2.1: Introduction

In this chapter we will begin our study of systems of differential equations. After defining first order systems, we will look at constant coefficient systems and the behavior of solutions for these systems. Also, most of the discussion will focus on planar, or two dimensional, systems. For such systems we will be able to look at a variety of graphical representations of the family of solutions and discuss the qualitative features of systems we can solve in preparation for the study of systems whose solutions cannot be found in an algebraic form.

A general form for first order systems in the plane is given by a system of two equations for unknowns \(x(t)\) and \(y(t)\):

\[\begin{aligned}
&x^{\prime}(t)=P(x, y, t) \\
&y^{\prime}(t)=Q(x, y, t)
\end{aligned} \label{2.1} \]

An autonomous system is one in which there is no explicit time dependence:

\[\begin{aligned}
x^{\prime}(t) &=P(x, y) \\
y^{\prime}(t) &=Q(x, y)
\end{aligned} \label{2.2} \]

Otherwise the system is called nonautonomous.
A linear system takes the form

\[\begin{aligned}
&x^{\prime}=a(t) x+b(t) y+e(t) \\
&y^{\prime}=c(t) x+d(t) y+f(t)
\end{aligned} \label{2.3} \]

A homogeneous linear system results when \(e(t)=0\) and \(f(t)=0\).
A linear, constant coefficient system of first order differential equations is given by

\[\begin{aligned}
&x^{\prime}=a x+b y+e \\
&y^{\prime}=c x+d y+f
\end{aligned} \label{2.4} \]

We will focus on linear, homogeneous systems of constant coefficient first order differential equations:

\[\begin{aligned}
&x^{\prime}=a x+b y \\
&y^{\prime}=c x+d y .
\end{aligned} \label{2.5} \]

As we will see later, such systems can result by a simple translation of the unknown functions. These equations are said to be coupled if either \(b \neq 0\) or \(c \neq 0\)

We begin by noting that the system (2.5) can be rewritten as a second order constant coefficient linear differential equation, which we already know how to solve. We differentiate the first equation in system system (2.5) and systematically replace occurrences of \(y\) and \(y^{\prime}\), since we also know from the first equation that \(y=\dfrac{1}{b}\left(x^{\prime}-a x\right)\). Thus, we have

\[\begin{aligned}
x^{\prime \prime} &=a x^{\prime}+b y^{\prime} \\
&=a x^{\prime}+b(c x+d y) \\
&=a x^{\prime}+b c x+d\left(x^{\prime}-a x\right)
\end{aligned} \label{2.6} \]

Rewriting the last line, we have

\[x^{\prime \prime}-(a+d) x^{\prime}+(a d-b c) x=0 \label{2.7} \]

This is a linear, homogeneous, constant coefficient ordinary differential equation. We know that we can solve this by first looking at the roots of the characteristic equation

\[r^{2}-(a+d) r+a d-b c=0 \label{2.8} \]

and writing down the appropriate general solution for \(x(t)\). Then we can find \(y(t)\) using Equation (2.5):

\[y=\dfrac{1}{b}\left(x^{\prime}-a x\right) \nonumber \]

We now demonstrate this for a specific example.

Example 2.1. Consider the system of differential equations

\[\begin{aligned}
&x^{\prime}=-x+6 y \\
&y^{\prime}=x-2 y
\end{aligned} \label{2.9} \]

Carrying out the above outlined steps, we have that \(x^{\prime \prime}+3 x^{\prime}-4 x=0\). This can be shown as follows:

\[\begin{aligned}
x^{\prime \prime} &=-x^{\prime}+6 y^{\prime} \\
&=-x^{\prime}+6(x-2 y) \\
&=-x^{\prime}+6 x-12\left(\dfrac{x^{\prime}+x}{6}\right) \\
&=-3 x^{\prime}+4 x
\end{aligned} \label{2.10} \]

The resulting differential equation has a characteristic equation of \(r^{2}+3 r-4=0\). The roots of this equation are \(r=1,-4\). Therefore, \(x(t)=c_{1} e^{t}+c_{2} e^{-4 t} .\) But, we still need \(y(t)\). From the first equation of the system we have

\[y(t)=\dfrac{1}{6}\left(x^{\prime}+x\right)=\dfrac{1}{6}\left(2 c_{1} e^{t}-3 c_{2} e^{-4 t}\right) \nonumber \]

Thus, the solution to our system is

\[\begin{aligned}
&x(t)=c_{1} e^{t}+c_{2} e^{-4 t} \\
&y(t)=\dfrac{1}{3} c_{1} e^{t}-\dfrac{1}{2} c_{2} e^{-4 t}
\end{aligned} \label{2.11} \]

Sometimes one needs initial conditions. For these systems we would specify conditions like \(x(0)=x_{0}\) and \(y(0)=y_{0}\). These would allow the determination of the arbitrary constants as before.

Example 2.2. Solve

\[\begin{aligned}
&x^{\prime}=-x+6 y \\
&y^{\prime}=x-2 y
\end{aligned} \label{2.12} \]

given \(x(0)=2, y(0)=0\).

We already have the general solution of this system in (2.11). Inserting the initial conditions, we have

\[\begin{aligned}
&2=c_{1}+c_{2} \\
&0=\dfrac{1}{3} c_{1}-\dfrac{1}{2} c_{2} .
\end{aligned} \label{2.13} \]

Solving for \(c_{1}\) and \(c_{2}\) gives \(c_{1}=6 / 5\) and \(c_{2}=4 / 5\). Therefore, the solution of the initial value problem is

\[\begin{aligned}
&x(t)=\dfrac{2}{5}\left(3 e^{t}+2 e^{-4 t}\right) \\
&y(t)=\dfrac{2}{5}\left(e^{t}-e^{-4 t}\right)
\end{aligned} \label{2.14} \]