{b}=cx+d\frac{x'-ax}{b}$$ Simplify.
Edit
If you consider that $a$ and $b$ depend on $t$, the approach is the same except that the expression of $y'$ will be slightly more complex.
A: You can also do this with the companion matrix of the characteristic polynomial of the matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
Let $c(x) = x^2 + a_1x + a_0$ be the (unique monic generator of the ideal in $\mathbb{R}[x]$ generated by the) characteristic polynomial of $A$, $c(x) = \det(xI - A) = x^2 -(a+b)x +(ad-bc)$. Then the system 
$$u' = v \hspace{1cm} v' = -a_0u - a_1v$$
will have the same solution $x(t) = u(t)$ as the previous system (although $y(t)$ will in general be different from $v(t)$). Moreover,
$$\ddot{u} = -a_0u - a_1\dot{u}$$
or 
$$\ddot{u} + a_1\dot{u} + a_0u = 0$$
$$\ddot{u} + (-a-d)\dot{u} + (ad-bc)u = 0$$
is the second order linear differential equation corresponding to above system, with solution $x(t) = u(t)$.
[To find $y(t)$, if $b \ne 0$, use the method suggested in the comment by Claude Leibovici, while if $b = 0$, $x(t) = x_0e^{at}$ and $y(t) = (y_0-\frac{cx_0}{a})e^{dt} +\frac{cx_0}{a}e^{at}$. (Or you could just solve the system $\ddot{w} + (-b-c)\dot{w} + (bc-ad)w = 0$ :) .)]
