INTRODUCTION

Recall that in Section 3.12 we illustrated how to solve systems of n linear differential equations in n unknowns of the form

(1)

where the P_ij were polynomials of various degrees in the differential operator D. In this chapter we confine our study to systems of first-order DEs that are special cases of systems that have the normal form

(2)

A system such as (2) of n first-order equations is called a first-order system.

Linear Systems

When each of the functions g₁, g₂, … , g_n in (2) is linear in the dependent variables x₁, x₂, … , x_n , we get the normal form of a first-order system of linear equations:

(3)

We refer to a system of the form given in (3) simply as a linear system. We assume that the coefficients a_{i j}(t) as well as the functions f_i(t) are continuous on a common interval I. When f_i(t) = 0, i = 1, 2, … , n, the linear system is said to be homogeneous; otherwise it is nonhomogeneous.

Matrix Form of a Linear System

If denote the respective matrices

then the system of linear first-order equations (3) can be written

or simply X′ = AX + F. (4)

If the system is homogeneous, then the matrix form is

X′ = AX. (5)

EXAMPLE 1 Systems Written in Matrix Notation

≡

EXAMPLE 2 Writing a Linear System Without Matrices

Write the linear system

without matrices.

SOLUTION

First, let and so the left-hand side of the given system is

Then, using matrix multiplication and addition, the terms on the right-hand side of the given linear system can be written as a single matrix:

Now two matrices are equal if their entries in the same position are equal. Therefore

≡

An Equation Written as a System

At times it is to your advantage to work with a system of equations rather than with a differential equation. Suppose

(6)

is a linear nth-order differential equation. If we let

(7)

then observe by taking the derivative of each of the terms in (7) we get

Thus a first-order system of n equations is

(8)

EXAMPLE 3 Writing an Equation as a First-Order System

Write the third-order differential equation

as a first-order system of equations using matrix notation.

SOLUTION

First, we write the differential equation as to agree with the form given in (6). Then, using the substitutions we see that

The result is a nonhomogeneous linear system

Using matrices, the foregoing system has the form or

≡

Solution of a Linear System

A solution of a linear system (4) is a set of n differentiable functions defined on a common interval that satisfy each of the differential equations in (3). Equivalently, we have the following definition in terms of matrices.

A solution vector on an interval I is any column matrix

whose entries are differentiable functions satisfying the system (4) on the interval.

A solution vector of (4) is equivalent to n scalar equations x₁ = Φ₁(t), x₂ = Φ₂(t), … , x_n = Φ_n(t), and can be interpreted geometrically as a set of parametric equations of a space curve. In the cases n = 2 and n = 3, the equations x₁ = Φ₁(t), x₂ = Φ₂(t), and x₁ = Φ₁(t), x₂ = Φ₂(t), x₃ = Φ₃(t) represent curves in 2-space and 3-space, respectively. It is common practice to call such a solution curve a trajectory. The plane is also called the phase plane. We will illustrate these concepts in the section that follows, as well as in Chapter 11.

EXAMPLE 4 Verification of Solutions

Verify that on the interval (−∞, ∞)

are solutions of (9)

SOLUTION

From we see that

and . ≡

Much of the theory of systems of n linear first-order differential equations is similar to that of linear nth-order differential equations.

Initial-Value Problem

Let t₀ denote a point on an interval I and

where the γ_i, i = 1, 2, … , n are given constants. Then the problem

(10)

is an initial-value problem on the interval.

Homogeneous Systems

In the next several definitions and theorems we are concerned only with homogeneous systems. Without stating it, we shall always assume that the a_{i j} and the f_i are continuous functions of t on some common interval I.

Superposition Principle

The following result is a superposition principle for solutions of linear systems.

It follows from Theorem 10.1.2 that a constant multiple of any solution vector of a homogeneous system of linear first-order differential equations is also a solution.

EXAMPLE 5 Using the Superposition Principle

You should practice by verifying that the two vectors

are solutions of the homogeneous system

(11)

By the superposition principle, Theorem 10.1.2, the linear combination

is yet another solution of the system. ≡

Linear Dependence and Linear Independence

We are primarily interested in linearly independent solutions of the homogeneous system (5).

The case when k = 2 should be clear; two solution vectors X₁ and X₂ are linearly dependent if one is a constant multiple of the other, and conversely. For k > 2, a set of solution vectors is linearly dependent if we can express at least one solution vector as a linear combination of the remaining vectors.

Wronskian

As in our earlier consideration of the theory of a single ordinary differential equation, we can introduce the concept of the Wronskian determinant as a test for linear independence. We state the following theorem without proof.

It can be shown that if X₁, X₂, … , X_n are solution vectors of (5), then for every t in I, either W(X₁, X₂, … , X_n) ≠ 0 or W(X₁, X₂, … , X_n) = 0. Thus if we can show that W ≠ 0 for some t₀ in I, then W ≠ 0 for every t; hence the set of solutions is linearly independent on the interval.

Notice that, unlike our definition of the Wronskian in Section 3.1, here the definition of the determinant (12) does not involve differentiation.

EXAMPLE 6 Linearly Independent Solutions

In Example 4 we saw that and are solutions of system (9).

Clearly X₁ and X₂ are linearly independent on the interval (−∞, ∞) since neither vector is a constant multiple of the other. In addition, we have

for all real values of t. ≡

The next two theorems are the linear system equivalents of Theorems 3.1.5 and 3.1.6.

EXAMPLE 7 General Solution of System (9)

EXAMPLE 8 General Solution of System (11)

The vectors

are solutions of the system (11) in Example 5 (see Problem 22 in Exercises 10.1). Now

for all real values of t. We conclude that X₁, X₂, and X₃ form a fundamental set of solutions on (−∞, ∞). Thus the general solution of the homogeneous system on the interval is the linear combination X = c₁X₁ + c₂X₂ + c₃X₃; that is,

≡

Nonhomogeneous Systems

For nonhomogeneous systems, a particular solution X_p on an interval I is any vector, free of arbitrary parameters, whose entries are functions that satisfy system (4).

EXAMPLE 9 General Solution—Nonhomogeneous System

The vector X_p = is a particular solution of the nonhomogeneous system

(14)

on the interval (−∞, ∞). (Verify this.) The complementary function of (14) on the same interval, or the general solution of X′ = X, was seen in (13) of Example 7 to be . Hence by Theorem 10.1.6

is the general solution of the nonhomogeneous system (14) on the interval (−∞, ∞). ≡

10.1 Exercises Answers to selected odd-numbered problems begin on page ANS-27.

In Problems 1–8, write the given linear system in matrix form.

1. = 3x − 5y
   = 4x + 8y
2. = 4x − 7y
   = 5x
3. = 5x − 14y + 9t
   = − x + 6y − 2
4. = −2x + 12y − 3 sin 4t
   = 8x − 3y + 10 cos 4t
5. = −3x + 4y − 9z
   = 6x − y
   = 10x + 4y + 3z
6. = x − y
   = x + 2z
   = −x + z
7. = x − y + z + t − 1
   = 2x + y − z − 3t²
   = x + y + z + t² − t + 2
8. = −3x + 4y + e^−t sin 2t
   = 5x + 9z + 4e^−t cos 2t
   = y + 6z − e^−t

In Problems 9–12, write the given linear system without the use of matrices.

In Problems 13–16, write the given differential equation as a first-order system using matrix notation.

In Problems 17–22, verify that the vector X is a solution of the given homogeneous linear system.

17. = 3x − 4y
    = 4x − 7y; X = e^−5t
18. = −2x + 5y
    = −2x + 4y; X = e^t
19. X′ = X; X = e^−3t/2
20. 
21. 
22. 

In Problems 23–26, the given vectors are solutions of a system X′ = AX. Determine whether the vectors form a fundamental set on the interval (−∞, ∞).

In Problems 27–30, verify that the vector X_p is a particular solution of the given nonhomogeneous linear system.

27. = x + 4y + 2t − 7
    
28. 
29. e^t;
    
30. 
    
31. Prove that the general solution of the homogeneous linear system
    

    

    on the interval (−∞, ∞) is

    X = c₁ e^−t + c₂ e^−2t + c₃ e^3t.

32. Prove that the general solution of the nonhomogeneous linear system
    

    

    on the interval (−∞, ∞) is