A first-order linear system of $n$ equations and $n$ variables is any system that can be written in the form \begin{align*} \frac{dx_1}{dt} & = a_{11}(t) x_1(t) + \cdots + a_{1n}(t) x_n(t) + f_1(t),\\ \frac{dx_2}{dt} & = a_{21}(t) x_1(t) + \cdots + a_{2n}(t) x_n(t) + f_2(t),\\ & \vdots\\ \frac{dx_n}{dt} & = a_{n1}(t) x_1(t) + \cdots + a_{nn}(t) x_n(t) + f_n(t). \end{align*} If each of the coefficients is constant and the functions $f_i$ vanish, then we have a homogeneous first-order linear system with constant coefficients, \begin{align*} \frac{dx_1}{dt} & = a_{11} x_1 + \cdots + a_{1n} x_n\\ \frac{dx_2}{dt} & = a_{21} x_1 + \cdots + a_{2n} x_n,\\ & \vdots\\ \frac{dx_n}{dt} & = a_{n1} x_1 + \cdots + a_{nn} x_n. \end{align*}

We will concentrate on $2 \times 2$ homogeneous first-order linear systems or planar systems for the time being, \begin{align} \frac{dx}{dt} & = a x + b y,\label{equation-linear02-2x2-1}\tag{3.3}\\ \frac{dy}{dt} & = c x + d y.\label{equation-linear02-2x2-2}\tag{3.4} \end{align}

# Subsection3.2.1Planar Systems and $2 \times 2$ Matrices¶ permalink

We will use linear systems of differential equations to illustrate how we can use systems of differential equations to model how subtances flow back and forth between two or more compartments. Suppose that we have two tanks ($A$ and $B$) between which a mixture of brine flows (Figure 3.7). Tank $A$ contains 300 liters of water in which 100 kilograms of salt has been dissolved and Tank $B$ contains 300 liters of pure water. Fresh water is pumped into Tank $A$ at the rate of 500 liters per hour, and brine is pumped into Tank $B$ from Tank $A$ at the rate of 900 liters per hour. Brine is also pumped back into Tank $A$ from Tank $B$ at the rate of 400 liters per hour, and an additional 500 liters of brine per hour is drained from Tank $B\text{.}$ All brine mixtures are well-stirred. If we let $x = x(t)$ be the amount of salt in Tank $A$ at time $t$ and $y = y(t)$ be the amount of salt in Tank $B$ at time $t\text{,}$ then we know that \begin{align*} x(0) & = 100\\ y(0) & = 0 \end{align*} We know that the salt concentrations in the two tanks are $x/300$ kilograms per liter and $y/300$ kilograms per liter. Thus, we can describe the rate of change in each tank with a differential equation, \begin{align*} \frac{dx}{dt} & = - 900 \cdot \frac{x}{300} + 400 \cdot \frac{y}{300} = - 3 x + \frac{4}{3} y,\\ \frac{dy}{dt} & = 900 \cdot \frac{x}{300} - 400 \cdot \frac{y}{300} - 500 \cdot \frac{y}{300} = 3x - 3 y. \end{align*}

Matrix notation gives us a convenient way of representing the $2 \times 2$ system (3.3)(3.3). If we let \begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \quad\text{and}\quad {\mathbf x}(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}, \end{equation*} then we can rewrite our system as \begin{equation*} \begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = \begin{pmatrix} ax(t) + b y(t) \\ cx(t) + d y(t) \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}. \end{equation*} In other words, we can write our system as \begin{equation*} \frac{d \mathbf x}{dt} = A {\mathbf x}, \end{equation*} where \begin{equation*} \mathbf x' = \frac{d \mathbf x}{dt} = \begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix}. \end{equation*}

# Subsection3.2.2Systems of Differential Equations¶ permalink

A linear planar system \begin{align*} x' & = ax + by\\ y' & = cx + dy \end{align*} has an equilibrium solution at $(x_0, y_0)$ if \begin{align*} a x_0 + b y_0 & = 0,\\ c x_0 + d y_0 & = 0. \end{align*} The following proposition tells us exactly where to find the equilibrium solutions of a linear system with constant coefficients.

Now let us attack the problem of finding all of the solutions of the system ${\mathbf x}' = A {\mathbf x}\text{.}$ Suppose that we can find a nonzero vector ${\mathbf v}_0$ such that $A {\mathbf v}_0 = \lambda {\mathbf v}_0$ for some real number $\lambda\text{.}$ In this case, the matrix $A$ just sends the vector ${\mathbf v}_0$ to a vector on the same line through the origin, $\lambda {\mathbf v}_0\text{.}$ This is a very special case of course; however, we claim that \begin{equation*} {\mathbf x}(t) = e^{\lambda t} {\mathbf v}_0 \end{equation*} is a solution for our linear system if we can find such a vector. To see that this is indeed the case, we compute \begin{align*} {\mathbf x}'(t) & = \lambda e^{\lambda t} {\mathbf v}_0\\ & = e^{\lambda t} (\lambda {\mathbf v}_0)\\ & = e^{\lambda t} (A {\mathbf v}_0 )\\ & = A( e^{\lambda t} {\mathbf v}_0)\\ & = A {\mathbf x}(t). \end{align*} In other words, the key to solving a linear system ${\mathbf x}' = A {\mathbf x}$ is to be able to find eigenvalues and eigenvectors for the matrix $A\text{.}$ We are now ready to state the results of our discussion in a theorem.

We say that the solution ${\mathbf x}(t) = e^{\lambda t}{\mathbf v}_0$ is a straight-line solution. The vector $e^{\lambda t}{\mathbf v}_0$ lies on the same line for each value of $t\text{.}$ Note that if ${\mathbf v}_0$ is an eigenvector for $A\text{,}$ then any nonzero multiple of ${\mathbf v}_0$ is also an eigenvector for $A\text{,}$ \begin{equation*} A(\alpha {\mathbf v}_0) = \alpha A{\mathbf v}_0 = \alpha (\lambda {\mathbf v}_0) = \lambda (\alpha {\mathbf v}_0). \end{equation*}

##### Example3.10

Consider the system \begin{align*} x' & = x + 3y\\ y' & = x -y. \end{align*} We can rewrite this system in matrix form as ${\mathbf x}' = A {\mathbf x}\text{,}$ where \begin{equation*} A = \begin{pmatrix} 1 & 3 \\ 1 & -1 \end{pmatrix}. \end{equation*} The matrix $A$ has an eigenvector ${\mathbf u} = (3, 1)$ with associated eigenvalue $\lambda = 2\text{,}$ since \begin{equation*} A \mathbf u = \begin{pmatrix} 1 & 3 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix} = 2 \begin{pmatrix} 3\\1 \end{pmatrix} = \lambda \mathbf u. \end{equation*} Similarly, ${\mathbf v} = (1, -1)$ is an eigenvalue for $A$ with associated eigenvector $\mu = -2\text{.}$ Thus, we have two solutions for our system: the equilibrium solution at the origin, the solution \begin{equation*} {\mathbf x}_1(t) = e^{2t} \begin{pmatrix} 3 \\ 1 \end{pmatrix}, \end{equation*} and the solution \begin{equation*} {\mathbf x}_2(t) = e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}. \end{equation*}

Since \begin{align*} \frac{d}{dt} (c_1 {\mathbf x}_1(t) + c_2 {\mathbf x}_2(t)) & = c_1\frac{d}{dt} {\mathbf x}_1(t) + c_2 \frac{d}{dt} {\mathbf x}_2(t)\\ & = c_1 A {\mathbf x}_1(t) + c_2 A {\mathbf x}_2(t)\\ & = A( c_1 {\mathbf x}_1(t) + c_2 {\mathbf x}_2(t)), \end{align*} any linear combination of solutions to a linear system is also a solution. Thus, a general solution to our system is \begin{equation*} \mathbf x(t) = c_1 e^{2t} \begin{pmatrix} 3 \\ 1 \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \end{equation*} or \begin{align*} x(t) & = 3 c_1 e^{2t} + c_2 e^{-2t}\\ y(t) & = c_1 e^{2t} - c_2 e^{-2t}. \end{align*} If we are given initial conditions, say $x(0) = 0$ and $y(0) = 1\text{,}$ then we can determine $c_1$ and $c_2$ by solving the linear system of equations \begin{align*} 3 c_1 + c_2 & = 0\\ c_1 - c_2 & = 1 \end{align*} to get $c_1 = 1/4$ and $c_2 = -3/4\text{.}$ Thus, the solution to our initial value problem is \begin{align*} x(t) & = \frac{3}{4} e^{2t} - \frac{3}{4} e^{-2t}\\ y(t) & = \frac{1}{4} e^{2t} + \frac{3}{4} e^{-2t}. \end{align*}

If ${\mathbf x}_1(t)$ and ${\mathbf x}_2(t)$ are solutions to the linear system ${\mathbf x}' = A {\mathbf x}\text{,}$ then \begin{align*} {\mathbf x}_1' & = A {\mathbf x}_1\\ {\mathbf x}_2' & = A {\mathbf x}_2. \end{align*} Thus, for any two real numbers $c_1$ and $c_2\text{,}$ \begin{align*} \frac{d}{dt} (c_1 {\mathbf x}_1(t) + c_2 {\mathbf x}_2(t)) & = \alpha \frac{d}{dt} {\mathbf x}_1(t) + c_2 \frac{d}{dt} {\mathbf x}_2(t)\\ & = c_1 A {\mathbf x}_1(t) + c_2 A {\mathbf x}_2(t)\\ & = A (c_1 {\mathbf x}_1(t) + c_2 {\mathbf x}_2(t) ). \end{align*} We state this result in the following theorem.

Revisiting the mixing problem that we posed at the beginning of this section, we have the following initial value problem, \begin{align*} \frac{dx}{dt} & = - 3 x + \frac{4}{3} y,\\ \frac{dy}{dt} & = 3 x - 3 y,\\ x(0) & = 100,\\ y(0) & = 0. \end{align*} If we write our system in matrix form, ${\mathbf x}' = A {\mathbf x}\text{,}$ then \begin{equation*} A = \begin{pmatrix} -3 & 4/3 \\ 3 & -3 \end{pmatrix}. \end{equation*} It is easy to check that we have eigenvectors ${\mathbf u} = (2, 3)$ and ${\mathbf v} = (-2, 3)$ with eigenvectors $\lambda = -1$ and $\mu = -5\text{,}$ respectively. Thus, we have two solutions to our system, \begin{align*} {\mathbf x}_1(t) & = e^{-t} {\mathbf u},\\ {\mathbf x}_2(t) & = e^{-5t} {\mathbf v}. \end{align*} Since any linear combination of solutions is also a solution, \begin{equation*} {\mathbf x}(t) = c_1 \begin{pmatrix} 2e^{-t} \\ 3e^{-t} \end{pmatrix} + c_2 \begin{pmatrix} -2e^{-5t} \\ 3e^{-5t} \end{pmatrix} \end{equation*} is a solution to our system. Using the initial values $x(0) = 100$ and $y(0) = 0\text{,}$ we can determine that $c_1 = 25$ and $c_2 = -25\text{.}$ We now have the solution that we seek, \begin{align*} x(t) & = 50 e^{-t} + 50 e^{-5t}\\ y(t) & = 75 e^{-t} - 75 e^{-5t}. \end{align*}

Our goal is to prove the following theorem.

We can now proceed to the proof of the theorem. Suppose that we have a linear system ${\mathbf x}' = A{\mathbf x}$ such that $A$ has a pair of distinct real eigenvalues, $\lambda_1$ and $\lambda_2\text{,}$ with associated eigenvectors ${\mathbf v}_1$ and ${\mathbf v}_2\text{.}$ By the Principle of Superposition, we know that \begin{equation*} {\mathbf x}(t) = c_1 e^{\lambda_1 t} {\mathbf v}_1 + c_2 e^{\lambda_2 t} {\mathbf v}_2. \end{equation*} is a solution to the linear system ${\mathbf x}' = A {\mathbf x}\text{.}$ To show that this is the general solution, we must show that we can choose $c_1$ and $c_2$ to satisfy a given initial condition ${\mathbf x}_0 = {\mathbf x}(0) = (x_0, y_0)\text{.}$ By Lemma 3.13, we know that ${\mathbf v}_1$ and ${\mathbf v}_2$ form a basis for ${\mathbb R}^2\text{.}$ That is, we can write ${\mathbf x}_0$ as a linear combination of ${\mathbf v}_1$ and ${\mathbf v}_2\text{.}$ In other words, we can find $c_1$ and $c_2$ such that \begin{equation*} {\mathbf x}_0 = {\mathbf x}(0) = c_1 {\mathbf v}_1 + c_2 {\mathbf v}_2. \end{equation*}

It remains to show that ${\mathbf x}(t) = c_1 e^{\lambda_1 t} {\mathbf v}_1 + c_2 e^{\lambda_2 t} {\mathbf v}_2$ is the unique solution to the system \begin{align*} {\mathbf x}'(t) & = A {\mathbf x}(t),\\ {\mathbf x}(0) & = {\mathbf x}_0. \end{align*} Suppose that there is another solution ${\mathbf y}(t)$ such that ${\mathbf y}(0) = {\mathbf x}_0\text{.}$ Then we can write \begin{equation*} {\mathbf y}(t) = f(t) {\mathbf v}_1 + g(t) {\mathbf v}_2, \end{equation*} where \begin{align*} f(0) & = c_1,\\ g(0) & = c_2. \end{align*} Since ${\mathbf y}(t)$ is a solution to our system of equations, we know that \begin{equation*} A {\mathbf y}(t) = {\mathbf y}'(t) = f'(t) {\mathbf v}_1 + g'(t) {\mathbf v}_2. \end{equation*} On the other hand, \begin{equation*} A {\mathbf y}(t) = f(t) A {\mathbf v}_1 + g(t) A {\mathbf v}_2 = \lambda_1 f(t) {\mathbf v}_1 + \lambda_2 g(t) {\mathbf v}_2. \end{equation*} Consequently, we have two first-order initial value problems, \begin{align*} f'(t) & = \lambda_1 f(t),\\ f(0) & = c_1, \end{align*} and \begin{align*} g'(t) & = \lambda_2 g(t),\\ g(0) & = c_2. \end{align*} The solutions of these initial value problems are \begin{align*} f(t) & = c_1 e^{\lambda_1 t},\\ g(t) & = c_2 e^{\lambda_2 t}, \end{align*} respectively. Thus, ${\mathbf y}(t) = {\mathbf x}(t)\text{,}$ and proof our theorem is complete.

• If ${\mathbf v}_1$ and ${\mathbf v}_2$ are eigenvectors of two distinct real eigenvalues of a matrix $A\text{,}$ then ${\mathbf v}_1$ and ${\mathbf v}_2$ are linearly independent.
• The Principle of Superposition tells us that any linear combination of solutions to the linear system ${\mathbf x}' = A {\mathbf x}$ is also a solution.
• Let $A$ be a $2 \times 2$ matrix. If $A$ has a pair of distinct real eigenvalues, $\lambda_1$ and $\lambda_2\text{,}$ with associated eigenvectors ${\mathbf v}_1$ and ${\mathbf v}_2\text{,}$ then the general solution of the linear system ${\mathbf x}' = A {\mathbf x}$ is given by \begin{equation*} {\mathbf x}(t) = \alpha e^{\lambda_1 t} {\mathbf v}_1 + \beta e^{\lambda_2 t} {\mathbf v}_2. \end{equation*}

##### 1

Solve each of the following linear systems for the given initial values.

1. \begin{align*} x' & = x + 2y\\ y' & = -x + 4y\\ x(0) & = 3\\ y(0) & = 2 \end{align*}
2. \begin{align*} x' & = -x + 3y\\ y' & = -3x - y\\ x(0) & = 3\\ y(0) & = 2 \end{align*}
3. \begin{align*} x' & = -2 x + y\\ y' & = -9x + 4y\\ x(0) & = 5\\ y(0) & = 3 \end{align*}
##### 2

Consider the nonhomogeneous system of linear differential equations \begin{align} x' & = a(t)x + b(t)y + f(t)\label{equation-exercise-linear02-nonhomogeneous-system}\tag{3.8}\\ y' & = c(t)x + d(t)y + g(t)\label{mrow-461}\tag{3.9} \end{align} and assume that the general solution of \begin{align*} x' & = a(t)x + b(t)y\\ y' & = c(t)x + d(t)y \end{align*} is given by \begin{equation*} {\mathbf x}_h = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 \begin{pmatrix} u_1(t) \\ u_2(t) \end{pmatrix} + c_2 \begin{pmatrix} v_1(t) \\ v_2(t) \end{pmatrix}. \end{equation*} If \begin{equation*} {\mathbf x}_p = \begin{pmatrix} \phi_1(t) \\ \phi_2(t) \end{pmatrix} \end{equation*} is a particular solution of ((3.8)), show that \begin{equation*} {\mathbf x}_h + {\mathbf x}_p = \begin{pmatrix} x(t) + \phi_1(t) \\ y(t) + \phi_2(t) \end{pmatrix} \end{equation*} is the general solution to the system. Thus, to solve a nonhomogeneous system of linear differential equations, we need to find the solution of the corresponding homogeneous system and one particular solution of the nonhomogeneous system.

##### 3

Consider the linear system \begin{align*} x' & = x + 3y + (t - 3t^2)\\ y' & = x - y + (2 - t + t^2)\\ x(0) & = 1\\ y(0) & = -1. \end{align*}

1. Find the general solution of the homogeneous system \begin{align*} x' & = x + 3y\\ y' & = x - y \end{align*}
2. Find a particular solution for \begin{align*} x' & = x + 3y + (t - 3t^2)\\ y' & = x - y + (2 - t + t^2) \end{align*}
3. Find the solution of \begin{align*} x' & = x + 3y + (t - 3t^2)\\ y' & = x - y + (2 - t + t^2)\\ x(0) & = 1\\ y(0) & = -1. \end{align*}
##### 4

Since 6000 B.C., when beer was already known in Babylonia and Sumeria, hundreds of brewing techniques have been used. One such technique requires two tanks, $A$ and $B\text{,}$ in which water mixes with two kinds of sugar, and a larger tank, $C\text{,}$ that collects the combined mixture for fermentation. Tanks $A$ and $B\text{,}$ initially contain 1000 liters and 2000 liters of water, respectively. From an external source 5 liters of liquid sugar flows into Tank $A$ every minute. From another external source, 10 liters of the second kind of liquid sugar flows into Tank $B$ every minute. The mixture from Tank $A$ flows into Tank $B$ at the rate of 10 liters per minute. The mixture from Tank $B$ is pumped back into Tank $A$ at the rate of 5 liters per minute. The mixture from from Tank $B$ is also pumped into Tank $C$ at the rate of 15 liters per minute. This technique requires that the amount of sugar in Tanks $A$ and $B$ be known for the duration of the the process. Assuming that the liquid in each tank is constantly stirred, derive the functions that give the amount of sugar in each tank at time $t\text{.}$

##### 5

Scottish economist Adam Smith (1723–1790) asserted that in a free market economy governed by the law of supply and demand, prices will eventually tend toward an equilibrium. Suppose that we wish to derive a system of differential equations that model the price change for one commodity. Let $d = d(t)$ and $s = s(t)$ be the time dependent functions that denote the demand and the supply of the commodity, respectively. The escess demand is $r = d - s\text{.}$ Let $p = p(t)$ be the price of the commodity. We will assume that both $p$ and $r$ have continuous derivatives.

Assume that the laws of supply and demand works as follows.

• If at a given time, demand is less than supply, prices decrease, so the demand increases. Thus, excess demand increases.
• If at a given time, demand is greater than supply, prices increase, so the supply increases. Thus, excess demand decreases.
• If at a given time, demand equals supply, prices are constant, so excess demand remains constant, namely zero.

1. Rewrite the laws of supply and demand in terms of $r\text{,}$ $p\text{,}$ and derivatives of $r$ and $p\text{.}$
2. Derive a linear system of equations in $p$ and $r$ that describe the law of supply and demand. That is, express both $p'$ and $r'$ as linear functions of $r$ and $p\text{.}$
3. From the system that you derived in part (1), show that prices must tend toward an equilibrium.
##### 6

Consider the system \begin{align*} x' & = ax + y\\ y' & = 2ax + 2y, \end{align*} where $a \in {\mathbb R}\text{.}$ For what values of $a$ do you find a bifurcation (a change in the type of phase portrait)? Sketch typical phase portraits for a values of $a$ above and below the bifurcation point.

##### 7

Prove that \begin{equation*} \alpha e^{\lambda t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta e^{\lambda t} \begin{pmatrix} t \\ 1 \end{pmatrix} \end{equation*} is the general solution of \begin{equation*} {\mathbf x}' = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix} {\mathbf x}. \end{equation*}