4.6 Systems of Linear Differential Equations

INTRODUCTION

When initial conditions are specified, the Laplace transform reduces a system of linear differential equations with constant coefficients to a set of simultaneous algebraic equations in the transformed functions.

Coupled Springs

In our first example we solve the model

(1)

that describes the motion of two masses m1 and m2 in the coupled spring/mass system shown in Figure 3.12.1 of Section 3.12.

EXAMPLE 1 Example 4 of Section 3.12 Revisited

Use the Laplace transform to solve

(2)

subject to . This is system (1) with k1 = 6, k2 = 4, m1 = 1, and m2 = 1.

SOLUTION

The transform of each equation is

where X1(s) = {x1(t)} and X2(s) = {x2(t)}. The preceding system is the same as

(3)

Solving (3) for X1(s) and using partial fractions on the result yields

and therefore

Substituting the expression for X1(s) into the first equation of (3) gives us

and

Finally, the solution to the given system (2) is

(4)

The solution (4) is the same as (14) of Section 3.12. ≡

Of course, if n masses are connected to n springs, then the system of simultaneous second-order differential equations (1) generalizes to

(5)

This system of equations serves as a model for another vibrating physical system. See the Remarks at the end of this section.

Networks

In (18) of Section 2.9 we saw that currents i1(t) and i2(t) in the network containing an inductor, a resistor, and a capacitor shown in FIGURE 4.6.1 were governed by the system of first-order differential equations

(6)

The electrical network consists of a voltage E, an inductor L, a resistor R, and a capacitor C. The voltage E, the resistor R, and the capacitor C are connected parallel to each other. The inductor L is connected in series between the voltage E and the resistor R. The resistor has blobs at both ends of its connection. A current i subscript 1 flows through the inductor, current i subscript 2 through the resistor, and current i subscript 3 through the capacitor.

FIGURE 4.6.1 Electrical network

We solve this system by the Laplace transform in the next example.

EXAMPLE 2 An Electrical Network

Solve the system in (6) under the conditions E(t) = 60 V, L = 1 h, R = 50 Ω, C = 10−4 F, and the currents i1 and i2 are initially zero.

SOLUTION

We must solve

subject to i1(0) = 0, i2(0) = 0.

Applying the Laplace transform to each equation of the system and simplifying gives

where I1(s) = {i1(t)} and I2(s) = {i2(t)}. Solving the system for I1 and I2 and decomposing the results into partial fractions gives

Taking the inverse Laplace transform, we find the currents to be

≡

Note that both i1(t) and i2(t) in Example 2 tend toward the value E/R = as t → ∞. Furthermore, since the current through the capacitor is i3(t) = i1(t)− i2(t) = 60te−100t, we observe that i3(t) → 0 as t → ∞.

Double Pendulum

As shown in FIGURE 4.6.2, a double pendulum oscillates in a vertical plane under the influence of gravity. For small displacements θ1(t) and θ2(t), it can be shown that the system of differential equations describing the motion is

(7)

An illustration of a double pendulum oscillating in a vertical plane. The pendulum is pivoted at the top with two masses m subscript 1 and m subscript 2 hanging downward. The length of the pendulum from the pivot to the center of mass m subscript 1 is labeled l subscript 1 and the length of the pendulum from the center of mass m subscript 1 to the center of mass m subscript 2 is labeled l subscript 2. The mass m subscript 1 is displaced to the right by an angle theta subscript 1 measured from a vertical line extending downward from the pivot of the system. The mass m subscript 2 is displaced to the right by an angle theta subscript 2 measured from a vertical line extending downward from the center of mass m subscript 1.

FIGURE 4.6.2 Double pendulum

As indicated in Figure 4.6.2, θ1 is measured (in radians) from a vertical line extending downward from the pivot of the system, and θ2 is measured from a vertical line extending downward from the center of mass m1. The positive direction is to the right, and the negative direction is to the left.

EXAMPLE 3 Double Pendulum

It is left as an exercise to fill in the details of using the Laplace transform to solve system (7) when m1 = 3, m2 = 1, l1 = l2 = 16, g = 32, θ1(0) = 1, θ2(0) = −1, , and . You should find that

(8)

With the aid of a CAS, the positions of the two masses at t = 0 and at subsequent times are shown in FIGURE 4.6.3. See Problem 24 in Exercises 4.6.

Positions of masses of a double pendulum at various times are shown in six parts. Part (a) has the positions at t equals 0. The first mass is displaced to the right of the pivot and the second mass to the left. Part (b) has the positions at t equals 1.4. The first mass is displaced to the left of the pivot and the second mass to the right. Part (c) has the positions at t equals 2.5. The first mass is displaced very slightly to the left of the pivot and the second mass to the left. Part (d) has the positions at t equals 4.3. The first mass is displaced to the left of the pivot and the second mass to the right. Part (e) has the positions at t equals 6.9. The first mass is displaced slightly to the right of the pivot and the second mass to the left. Part (f) has the positions at t equals 8.5. The first mass is displaced to the right of the pivot and the second mass also to the right.

FIGURE 4.6.3 Positions of masses at various times in Example 3 ≡

REMARKS

System (5) can also be interpreted as a mathematical model for the horizontal displacements of successive floors of a multistory building due to an earthquake. If the horizontal position of the center of the floor and its mass are denoted by and respectively, then is the horizontal displacement of the floor relative to the floor. Assuming Hooke’s law holds, the restoring force between those two floors is , where is a constant. In the case of a three-story building, the displacements are given by a system of three second-order differential equations

(9)

See FIGURE 4.6.4.

The illustrations are shown in two parts. Part (a) has the three-storey building on the ground. The first floor is labeled m subscript 1 and k subscript 0. The second floor is labeled m subscript 2 and k subscript 1. The third floor is labeled m-subscript3 and k subscript 2. Part (b) has the horizontal forces acting on a floor. The floor is labeled m subscript i. A horizontal force k subscript i times (x subscript (i + 1) minus x subscript i) acts to the right on the floor’s top surface. Another horizontal force k subscript (i minus 1) times (x subscript i minus x subscript (i minus 1)) acts to the left at the floor’s bottom surface.

FIGURE 4.6.4 Three-story building

A photo of a building destroyed by an earthquake.

© Nigel Spiers/Shutterstock

Building destroyed by an earthquake

4.6 Exercises Answers to selected odd-numbered problems begin on page ANS-11.

In Problems 1–12, use the Laplace transform to solve the given system of differential equations.

  1. = −x + y
    = 2x
    x(0) = 0, y(0) = 1
  2. = 2y + et
    = 8x − t
    x(0) = 1, y(0) = 1

  3. = 5x − y
    x(0) = −1, y(0) = 2
  4. = 1

    x(0) = 0, y(0) = 0
  5. 2 + − 2x = 1
    + − 3x − 3y = 2
    x(0) = 0, y(0) = 0

  6. + + 2y = 0
    x(0) = 0, y(0) = 1
  7. + x − y = 0
    + y − x = 0
    x(0) = 0, x′(0)= −2,
    y(0) = 0, y′(0) = 1
  8. = 0
    = 0
    x(0) = 1, x′(0) = 0,
    y(0)= −1, y′(0) = 5
  9. +
    − = 4t
    x(0) = 8, x′(0) = 0,
    y(0) = 0, y′(0) = 0
  10. − 4x + = 6 sin t
    + 2x − 2 = 0
    x(0) = 0, y(0) = 0,
    y′(0) = 0, y″(0) = 0
  11. + 3 + 3y = 0
    + 3y = te−t
    x(0) = 0, x′(0) = 2,
    y(0) = 0
  12. = 4x − 2y + 2 (t − 1)
    = 3x − y + (t − 1)
    x(0) = 0, y(0) =
  13. Solve system (1) when k1 = 3, k2 = 2, m1 = 1, m2 = 1 and x1(0) = 0, , x2(0) = 1, .
  14. Derive the system of differential equations describing the straight-line vertical motion of the coupled springs shown in equilibrium in FIGURE 4.6.5. Use the Laplace transform to solve the system when k1 = 1, k2 = 1, k3 = 1, m1 = 1, m2 = 1 and x1(0) = 0, (0) = −1, x2(0) = 0, (0) = 1.
    Three springs are coupled in series vertically with two masses m subscript 1 and m subscript 2 between them. The springs are coupled between two fixed supports. The springs are labeled k subscript 1, k subscript 2, and k subscript 3 from top to bottom, respectively. The top of the first spring is labeled x subscript 1 equals 0. The distance of the spring from the top to the center of mass m subscript 1 is labeled x subscript 1. The top of the second spring is labeled x subscript 2 equals 0. The distance of the spring from the top to the center of mass m subscript 2 is labeled x subscript 2.

    FIGURE 4.6.5 Coupled springs in Problem 14

  15. (a) Show that the system of differential equations for the currents i2(t) and i3(t) in the electrical network shown in FIGURE 4.6.6 is

    L1 + Ri2 + Ri3 = E(t)

    L2 + Ri2 + Ri3 = E(t).

    (b) Solve the system in part (a) if R = 5 Ω, L1 = 0.01 h, L2 = 0.0125 h, E = 100 V, i2(0) = 0, and i3(0) = 0.

    (c) Determine the current i1(t).

    The electrical network consists of a voltage E, two inductors L subscript 1 and L subscript 2, and a resistor R. The voltage E, and the inductors L subscript 1 and L subscript 2 are connected parallel to each other. The inductors L subscript 1 and L subscript 2 are also connected parallel to each other. The resistor is connected in series between the voltage E and the inductor L subscript 1. The inductor L subscript 1 has blobs at both ends of its connection. A current i subscript 1 from the voltage E flows through the resistor, current i subscript 2 through the inductor L subscript 1, and current i subscript 3 through the inductor L subscript 2.

    FIGURE 4.6.6 Network in Problem 15

  16. (a) In Problem 14 in Exercises 2.9 you were asked to show that the currents i2(t) and i3(t) in the electrical network shown in FIGURE 4.6.7 satisfy

    Solve the system if R1 = 10 Ω, R2 = 5 Ω, L = 1 h, C = 0.2 F,

    E(t) =

    i2(0) = 0, and i3(0) = 0.

    (b) Determine the current i1(t).

    The electrical network consists of a voltage E, an inductor L, two resistors R subscript 1 and R subscript 2, and a capacitor C. The voltage E, the resistor R subscript 1, and the capacitor C are connected parallel to each other. The inductor L and the resistor R subscript 2 are connected in series to the voltage E, the resistor R subscript 1, and the capacitor C. The resistor R subscript 1 is connected between L and R subscript 2. The resistor R subscript 1 has blobs at both ends of its connection. A current i subscript 1 from the voltage E flows through the inductor L, current i subscript 2 through the resistor R subscript 1, and current i subscript 3 through the resistor R subscript 2.

    FIGURE 4.6.7 Network in Problem 16

  17. Solve the system given in (17) of Section 2.9 when R1 = 6 Ω, R2 = 5 Ω, L1 = 1 h, L2 = 1 h, E(t) = 50 sin t V, i2(0) = 0, and i3(0) = 0.
  18. Solve (6) when E = 60 V, L = h, R = 50 Ω, C = 10−4 F, i1(0) = 0, and i2(0) = 0.
  19. Solve (6) when E = 60 V, L = 2 h, R = 50 Ω, C = 10−4 F, i1(0) = 0, and i2(0) = 0.
    1. Show that the system of differential equations for the charge on the capacitor q(t) and the current i3(t) in the electrical network shown in FIGURE 4.6.8 is

    2. Find the charge on the capacitor when L = 1 h, R1 = 1 Ω, R2 = 1 Ω, C = 1 F,

      E(t) = ,

      i3(0) = 0, and q(0) = 0.

      The electrical network consists of a voltage E, an inductor L, two resistors R subscript 1 and R subscript 2, and a capacitor C. The voltage E, the inductor L, and the capacitor C are connected parallel to each other. The resistor R subscript 1 is connected in series between the voltage E and the capacitor C. The resistor R subscript 2 is connected in series between the inductor L and the capacitor C. The capacitor C has blobs at both ends of its connection. A current i subscript 1 from the voltage E flows through the resistor R subscript 1, current i subscript 2 through the capacitor C, and current i subscript 3 through the inductor L.

      FIGURE 4.6.8 Network in Problem 20

Mathematical Models

  1. Range of a Projectile—No Air Resistance If you worked Problem 23 in Exercises 3.12, you saw that when air resistance and all other forces except its weight w = mg are ignored, the path of motion of a ballistic projectile, such as a cannon shell, is described by the system of linear differential equations

    (10)

    1. If the projectile is launched from level ground with an initial velocity v0 assumed to be tangent to its path of motion or trajectory, then the initial conditions accompanying the system are x(0) = 0, x′(0) = v0 cos θ, y(0) = 0, y′(0) = v0 sin θ where the initial speed is constant and θ is the constant angle of elevation of the cannon. See Figure 3.R.5 in The Paris Guns problem on page 212. Use the Laplace transform to solve system (10).
    2. The solutions x(t) and y(t) of the system in part (a) are parametric equations of the trajectory of the projectile. By using x(t) to eliminate the parameter t in y(t) show that the trajectory is parabolic.
    3. Use the results of part (b) to show that the horizontal range R of the projectile is given by

      (11)

      From (11) we not only see that R is a maximum when θ = π/4 but also that a projectile launched at distinct complementary angles θ and π/2 − θ has the same submaximum range. See FIGURE 4.6.9. Use a trigonometric identity to prove this last result.

    4. Show that the maximum height H of the projectile is given by

      (12)

    5. Suppose g = 32 ft/s2, θ = 38°, and v0 = 300 ft/s. Use (11) and (12) to find the horizontal range and maximum height of the projectile. Repeat with θ = 52° and v0 = 300 ft/s.
    6. Because formulas (11) and (12) are not valid in all cases (see Problem 22), it is advantageous to you to remember that the range and maximum height of a ballistic projectile can be obtained by working directly with x(t) and y(t), that is, by solving y(t) = 0 and y′(t) = 0. The first equation gives the time when the projectile hits the ground and the second gives the time when y(t) is a maximum. Find these times and verify the range and maximum height obtained in part (e) for the trajectory with θ = 38° and v0 = 300 ft/s. Repeat with θ = 52°.
    7. With g = 32 ft/s2, θ = 38° and v0 = 300 ft/s use a graphing utility or CAS to plot the trajectory of the projectile defined by the parametric equations x(t) and y(t) in part (a). Repeat with θ = 52°. Using different colors superimpose both curves on the same coordinate system.
      Five curves are graphed on an x y plane. The curve labeled theta equals 15 degrees starts from the origin, goes slightly up and to the right, then goes down and to the right, and ends at a point almost midway on the positive x axis. The curve labeled theta equals 30 degrees starts from the origin, goes up and to the right, then goes down and to the right, and ends at a point on the positive x axis away from the endpoint of the curve labeled theta equals 15 degrees. The curve labeled theta equals 45 degrees starts from the origin, goes up above the Two graphed curves and to the right, then goes down and to the right, and ends at a point on the positive x axis away from the endpoint of the curve labeled theta equals 30 degrees. The curve labeled theta equals 60 degrees starts from the origin, goes up above the three graphed curves and to the right, then goes down and to the right, and ends at a point on the positive x axis which is the endpoint of the curve labeled theta equals 30 degrees. The curve labeled theta equals 75 degrees starts from the origin, goes up above the four graphed curves and to the right, then goes down and to the right, and ends at a point on the positive x axis which is the endpoint of the curve labeled theta equals 15 degrees. The distance on the positive x axis from the origin to the endpoint of the curve labeled theta equals 15 degrees is labeled range R.

      FIGURE 4.6.9 Projectiles in Problem 21

  2. Range of a Projectile—With Linear Air Resistance In The Paris Guns problem on pages 209–214, the effect that nonlinear air resistance has on the trajectory of a cannon shell was examined numerically. In this problem we consider linear air resistance on a projectile.
    1. Suppose that air resistance is a retarding force tangent to the path of the projectile but acts opposite to the motion. If we take air resistance to be proportional to the velocity of the projectile, then from Problem 24 of Exercises 3.12 the motion of the projectile is described by the system of linear differential equations

      (13)

      where β > 0 is a constant. Use the Laplace transform to solve system (13) subject to the initial conditions in part (a) of Problem 21.

    2. Suppose g = 32 ft/s2, β = 0.02, θ = 38°, and v0 = 300 ft/s. Use a calculator or CAS to approximate the time when the projectile hits the ground and then compute x(t) to find its corresponding horizontal range.
    3. The complementary-angle property in part (c) of Problem 21 does not hold when air resistance is taken into consideration. To show this, repeat part (b) using the complementary angle θ = 52° and compare the horizontal range with that found in part (b).
    4. With , g = 32 ft/s2, β = 0.02, θ = 38°, and v0 = 300 ft/s, use a graphing utility or CAS to plot the trajectory of the projectile defined by the parametric equations x(t) and y(t). Repeat with θ = 52°. Using different colors, superimpose both of these curves along with the two curves in part (g) of Problem 21 on the same coordinate system.
  3. Spring-Coupled Pendulums
    1. Suppose two identical pendulums are coupled by means of a spring with constant k. See FIGURE 4.6.10. Under the same assumptions made in the discussion preceding Example 3 in Section 4.6, it can be shown that when the displacement angles and are small, the system of differential equations describing the motion is

      Use the Laplace transform to solve the system when where and are constants. For convenience, let and

    2. Use the solution in part (a) to discuss the motion of the coupled pendulums in the special case when the initial conditions are
    3. Repeat part (b) when the initial conditions are
      Two identical pendulums hung vertically are coupled by a spring. The length of each pendulum is labeled l. The mass at the bottom of each pendulum is labeled m. The first pendulum is displaced to the right by an angle theta subscript 1 measured from a vertical line extending downward from the pivot of the pendulum. The second pendulum is displaced to the right by an angle theta subscript 2 measured from a vertical line extending downward from the pivot of the pendulum.

      FIGURE 4.6.10 Coupled pendulums in Problem 23

Computer Lab Assignment

    1. Use the Laplace transform and the information given in Example 3 to obtain the solution (8) of the system given in (7).
    2. Use a graphing utility to plot the graphs of θ1(t) and θ2(t) in the tθ-plane. Which mass has extreme displacements of greater magnitude? Use the graphs to estimate the first time that each mass passes through its equilibrium position. Discuss whether the motion of the pendulums is periodic.
    3. As parametric equations, graph θ1(t) and θ2(t) in the θ1θ2-plane. The curve defined by these parametric equations is called a Lissajous curve.
    4. The position of the masses at t = 0 is given in Figure 4.6.3(a). Note that we have used 1 radian ≈ 57.3°. Use a calculator or a table application in a CAS to construct a table of values of the angles θ1 and θ2 for t = 1, 2, . . ., 10 seconds. Then plot the positions of the two masses at these times.
    5. Use a CAS to find the first time that θ1(t) = θ2(t) and compute the corresponding angular value. Plot the positions of the two masses at these times.
    6. Utilize a CAS to also draw appropriate lines to simulate the pendulum rods as in Figure 4.6.3. Use the animation capability of your CAS to make a “movie” of the motion of the double pendulum from t = 0 to t = 10 using a time increment of 0.1. [Hint: Express the coordinates (x1(t), y1(t)) and (x2(t), y2(t)) of the masses m1 and m2, respectively, in terms of θ1(t) and θ2(t).]