3.12 Solving Systems of Linear DEs

INTRODUCTION

We conclude this chapter as we did in Chapter 2 with systems of differential equations. But unlike Section 2.9, we will actually solve systems in the discussion that follows.

Coupled Systems/Coupled DEs

In Section 2.9 we briefly examined some mathematical models that were systems of linear and nonlinear first-order ODEs. In Section 3.8 we saw that the mathematical model describing the displacement of a mass on a single spring, current in a series circuit, and charge on a capacitor in a series circuit consisted of a single differential equation. When physical systems are coupled—for example, when two or more mixing tanks are connected, when two or more spring/mass systems are attached, or when circuits are joined to form a network—the mathematical model of the system usually consists of a set of coupled differential equations, in other words, a system of differential equations.

We did not attempt to solve any of the systems considered in Section 2.9. The same remarks made in Sections 3.7 and 3.11 pertain as well to systems of nonlinear ODEs; that is, it is nearly impossible to solve such systems analytically. However, linear systems with constant coefficients can be solved. The method that we shall examine in this section for solving linear systems with constant coefficients simply uncouples the system into distinct linear ODEs in each dependent variable. Thus, this section gives you an opportunity to practice what you learned earlier in the chapter.

Before proceeding, let us continue in the same vein as Section 3.8 by considering a spring/mass system, but this time we derive a mathematical model that describes the vertical displacements of two masses in a coupled spring/mass system.

Coupled Spring/Mass System

Suppose two masses m1 and m2 are connected to two springs A and B of negligible mass having spring constants k1 and k2, respectively. As shown in FIGURE 3.12.1(a), spring A is attached to a rigid support and spring B is attached to the bottom of mass m1. Let x1(t) and x2(t) denote the vertical displacements of the masses from their equilibrium positions. When the system is in motion, Figure 3.12.1(b), spring B is subject to both an elongation and a compression; hence its net elongation is x2 − x1. Therefore it follows from Hooke’s law that springs A and B exert forces −k1x1 and k2(x2 − x1), respectively, on m1. If no damping is present and no external force is impressed on the system, then the net force on m1 is −k1x1 + k2(x2 − x1). By Newton’s second law we can write

Similarly, the net force exerted on mass m2 is due solely to the net elongation of spring B, that is, −k2(x2 − x1). Hence we have

A coupled spring over mass system is illustrated in three parts. Part (a) has the system at equilibrium. A spring A is suspended vertically from a rigid support. It is labeled k subscript 1. A mass m subscript 1 is attached to the free end of the spring A. A horizontal dashed line through the mass m subscript 1 is labeled x subscript 1 = 0. Another spring B is attached vertically to the bottom of the mass m subscript 1. It is labeled k subscript 2. A mass m subscript 2 is attached to the free end of the spring B. A horizontal dashed line through the mass m subscript 2 is labeled x subscript 2 = 0. Part (b) has the system at motion. The spring A is elongated and the mass m subscript 1 is displaced vertically to a distance x subscript 1 from the line x subscript 1 = 0. The spring B is also elongated and the mass m subscript 2 is displaced vertically to a distance x subscript 2 from the line x subscript 2 = 0. Part (c) has forces acting on the masses. An upward arrow from the mass m subscript 1 is labeled k subscript 1 times x subscript 1. A downward arrow from the mass m subscript 1 is labeled k subscript 2 times (x subscript 2 minus x subscript 1). An upward arrow from the mass m subscript 2 is labeled k subscript 2 times (x subscript 2 minus x subscript 1).

FIGURE 3.12.1 Coupled spring/mass system

In other words, the motion of the coupled system is represented by the system of linear second-order equations

           

(1)

After we have illustrated the main idea of this section, we will return to system (1).

Systematic Elimination

The method of systematic elimination for solving systems of linear equations with constant coefficients is based on the algebraic principle of elimination of variables. The analogue of multiplying an algebraic equation by a constant is operating on an ODE with some combination of derivatives. The elimination process is expedited by rewriting each equation in a system using differential operator notation. Recall from Section 3.1 that a single linear equation

any(n) + an−1y(n−1) + … + a1y′ + a0y = g(t),

where the coefficients ai, i = 0, 1, ..., n are constants, can be written as

(anDn + an−1Dn−1 + … + a1D + a0)y = g(t).

If an nth-order differential operator anDn + an−1Dn−1 + ... + a1D + a0 factors into differential operators of lower order, then the factors commute. Now, for example, to rewrite the system

x″ + 2x′ + y″ = x + 3y + sin t

x′ + y′ = −4x + 2y + e−t

in terms of the operator D, we first bring all terms involving the dependent variables to one side and group the same variables:

Solution of a System

A solution of a system of differential equations is a set of sufficiently differentiable functions x = φ1(t), y = φ2(t), z = φ3(t), and so on, that satisfies each equation in the system on some common interval I.

Method of Solution

Consider the simple system of linear first-order equations

or, equivalently, (2)

Operating on the first equation in (2) by D while multiplying the second by −3 and then adding eliminates y from the system and gives D2x − 6x = 0. Since the roots of the auxiliary equation of the last DE are m1 = and m2 = −, we obtain

(3)

Multiplying the first equation in (2) by 2 while operating on the second by D and then subtracting gives the differential equation for y, D2y − 6y = 0. It follows immediately that

(4)

This is important.

Now, (3) and (4) do not satisfy the system (2) for every choice of c1, c2, c3, and c4 because the system itself puts a constraint on the number of parameters in a solution that can be chosen arbitrarily. To see this, observe that after substituting x(t) and y(t) into the first equation of the original system, (2) gives, after simplification,

Since the latter expression is to be zero for all values of t, we must have −c1 − 3c3 = 0 and c2 − 3c4 = 0. Thus we can write c3 as a multiple of c1 and c4 as a multiple of c2:

(5)

Hence we conclude that a solution of the system must be

You are urged to substitute (3) and (4) into the second equation of (2) and verify that the same relationship (5) holds between the constants.

EXAMPLE 1 Solution by Elimination

Solve

(6)

SOLUTION

Operating on the first equation by D − 3 and on the second by D and then subtracting eliminates x from the system. It follows that the differential equation for y is

[(D − 3)(D + 2) + 2D]y = 0 or (D2 + D − 6)y = 0.

Since the characteristic equation of this last differential equation is m2 + m − 6 = (m − 2)(m + 3) = 0, we obtain the solution

y(t) = c1e2t + c2e−3t. (7)

Eliminating y in a similar manner yields (D2 + D − 6)x = 0, from which we find

x(t) = c3e2t + c4e−3t. (8)

As we noted in the foregoing discussion, a solution of (6) does not contain four independent constants. Substituting (7) and (8) into the first equation of (6) gives

(4c1 + 2c3)e2t + (− c2 − 3c4)e−3t = 0.

From 4c1 + 2c3 = 0 and − c2 − 3c4 = 0 we get c3 = −2c1 and c4 = −c2. Accordingly, a solution of the system is

x(t) = −2c1e2t − c2e−3t, y(t) = c1e2t + c2e−3t. ≡

Since we could just as easily solve for c3 and c4 in terms of c1 and c2, the solution in Example 1 can be written in the alternative form

x(t) = c3e2t + c4e−3t, y(t) = −c3e2t − 3c4e−3t.

Watch for a shortcut.

It sometimes pays to keep one’s eyes open when solving systems. Had we solved for x first, then y could be found, along with the relationship between the constants, by using the last equation in (6). You should verify that substituting x(t) into y = (Dx − 3x) yields y = −c3e2t − 3c4e−3t.

EXAMPLE 2 Solution by Elimination

Solve

x′ − 4x + y″ = t2

x′ + x + y′ = 0. (9)

SOLUTION

First we write the system in differential operator notation:

(D − 4)x + D2y = t2

(D + 1)x + Dy = 0. (10)

Then, by eliminating x, we obtain

[(D + 1)D2 − (D − 4)D]y = (D + 1)t2 − (D − 4)0

or

(D3 + 4D)y = t2 + 2t.

Since the roots of the auxiliary equation m(m2 + 4) = 0 are m1 = 0, m2 = 2i, and m3 = −2i, the complementary function is

yc = c1 + c2 cos 2t + c3 sin 2t.

To determine the particular solution yp we use undetermined coefficients by assuming yp = At3 + Bt2 + Ct. Therefore

The last equality implies 12A = 1, 8B = 2, 6A + 4C = 0, and hence A = , B = , C = −. Thus

y = yc + yp = c1 + c2 cos 2t + c3 sin 2t + t3 + t2 − t. (11)

Eliminating y from the system (9) leads to

[(D − 4) − D(D + 1)]x = t2 or (D2 + 4)x = −t2.

It should be obvious that

xc = c4 cos 2t + c5 sin 2t

and that undetermined coefficients can be applied to obtain a particular solution of the form xp = At2 + Bt + C. In this case the usual differentiations and algebra yield xp = −t2 + , and so

x = xc + xp = c4 cos 2t + c5 sin 2t − t2 + . (12)

Now c4 and c5 can be expressed in terms of c2 and c3 by substituting (11) and (12) into either equation of (9). By using the second equation, we find, after combining terms,

(c5 − 2c4 − 2c2) sin 2t + (2c5 + c4 + 2c3) cos 2t = 0

so that c5 − 2c4 − 2c2 = 0 and 2c5 + c4 + 2c3 = 0. Solving for c4 and c5 in terms of c2 and c3 gives c4 = −(4c2 + 2c3) and c5 = (2c2 − 4c3). Finally, a solution of (9) is found to be

x(t) = − (4c2 + 2c3) cos 2t + (2c2 − 4c3) sin 2t − t2 + ,

y(t) = c1 + c2 cos 2t + c3 sin 2t + t3 + t2 − t. ≡

EXAMPLE 3 A Mathematical Model Revisited

In (3) of Section 2.9 we saw that a system of linear first-order differential equations described the number of pounds of salt x1(t) and x2(t) of a brine mixture that flows between two tanks. See Figure 2.9.1. At that time we were not able to solve the system. But now, in terms of differential operators, the system is

Operating on the first equation by D + , multiplying the second equation by , adding, and then simplifying, gives

(625D2 + 100D + 3)x1 = 0.

From the auxiliary equation 625m2 + 100m + 3 = (25m + 1)(25m + 3) = 0 we see immediately that

x1(t) = c1e−t/25 + c2e−3t/25.

In like manner we find

(625D2 + 100D + 3)x2 = 0

and so

x2(t) = c3e−t/25 + c4e−3t/25.

Substituting x1(t) and x2(t) into, say, the first equation of the system then gives

(2c1 − c3)e−t/25 + (−2c2 − c4)e−3t/25 = 0.

From this last equation we find c3 = 2c1 and c4 = −2c2. Thus a solution of the system is

x1(t) = c1e−t/25 + c2e−3t/25, x2(t) = 2c1e−t/25 − 2c2e−3t/25.

In the original discussion we assumed that initial conditions were x1(0) = 25 and x2(0) = 0. Applying these conditions to the solution yields c1 + c2 = 25 and 2c1 − 2c2 = 0. Solving these equations simultaneously gives c1 = c2 = . Finally, a solution of the initial-value problem is

x1(t) = e−t/25 + e−3t/25, x2(t) = 25e−t/25 − 25e−3t/25.

The graphs of x1(t) and x2(t) are given in FIGURE 3.12.2. Notice that even though the number of pounds x2(t) of salt in tank B starts initially at 0 lb it quickly increases and surpasses the number of pounds x1(t) of salt in tank A. ≡

Two curves are graphed on the t x coordinate plane. The first curve labeled x subscript 1 (t) starts at the approximate point (0, 25), goes down and to the right gradually, and ends at the approximate point (45, 2.5). The curve is slightly flat at the end. The second curve labeled x subscript 2 (t) starts at the point (0, 0), goes up and to the right quickly, crosses the first curve, then goes slightly down and to the right, and ends at the approximate point (45, 4). The curve is flat at the end.

FIGURE 3.12.2 Pounds of salt in tanks in Example 3

In our next example we solve system (1) under the assumption that k1 = 6, k2 = 4, m1 = 1, and m2 = 1.

EXAMPLE 4 A Special Case of System (1)

Solve

(13)

subject to

SOLUTION

Using elimination on the equivalent form of the system

(5)

we find that x1 and x2 satisfy, respectively,

(D2 + 2)(D2 + 12)x1 = 0 and (D2 + 2)(D2 + 12)x2 = 0.

Thus we find

x1(t) = c1 cos + c2 sin + c3 cos 2 + c4 sin 2

x2(t) = c5 cos + c6 sin + c7 cos 2 + c8 sin 2.

Substituting both expressions into the first equation of (13) and simplifying eventually yields c5 = 2c1, c6 = 2c2, c7 = −c3, c8 = −c4. Thus, a solution of (13) is

x1(t) = c1 cos + c2 sin + c3 cos 2 + c4 sin 2

x2(t) = 2c1 cos + 2c2 sin − c3 cos 2 − c4 sin 2.

The stipulated initial conditions then imply c1 = 0, c2 = −/10, c3 = 0, c4 = /5. And so the solution of the initial-value problem is

(14)

The graphs of x1 and x2 in FIGURE 3.12.3 reveal the complicated oscillatory motion of each mass.

The graph of x subscript 1 (t) is shown in part (a). A curve is graphed on the t x subscript 1 plane. It follows an oscillatory pattern. It starts at a point on the positive x subscript 1 axis just above the origin, goes up and to the right, then goes down and to the right through the positive t axis, and reaches a low point in the fourth quadrant. It goes up and down and to the right following the oscillatory pattern, and ends at a point in the fourth quadrant. The curve reaches a high point corresponding to the t value of 7.5. The graph of x subscript 2 (t) is shown in part (b). A curve is graphed on the t x subscript 2 plane. It follows an oscillatory pattern. It starts at a point on the negative x subscript 2 axis just below the origin, goes down and to the right, then goes up and to the right, again goes slightly down and to the right, goes up and to the right through the positive t axis, and reaches a high point in the first quadrant. It goes up and down and to the right following the oscillatory pattern, and ends at a point in the first quadrant.

FIGURE 3.12.3 Displacements of the two masses in Example 4

≡

We will revisit Example 4 in Section 4.6, where we will solve the system in (13) by means of the Laplace transform.

3.12 Exercises Answers to selected odd-numbered problems begin on page ANS-8.

In Problems 1−20, solve the given system of differential equations by systematic elimination.







  9. Dx + D2y = e3t
    (D + 1)x + (D − 1)y = 4e3t
  10. D2x − Dy = t
    (D + 3)x + (D + 3)y = 2
  11. (D2 − 1)x − y = 0
    (D − 1)x + Dy = 0
  15. (D − 1)x + (D2 + 1)y = 1
    (D2 − 1)x + (D + 1)y = 2
  17. Dx = y
    Dy = z
    Dz = x




In Problems 21 and 22, solve the given initial-value problem.





Mathematical Models

  1. Projectile Motion A projectile shot from a gun has weight w = mg and velocity v tangent to its path of motion or trajectory. Ignoring air resistance and all other forces acting on the projectile except its weight, determine a system of differential equations that describes its path of motion. See FIGURE 3.12.4. Solve the system. [Hint: Use Newton’s second law of motion in the x and y directions.]
    A curve is graphed on an x y coordinate plane. It starts on the positive y axis just above the origin, goes up and to the right, reaches a high point, then goes down and to the right, and ends at a point on the positive x axis. A point is marked with a heavy dot on the curve before the high point. A downward arrow from the marked point is labeled mg. An arrow tangential to the curve starts from the marked point and is labeled v.

    FIGURE 3.12.4 Path of projectile in Problem 23

  2. Projectile Motion with Air Resistance Determine a system of differential equations that describes the path of motion in Problem 23 if linear air resistance is a retarding force k (of magnitude k) acting tangent to the path of the projectile but opposite to its motion. See FIGURE 3.12.5. Solve the system. [Hint: k is a multiple of velocity, say, βv.]
    A right triangle. A point is marked with a heavy dot on the hypotenuse. An arrow along the hypotenuse pointing upward is labeled v. An arrow along the hypotenuse pointing downward is labeled k. The angle between the hypotenuse and the horizontal side is labeled theta.

    FIGURE 3.12.5 Graph of y(x) in Example 6

Computer Lab Assignments

  1. Consider the solution x1(t) and x2(t) of the initial-value problem given at the end of Example 3. Use a CAS to graph x1(t) and x2(t) in the same coordinate plane on the interval [0, 100]. In Example 3, x1(t) denotes the number of pounds of salt in tank A at time t, and x2(t) the number of pounds of salt in tank B at time t. See Figure 2.9.1. Use a root-finding application to determine when tank B contains more salt than tank A.
    1. Reread Problem 10 of Exercises 2.9. In that problem you were asked to show that the system of differential equations

      is a model for the amounts of salt in the connected mixing tanks A, B, and C shown in Figure 2.9.7. Solve the system subject to x1(0) = 15, x2(0) = 10, x3(0) = 5.

    2. Use a CAS to graph x1(t), x2(t), and x3(t) in the same coordinate plane on the interval [0, 200].
    3. Since only pure water is pumped into tank A, it stands to reason that the salt will eventually be flushed out of all three tanks. Use a root-finding application of a CAS to determine the time when the amount of salt in each tank is less than or equal to 0.5 pounds. When will the amounts of salt x1(t), x2(t), and x3(t) be simultaneously less than or equal to 0.5 pounds?
    1. Use systematic elimination to solve the system (1) for the coupled spring/mass system when k1 = 4, k2 = 2, m1 = 2, and m2 = 1 and with initial conditions x1(0) = 2, x′1(0) = 1, x2(0) = −1, x′2(0) = 1.
    2. Use a CAS to plot the graphs of x1(t) and x2(t) in the tx-plane. What is the fundamental difference in the motions of the masses m1 and m2 in this problem and that of the masses illustrated in Figure 3.12.3?
    3. As parametric equations, plot x1(t) and x2(t) in the x1x2-plane. The curve defined by these parametric equations is called a Lissajous curve.