Section 5.4. Solve a system of homogeneous differential equations using the coefficient matrix

Objective:

1. Solve a system of homogeneous differential equations using the coefficient matrix with distinct real eigenvalue

2. Understand connection between the phase plan and the solution

We are using what we learn from the matrix theory to solve a system of differential equation. Recall that $y^{\prime }=ay$ is a first order differential equation. If $a$ is not zero then $y=e^{at}$ is a solution of the differential equation. The only equilibrium solution of $y^{\prime }=ay=0$ is when $y=0$. When $a>0$, then $y=0$ is an unstable solution, and when $a<0$, then $y=0$ is a stable solution. We are using the same principal for a system of equation $$\begin{array}{rl}{x}_1^{\prime }& =a_{11}{x}_1+a_{12}{x}_2\\ x_2^{\prime }& =a_{21}{x}_1+a_{22}{x}_2\end{array}$$  which could be rewrite as  

$$\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}.$$

We assume $A$ is nonsingular, i.e. $\text{det}A\ne 0$, hence the only equilibrium solution is $\overrightarrow{x}=\overrightarrow{0}$. We now have to determining if $\overrightarrow{x}=0$ is a stable solution or unstable solution. We can visualize the solution trajectories using the direction field, called phase plane. How do we find the solution like $y=e^{at}?$ The answer is eigenvalues and the eigenvectors. 

Example 1: Given $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}5& -1\\ 3& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}$, find the general solution of the system of equations. Sketch its phase plane.

Exercise 1: Given $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}4& 1\\ -2& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}$, find the general solution of the system of equations. Sketch its phase plane.

Example 2: Given $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}3& -1\\ 5& -3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}$, find the general solution of the system of equations. Sketch its phase plane.

Exercise 2: Given $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}-2& 1\\ -5& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}$, find the general solution of the system of equations. Sketch its phase plane.

Example 3: Given $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}-2& 5\\ 0& -1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}$, find the general solution of the system of equations. Sketch its phase plane.

Exercise 3: Given $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}-3& 2\\ 0& -4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}$, find the general solution of the system of equations. Sketch its phase plane.

Example 4: Given $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}2& 5\\ 1& -2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}$, find the general solution of the system of equations. Sketch its phase plane.

Exercise 4: Given $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 4& 5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=A\overrightarrow{x}$, find the general solution of the system of equations. Sketch its phase plane.

Group Work: 

1. Solve the initial value problem and describe the behavior of the solution as $t$ approaches infinity.

(a) $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}2& 5\\ 1& -2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right],$ $\left[\begin{array}{c}{x}_1(0)\\ x_2(0)\end{array}\right]=\left[\begin{array}{c}2\\ -1\end{array}\right]$. 

(b) $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}5& -1\\ 3& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right],$ $\left[\begin{array}{c}{x}_1(0)\\ x_2(0)\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]$. 

(c) $\left[\begin{array}{c}{x}_1^{\prime }\\ x_2^{\prime }\end{array}\right]=\left[\begin{array}{cc}-2& 0\\ 5& -3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right],$ $\left[\begin{array}{c}{x}_1(0)\\ x_2(0)\end{array}\right]=\left[\begin{array}{c}0\\ -1\end{array}\right]$.