From a 2-dimensional system of homogenous first-order linear ODEs to a homogeneous second-order linear ODE We consider the following system of coupled homogeneous first-order linear differential equations:

$$\left\{\begin{aligned} \frac{\dd x}{\dd t} &= a\, x+ b\, y\\ \frac{\dd y}{\dd t} &= c\, x + d\, y\end{aligned}\right.$$ We can rewrite the system to a homogeneous second-order linear differential equation with constant coefficients for $x(t)$ and a similar differential equation for $y(t)$. This means that we can solve the system!

Finding the corresponding ODE Let's see how this works $x(t)$. When we differentiate the left- and right-hand side of the first equation of the system, we get

$$\frac{\dd^2x}{\dd t^2}=a\frac{\dd x}{\dd t}+b\frac{\dd y}{\dd t}$$ and after substitution of the second equation of the system $$\frac{\dd^2x}{\dd t^2}=a\frac{\dd x}{\dd t}+b\,c\,x + b\,d\,y$$ The first equation of the system can be written as $b\,y =\frac{\dd x}{\dd t}-a\, x$ and substitution leads then to the desired second-order ODE, which can be written as $$\frac{\dd^2x}{\dd t^2}-(a+d)\frac{\dd x}{\dd t}+(ad-bc)x =0$$

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We already know how we can solve the found two-order ODE and that the behaviour depends on the discriminant of the corresponding characteristic polynomial

$$\lambda^2 -(a+d)\lambda+(ad-bc)$$ This discriminant $D$ can be rewritten as: $$D=(a+d)^2-4(ad-bc)=(a-d)^2+4bc$$ We distinguish three cases:

1. If $(a-d)^2+4bc>0$ then there are two real root of the characteristic polynomial: $$\lambda_{1,2}=\frac{a+d\pm\sqrt{(a-d)^2+4bc}}{2}$$ and the general solution for $x(t)$ is $$x(t)=c_1e^{\lambda_1t}+c_1e^{\lambda_2t}$$
2. If $(a-d)^2+4bc=0$ then there exist one real root of the characteristic polynomial: $$\lambda=\frac{a+d}{2}$$ and the general solution for $x(t)$ is $$x(t)=(c_1+c_2 t)e^{\lambda t}$$
3. If $(a-d)^2+4bc<0$ then there are two imaginary roots of the characteristic polynomial: $$\lambda_{1,2}=\alpha\pm\mathrm{i}\,\beta$$ with $$\alpha=\frac{a+d}{2}\quad\text{and}\quad\beta=\frac{\sqrt{-(a-d)^2-4bc}}{2}$$ and the general solution for $x(t)$ is $$x(t)=e^{\alpha t}\bigl(c_1\cos(\beta t)+c_2\sin(\beta t)\bigr)$$

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For $y$ you can do similar work.

Solution curves for a two-dimensional system of homogeneous first-order linear ODEs To get more insight into the behaviour of the solutions of the system of coupled homogeneous first-order linear differential equations one uses the so-called phase portrait. One plots parameter curves $t\mapsto (x(t), y(t))$ for several initial conditions of the differential equation (and sometimes with the so-called direction field as background, which we have omitted here): the solutions $x(t)$ and $y(t)$ serve here as coordinate functions of a curve in the plane. The solution curves give a good impression of the behaviour of the two solutions together. Without proof, we show some pictures of various types of phase portraits that can occur (note that these examples are not exhaustive and that we do not show in which direction the curves go in time).

• negative and positive real roots of the characteristic polynomial:
• one real root of the characteristic polynomial:
• two complex roots of the characteristic polynomial: