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The Ordinary Differential Equations Project Instructor’s Version

Thomas W. Judson

Section 3.2 Planar Systems

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*}
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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,\tag{3.2.1}\\ \frac{dy}{dt} & = c x + d y.\tag{3.2.2} \end{align}
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Subsection 3.2.1 Planar Systems and \(2 \times 2\) Matrices

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 3.2.1). 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*}
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Figure 3.2.1. Mixing example with two tanks
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Matrix notation gives us a convenient way of representing the \(2 \times 2\) system (3.2.1)–(3.2.1). 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*}
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Subsection 3.2.2 Systems of Differential Equations

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.
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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.
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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*}
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Example 3.2.4.

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 eigenvector for \(A\) with associated eigenvalue \(\mu = -2\text{.}\) 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*}
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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*}
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Activity 3.2.1. Planar Systems with Distinct Real Eigenvalues.

Consider the initial value problem \(d\mathbf x/dt = A \mathbf x\text{,}\) where
\begin{align*} A \amp = \begin{pmatrix} 23 \amp 10 \\-50 \amp -22 \end{pmatrix}\\ \mathbf x(0) \amp = \begin{pmatrix} 3 \\ 3 \end{pmatrix}. \end{align*}
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(a)
Find the eigenvalues of \(A\text{.}\) You should find distinct real eigenvalues \(\lambda\) and \(\mu\text{.}\)
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(b)
Find eigenvectors \(\mathbf v_1\) and \(\mathbf v_2\) for the eigenvalues \(\lambda\) and \(\mu\text{,}\) respectively.
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(c)
Find the straight-line solutions of \(d\mathbf x/dt = A \mathbf x\text{.}\) Plot the solutions in the \(xy\)-plane.
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(d)
Sketch the solution curve to the initial value problem in the \(xy\)-plane.
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(e)
Show that \(\mathbf x(t) = c_1 e^{\lambda t} \mathbf v_1 + c_2 e^{\mu t} \mathbf v_2\) is a solution to the linear system \(d\mathbf x/dt = A \mathbf x\text{.}\)
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(f)
Use the fact that
\begin{equation*} \mathbf x(0) = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \end{equation*}
to find \(c_1\) and \(c_2\) such that \(\mathbf x(t) = c_1 e^{\lambda t} \mathbf v_1 + c_2 e^{\mu t} \mathbf v_2\) is a solution to the initial value problem. Does this solution agree with the solution that you plotted in Task 3.2.1.d?
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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.
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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 eigenvalues \(\lambda = -1\) and \(\mu = -5\) with eigenvectors \({\mathbf u} = (2, 3)\) and \({\mathbf v} = (-2, 3)\text{,}\) 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*}
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Subsection 3.2.3 Solving Linear Systems

Our goal is to prove the following theorem.
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Proof.

If \({\mathbf v}_1\) and \({\mathbf v}_2\) are linearly dependent, then there exists \(\alpha \neq 0\) such that
\begin{equation} {\mathbf v}_1 = \alpha {\mathbf v}_2.\tag{3.2.3} \end{equation}
Multiplying both sides of this equation by \(A\text{,}\) we have
\begin{equation} \lambda_1 {\mathbf v}_1 = A{\mathbf v}_1 = \alpha A {\mathbf v}_2 = \alpha \lambda_2 {\mathbf v}_2.\tag{3.2.4} \end{equation}
On the other hand, we obtain
\begin{equation} \lambda_2 {\mathbf v}_1 = \alpha \lambda_2 {\mathbf v}_2\tag{3.2.5} \end{equation}
if we multiply both sides of (3.2.3) by \(\lambda_2\text{.}\) Using (3.2.4) and (3.2.5), we can conclude that
\begin{equation*} (\lambda_1 - \lambda_2) {\mathbf v}_1 = \alpha(\lambda_2 - \lambda_2 ){\mathbf v}_2 = 0 {\mathbf v}_2 = {\mathbf 0}. \end{equation*}
However, this contradicts the assumption that \(\lambda_1\) and \(\lambda_2\) are distinct.
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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 3.2.7, 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*}
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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.
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Subsection 3.2.4 Important Lessons

  • 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.
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  • 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.
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  • 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*}
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Reading Questions 3.2.5 Reading Questions

1.

What is a \(2 \times 2\) linear system of differential equations?
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2.

What is the Principle of Superposition?
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Exercises 3.2.6 Exercises

Solving Linear Systems with Distinct Real Eigenvalues.

Find the general solution of each of the linear systems in Exercise Group 3.2.6.1–4.
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1.
\begin{align*} x' & = x + 2y\\ y' & = -x + 4y \end{align*}
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Solution.
Write the system in matrix form
\begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}' = A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}1 & 2\\ -1 & 4\end{pmatrix}. \end{equation*}
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The characteristic polynomial is
\begin{equation*} \det(A-\lambda I) =\det\begin{pmatrix}1-\lambda & 2\\ -1 & 4-\lambda\end{pmatrix} =(1-\lambda)(4-\lambda)+2 =\lambda^2-5\lambda+6 =(\lambda-2)(\lambda-3). \end{equation*}
Hence the eigenvalues are \(\lambda_1=2\) and \(\lambda_2=3\text{.}\)
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For \(\lambda=2\text{,}\)
\begin{equation*} (A-2I)=\begin{pmatrix}-1 & 2\\ -1 & 2\end{pmatrix}, \end{equation*}
so \(-x+2y=0\) and an eigenvector is
\begin{equation*} \mathbf v_1=\begin{pmatrix}2\\1\end{pmatrix}. \end{equation*}
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For \(\lambda=3\text{,}\)
\begin{equation*} (A-3I)=\begin{pmatrix}-2 & 2\\ -1 & 1\end{pmatrix}, \end{equation*}
so \(-2x+2y=0\) and an eigenvector is
\begin{equation*} \mathbf v_2=\begin{pmatrix}1\\1\end{pmatrix}. \end{equation*}
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Therefore the general solution is
\begin{equation*} \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =c_1 e^{2t}\begin{pmatrix}2\\1\end{pmatrix} +c_2 e^{3t}\begin{pmatrix}1\\1\end{pmatrix}. \end{equation*}
Equivalently,
\begin{equation*} x(t)=2c_1 e^{2t}+c_2 e^{3t}, \qquad y(t)=c_1 e^{2t}+c_2 e^{3t}. \end{equation*}
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2.
\begin{align*} x' & = 4x + 2y\\ y' & = x + 3y \end{align*}
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Solution.
Write the system in matrix form
\begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}'= A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}4 & 2\\ 1 & 3\end{pmatrix}. \end{equation*}
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The characteristic polynomial is
\begin{equation*} \det(A-\lambda I) =\det\begin{pmatrix}4-\lambda & 2\\ 1 & 3-\lambda\end{pmatrix} =(4-\lambda)(3-\lambda)-2 =\lambda^2-7\lambda+10 =(\lambda-5)(\lambda-2). \end{equation*}
Hence the eigenvalues are \(\lambda_1=5\) and \(\lambda_2=2\text{.}\)
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For \(\lambda=5\text{,}\)
\begin{equation*} A-5I=\begin{pmatrix}-1 & 2\\ 1 & -2\end{pmatrix}, \end{equation*}
so \(-x+2y=0\) and an eigenvector is
\begin{equation*} \mathbf v_1=\begin{pmatrix}2\\1\end{pmatrix}. \end{equation*}
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For \(\lambda=2\text{,}\)
\begin{equation*} A-2I=\begin{pmatrix}2 & 2\\ 1 & 1\end{pmatrix}, \end{equation*}
so \(x+y=0\) and an eigenvector is
\begin{equation*} \mathbf v_2=\begin{pmatrix}1\\-1\end{pmatrix}. \end{equation*}
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Therefore the general solution is
\begin{equation*} \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =c_1 e^{5t}\begin{pmatrix}2\\1\end{pmatrix} +c_2 e^{2t}\begin{pmatrix}1\\-1\end{pmatrix}. \end{equation*}
Equivalently,
\begin{equation*} x(t)=2c_1 e^{5t}+c_2 e^{2t}, \qquad y(t)=c_1 e^{5t}-c_2 e^{2t}. \end{equation*}
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3.
\begin{align*} x' & = -3 x + 4y\\ y' & = 3x - 2y \end{align*}
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Solution.
Write the system in matrix form
\begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}'= A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}-3 & 4\\ 3 & -2\end{pmatrix}. \end{equation*}
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The characteristic polynomial is
\begin{equation*} \det(A-\lambda I) =\det\begin{pmatrix}-3-\lambda & 4\\ 3 & -2-\lambda\end{pmatrix} =(-3-\lambda)(-2-\lambda)-12 =(\lambda+3)(\lambda+2)-12 =\lambda^2+5\lambda-6. \end{equation*}
Hence
\begin{equation*} \lambda^2+5\lambda-6=(\lambda-1)(\lambda+6)=0, \end{equation*}
so the eigenvalues are \(\lambda_1=1\) and \(\lambda_2=-6\text{.}\)
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For \(\lambda=1\text{,}\)
\begin{equation*} A-I=\begin{pmatrix}-4 & 4\\ 3 & -3\end{pmatrix}, \end{equation*}
giving \(-4x+4y=0\) and eigenvector
\begin{equation*} \mathbf v_1=\begin{pmatrix}1\\1\end{pmatrix}. \end{equation*}
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For \(\lambda=-6\text{,}\)
\begin{equation*} A+6I=\begin{pmatrix}3 & 4\\ 3 & 4\end{pmatrix}, \end{equation*}
giving \(3x+4y=0\) and eigenvector
\begin{equation*} \mathbf v_2=\begin{pmatrix}4\\-3\end{pmatrix}. \end{equation*}
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Therefore the general solution is
\begin{equation*} \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =c_1 e^{t}\begin{pmatrix}1\\1\end{pmatrix} +c_2 e^{-6t}\begin{pmatrix}4\\-3\end{pmatrix}. \end{equation*}
Equivalently,
\begin{equation*} x(t)=c_1 e^{t}+4c_2 e^{-6t}, \qquad y(t)=c_1 e^{t}-3c_2 e^{-6t}. \end{equation*}
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4.
\begin{align*} x' & = 6x + 4y\\ y' & = -8x - 6y \end{align*}
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Solution.
Write the system in matrix form
\begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}'= A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}6 & 4\\ -8 & -6\end{pmatrix}. \end{equation*}
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The characteristic polynomial is
\begin{equation*} \det(A-\lambda I) =\det\begin{pmatrix}6-\lambda & 4\\ -8 & -6-\lambda\end{pmatrix} =(6-\lambda)(-6-\lambda)+32 =\lambda^2-4. \end{equation*}
Hence the eigenvalues are
\begin{equation*} \lambda=\pm 2. \end{equation*}
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For \(\lambda=2\text{,}\)
\begin{equation*} A-2I=\begin{pmatrix}4 & 4\\ -8 & -8\end{pmatrix}, \end{equation*}
giving \(x+y=0\text{.}\) An eigenvector is
\begin{equation*} \mathbf v_1=\begin{pmatrix}1\\-1\end{pmatrix}. \end{equation*}
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For \(\lambda=-2\text{,}\)
\begin{equation*} A+2I=\begin{pmatrix}8 & 4\\ -8 & -4\end{pmatrix}, \end{equation*}
giving \(2x+y=0\text{.}\) An eigenvector is
\begin{equation*} \mathbf v_2=\begin{pmatrix}1\\-2\end{pmatrix}. \end{equation*}
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Therefore the general solution is
\begin{equation*} \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =c_1 e^{2t}\begin{pmatrix}1\\-1\end{pmatrix} +c_2 e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}. \end{equation*}
Equivalently,
\begin{equation*} x(t)=c_1 e^{2t}+c_2 e^{-2t}, \qquad y(t)=-c_1 e^{2t}-2c_2 e^{-2t}. \end{equation*}
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Solving Initial Value Problems.

Solve each of the following linear systems for the given initial values. in Exercise Group 3.2.6.5–8.
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5.
\begin{align*} x' & = x + 2y\\ y' & = -x + 4y\\ x(0) & = 3\\ y(0) & = 2 \end{align*}
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Solution.
Write the system in matrix form
\begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}'= A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}1 & 2\\ -1 & 4\end{pmatrix}. \end{equation*}
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From the eigenvalue/eigenvector calculation, the eigenvalues are \(\lambda_1=2\) with eigenvector \(\mathbf v_1=\begin{pmatrix}2\\1\end{pmatrix}\text{,}\) and \(\lambda_2=3\) with eigenvector \(\mathbf v_2=\begin{pmatrix}1\\1\end{pmatrix}\text{.}\) Hence the general solution is
\begin{equation*} \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =c_1 e^{2t}\begin{pmatrix}2\\1\end{pmatrix} +c_2 e^{3t}\begin{pmatrix}1\\1\end{pmatrix}. \end{equation*}
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Therefore
\begin{equation*} x(t)=2c_1 e^{2t}+c_2 e^{3t},\qquad y(t)=c_1 e^{2t}+c_2 e^{3t}. \end{equation*}
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Apply the initial conditions:
\begin{equation*} x(0)=2c_1+c_2=3,\qquad y(0)=c_1+c_2=2. \end{equation*}
Subtracting the second equation from the first gives \(c_1=1\text{,}\) and then \(c_2=1\text{.}\)
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Thus the solution of the initial value problem is
\begin{equation*} x(t)=2e^{2t}+e^{3t},\qquad y(t)=e^{2t}+e^{3t}. \end{equation*}
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6.
\begin{align*} x' & = 4x + 2y\\ y' & = x + 3y\\ x(0) & = 3\\ y(0) & = 2 \end{align*}
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Solution.
Write the system in matrix form
\begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}'= A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}4 & 2\\ 1 & 3\end{pmatrix}. \end{equation*}
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The eigenvalues of \(A\) satisfy
\begin{equation*} \det(A-\lambda I) =\det\begin{pmatrix}4-\lambda & 2\\ 1 & 3-\lambda\end{pmatrix} =(4-\lambda)(3-\lambda)-2 =\lambda^2-7\lambda+10 =(\lambda-5)(\lambda-2). \end{equation*}
Hence \(\lambda_1=5\) and \(\lambda_2=2\text{.}\)
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For \(\lambda=5\text{,}\)
\begin{equation*} A-5I=\begin{pmatrix}-1 & 2\\ 1 & -2\end{pmatrix}, \end{equation*}
so \(-x+2y=0\) and an eigenvector is
\begin{equation*} \mathbf v_1=\begin{pmatrix}2\\1\end{pmatrix}. \end{equation*}
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For \(\lambda=2\text{,}\)
\begin{equation*} A-2I=\begin{pmatrix}2 & 2\\ 1 & 1\end{pmatrix}, \end{equation*}
so \(x+y=0\) and an eigenvector is
\begin{equation*} \mathbf v_2=\begin{pmatrix}1\\-1\end{pmatrix}. \end{equation*}
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Therefore the general solution is
\begin{equation*} \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =c_1 e^{5t}\begin{pmatrix}2\\1\end{pmatrix} +c_2 e^{2t}\begin{pmatrix}1\\-1\end{pmatrix}. \end{equation*}
That is,
\begin{equation*} x(t)=2c_1 e^{5t}+c_2 e^{2t}, \qquad y(t)=c_1 e^{5t}-c_2 e^{2t}. \end{equation*}
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Apply the initial conditions:
\begin{equation*} x(0)=2c_1+c_2=3,\qquad y(0)=c_1-c_2=2. \end{equation*}
Solving, from the second equation \(c_1=2+c_2\text{.}\) Substitute into the first:
\begin{equation*} 2(2+c_2)+c_2=3 \;\Rightarrow\; 4+3c_2=3 \;\Rightarrow\; c_2=-\frac13, \end{equation*}
hence \(c_1=2-\frac13=\frac53\text{.}\)
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Thus the solution of the initial value problem is
\begin{equation*} x(t)=\frac{10}{3}e^{5t}-\frac{1}{3}e^{2t}, \qquad y(t)=\frac{5}{3}e^{5t}+\frac{1}{3}e^{2t}. \end{equation*}
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7.
\begin{align*} x' & = -3 x + 4y\\ y' & = 3x - 2y\\ x(0) & = 1\\ y(0) & = 3 \end{align*}
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Solution.
Write the system in matrix form
\begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}'= A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}-3 & 4\\ 3 & -2\end{pmatrix}. \end{equation*}
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The eigenvalues satisfy
\begin{equation*} \det(A-\lambda I) =\det\begin{pmatrix}-3-\lambda & 4\\ 3 & -2-\lambda\end{pmatrix} =(-3-\lambda)(-2-\lambda)-12 =\lambda^2+5\lambda-6 =(\lambda-1)(\lambda+6). \end{equation*}
Hence \(\lambda_1=1\) and \(\lambda_2=-6\text{.}\)
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For \(\lambda=1\text{,}\) an eigenvector is \(\mathbf v_1=\begin{pmatrix}1\\1\end{pmatrix}\text{.}\) For \(\lambda=-6\text{,}\) an eigenvector is \(\mathbf v_2=\begin{pmatrix}4\\-3\end{pmatrix}\text{.}\) Therefore the general solution is
\begin{equation*} \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =c_1 e^{t}\begin{pmatrix}1\\1\end{pmatrix} +c_2 e^{-6t}\begin{pmatrix}4\\-3\end{pmatrix}. \end{equation*}
That is,
\begin{equation*} x(t)=c_1 e^{t}+4c_2 e^{-6t}, \qquad y(t)=c_1 e^{t}-3c_2 e^{-6t}. \end{equation*}
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Apply the initial conditions:
\begin{equation*} x(0)=c_1+4c_2=1,\qquad y(0)=c_1-3c_2=3. \end{equation*}
Subtracting the second from the first gives \(7c_2=-2\text{,}\) so \(c_2=-\tfrac{2}{7}\text{.}\) Then \(c_1=3+3c_2=3-\tfrac{6}{7}=\tfrac{15}{7}\text{.}\)
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Thus the solution of the initial value problem is
\begin{equation*} x(t)=\frac{15}{7}e^{t}-\frac{8}{7}e^{-6t}, \qquad y(t)=\frac{15}{7}e^{t}+\frac{6}{7}e^{-6t}. \end{equation*}
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8.
\begin{align*} x' & = 6x + 4y\\ y' & = -8x - 6y\\ x(0) & = 1\\ y(0) & = 3 \end{align*}
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Solution.
Write the system in matrix form
\begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}'= A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}6 & 4\\ -8 & -6\end{pmatrix}. \end{equation*}
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The characteristic polynomial is
\begin{equation*} \det(A-\lambda I) =\det\begin{pmatrix}6-\lambda & 4\\ -8 & -6-\lambda\end{pmatrix} =(6-\lambda)(-6-\lambda)+32 =\lambda^2-4, \end{equation*}
so the eigenvalues are \(\lambda_1=2\) and \(\lambda_2=-2\text{.}\)
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For \(\lambda=2\text{,}\) solve \((A-2I)\mathbf{x}=\mathbf{0}\text{:}\)
\begin{equation*} A-2I=\begin{pmatrix}4 & 4\\ -8 & -8\end{pmatrix}, \end{equation*}
giving \(x+y=0\text{.}\) An eigenvector is
\begin{equation*} \mathbf v_1=\begin{pmatrix}1\\-1\end{pmatrix}. \end{equation*}
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For \(\lambda=-2\text{,}\) solve \((A+2I)\mathbf{x}=\mathbf{0}\text{:}\)
\begin{equation*} A+2I=\begin{pmatrix}8 & 4\\ -8 & -4\end{pmatrix}, \end{equation*}
giving \(2x+y=0\text{.}\) An eigenvector is
\begin{equation*} \mathbf v_2=\begin{pmatrix}1\\-2\end{pmatrix}. \end{equation*}
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Therefore the general solution is
\begin{equation*} \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =c_1 e^{2t}\begin{pmatrix}1\\-1\end{pmatrix} +c_2 e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}. \end{equation*}
Equivalently,
\begin{equation*} x(t)=c_1 e^{2t}+c_2 e^{-2t}, \qquad y(t)=-c_1 e^{2t}-2c_2 e^{-2t}. \end{equation*}
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Apply the initial conditions:
\begin{equation*} x(0)=c_1+c_2=1,\qquad y(0)=-c_1-2c_2=3. \end{equation*}
From \(c_1=1-c_2\text{,}\) substitute into the second equation:
\begin{equation*} -(1-c_2)-2c_2=3 \;\Rightarrow\; -1-c_2=3 \;\Rightarrow\; c_2=-4. \end{equation*}
Hence \(c_1=1-(-4)=5\text{.}\)
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Thus the solution of the initial value problem is
\begin{equation*} x(t)=5e^{2t}-4e^{-2t}, \qquad y(t)=-5e^{2t}+8e^{-2t}. \end{equation*}
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9.

Consider the nonhomogeneous system of linear differential equations
\begin{align} x' & = a(t)x + b(t)y + f(t)\tag{3.2.6}\\ y' & = c(t)x + d(t)y + g(t)\tag{3.2.7} \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 (3.2.6), 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.
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Solution.
Write the nonhomogeneous system in vector form:
\begin{equation*} \mathbf{x}'=A(t)\mathbf{x}+\mathbf{F}(t), \qquad A(t)=\begin{pmatrix}a(t) & b(t)\\ c(t) & d(t)\end{pmatrix}, \quad \mathbf{F}(t)=\begin{pmatrix}f(t)\\ g(t)\end{pmatrix}. \end{equation*}
The corresponding homogeneous system is \(\mathbf{x}'=A(t)\mathbf{x}\text{.}\)
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Let \(\mathbf{x}_p=\begin{pmatrix}\phi_1\\ \phi_2\end{pmatrix}\) be a particular solution of the nonhomogeneous system, so
\begin{equation*} \mathbf{x}_p' = A(t)\mathbf{x}_p+\mathbf{F}(t). \end{equation*}
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First, show that \(\mathbf{x}_h+\mathbf{x}_p\) is a solution. Since \(\mathbf{x}_h\) satisfies \(\mathbf{x}_h'=A(t)\mathbf{x}_h\text{,}\) we have
\begin{equation*} (\mathbf{x}_h+\mathbf{x}_p)' =\mathbf{x}_h'+\mathbf{x}_p' =A(t)\mathbf{x}_h+\bigl(A(t)\mathbf{x}_p+\mathbf{F}(t)\bigr) =A(t)(\mathbf{x}_h+\mathbf{x}_p)+\mathbf{F}(t). \end{equation*}
Hence \(\mathbf{x}_h+\mathbf{x}_p\) solves the nonhomogeneous system.
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Next, show that every solution has this form. Let \(\mathbf{x}\) be any solution of the nonhomogeneous system and define
\begin{equation*} \mathbf{z}=\mathbf{x}-\mathbf{x}_p. \end{equation*}
Then
\begin{equation*} \mathbf{z}'=\mathbf{x}'-\mathbf{x}_p' =\bigl(A(t)\mathbf{x}+\mathbf{F}(t)\bigr)-\bigl(A(t)\mathbf{x}_p+\mathbf{F}(t)\bigr) =A(t)(\mathbf{x}-\mathbf{x}_p) =A(t)\mathbf{z}. \end{equation*}
Thus \(\mathbf{z}\) is a solution of the homogeneous system, so
\begin{equation*} \mathbf{z}=\mathbf{x}_h =c_1\begin{pmatrix}u_1\\ u_2\end{pmatrix} +c_2\begin{pmatrix}v_1\\ v_2\end{pmatrix}. \end{equation*}
Therefore
\begin{equation*} \mathbf{x}=\mathbf{z}+\mathbf{x}_p=\mathbf{x}_h+\mathbf{x}_p. \end{equation*}
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Hence the general solution of the nonhomogeneous system is
\begin{equation*} \mathbf{x}(t)=\mathbf{x}_h(t)+\mathbf{x}_p(t) = \begin{pmatrix} x(t)+\phi_1(t)\\ y(t)+\phi_2(t) \end{pmatrix}. \end{equation*}
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10.

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*}
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  2. Find a particular solution for
    \begin{align*} x' & = x + 3y + (t - 3t^2)\\ y' & = x - y + (2 - t + t^2) \end{align*}
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  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*}
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Hint.
Assume that your solution must be of the form
\begin{equation*} {\mathbf x}_p = \begin{pmatrix} a_2 t^2 + a_1 t + a_0 \\ b_2 t^2 + b_1 t + b_0. \end{pmatrix} \end{equation*}
This is called the method of undetermined coefficients.
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Solution.
  1. The homogeneous system is
    \begin{equation*} \begin{pmatrix}x\\y\end{pmatrix}'= A\begin{pmatrix}x\\y\end{pmatrix}, \qquad A=\begin{pmatrix}1 & 3\\ 1 & -1\end{pmatrix}. \end{equation*}
    The characteristic polynomial is
    \begin{equation*} \det(A-\lambda I) =\det\begin{pmatrix}1-\lambda & 3\\ 1 & -1-\lambda\end{pmatrix} =(1-\lambda)(-1-\lambda)-3 =\lambda^2-4, \end{equation*}
    so \(\lambda=\pm 2\text{.}\)
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    For \(\lambda=-2\text{,}\) an eigenvector is \(\begin{pmatrix}-1\\1\end{pmatrix}\text{.}\) For \(\lambda=2\text{,}\) an eigenvector is \(\begin{pmatrix}3\\1\end{pmatrix}\text{.}\) Hence
    \begin{equation*} \mathbf{x}_h(t) =c_1 e^{-2t}\begin{pmatrix}-1\\1\end{pmatrix} +c_2 e^{2t}\begin{pmatrix}3\\1\end{pmatrix}. \end{equation*}
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  2. Seek a polynomial particular solution
    \begin{equation*} x_p(t)=at^2+bt+c,\qquad y_p(t)=dt^2+et+f. \end{equation*}
    Substitute into
    \begin{equation*} x' = x+3y+(t-3t^2),\qquad y' = x-y+(2-t+t^2), \end{equation*}
    and match coefficients. This gives
    \begin{equation*} a=0,\; b=2,\; c=-\frac{7}{4},\qquad d=1,\; e=-1,\; f=\frac{5}{4}. \end{equation*}
    Therefore one particular solution is
    \begin{equation*} \mathbf{x}_p(t)= \begin{pmatrix} 2t-\frac{7}{4}\\[4pt] t^2-t+\frac{5}{4} \end{pmatrix}. \end{equation*}
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  3. The general solution is \(\mathbf{x}=\mathbf{x}_h+\mathbf{x}_p\text{,}\) so
    \begin{equation*} \begin{pmatrix}x\\y\end{pmatrix} =c_1 e^{-2t}\begin{pmatrix}-1\\1\end{pmatrix} +c_2 e^{2t}\begin{pmatrix}3\\1\end{pmatrix} + \begin{pmatrix} 2t-\frac{7}{4}\\[4pt] t^2-t+\frac{5}{4} \end{pmatrix}. \end{equation*}
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    Apply \(x(0)=1\text{,}\) \(y(0)=-1\text{:}\)
    \begin{equation*} -c_1+3c_2-\frac{7}{4}=1,\qquad c_1+c_2+\frac{5}{4}=-1, \end{equation*}
    which yields \(c_1=-\frac{19}{8}\text{,}\) \(c_2=\frac{1}{8}\text{.}\)
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    Thus the solution of the initial value problem is
    \begin{equation*} x(t)=\frac{19}{8}e^{-2t}+\frac{3}{8}e^{2t}+2t-\frac{7}{4}, \qquad y(t)=-\frac{19}{8}e^{-2t}+\frac{1}{8}e^{2t}+t^2-t+\frac{5}{4}. \end{equation*}
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11.

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.
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12.

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*}
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Solution.
Let
\begin{equation*} A=\begin{pmatrix}\lambda & 1\\ 0 & \lambda\end{pmatrix}, \qquad \mathbf{x}(t)=\begin{pmatrix}x_1(t)\\ x_2(t)\end{pmatrix}. \end{equation*}
Then \(\mathbf{x}'=A\mathbf{x}\) is the system
\begin{equation*} x_1'=\lambda x_1+x_2,\qquad x_2'=\lambda x_2. \end{equation*}
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First solve the second equation:
\begin{equation*} x_2'=\lambda x_2 \;\Rightarrow\; x_2(t)=Ce^{\lambda t}. \end{equation*}
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Substitute \(x_2(t)=Ce^{\lambda t}\) into the first equation:
\begin{equation*} x_1'-\lambda x_1=Ce^{\lambda t}. \end{equation*}
Multiply by the integrating factor \(e^{-\lambda t}\text{:}\)
\begin{equation*} \frac{d}{dt}\!\bigl(e^{-\lambda t}x_1(t)\bigr)=C. \end{equation*}
Integrate:
\begin{equation*} e^{-\lambda t}x_1(t)=Ct+D \;\Rightarrow\; x_1(t)=(Ct+D)e^{\lambda t}. \end{equation*}
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Therefore
\begin{equation*} \mathbf{x}(t)= \begin{pmatrix} (Ct+D)e^{\lambda t}\\ Ce^{\lambda t} \end{pmatrix} =De^{\lambda t}\begin{pmatrix}1\\0\end{pmatrix} +Ce^{\lambda t}\begin{pmatrix}t\\1\end{pmatrix}. \end{equation*}
Renaming constants \(\alpha=D\) and \(\beta=C\text{,}\) we obtain
\begin{equation*} \mathbf{x}(t)=\alpha e^{\lambda t}\begin{pmatrix}1\\0\end{pmatrix} +\beta e^{\lambda t}\begin{pmatrix}t\\1\end{pmatrix}. \end{equation*}
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Conversely, a direct differentiation shows each term satisfies \(\mathbf{x}'=A\mathbf{x}\text{,}\) and since the system is linear with two constants of integration, this family gives the general solution.
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