Answer Problems 1−8 without referring back to the text. Fill in the blank or answer true/false.

1. The only solution of the initial-value problem y″ + x²y = 0, y(0) = 0, y′(0) = 0 is .
2. For the method of undetermined coefficients, the assumed form of the particular solution y_p for y″ − y = 1 + e^x is .
3. A constant multiple of a solution of a linear differential equation is also a solution.
4. If f₁ and f₂ are linearly independent functions on an interval I, then their Wronskian W(f₁, f₂) ≠ 0 for all x in I.
5. If a 10-pound weight stretches a spring 2.5 feet, a 32-pound weight will stretch it feet.
6. The period of simple harmonic motion of an 8-pound weight attached to a spring whose constant is 6.25 lb/ft is seconds.
7. The differential equation describing the motion of a mass attached to a spring is x″ + 16x = 0. If the mass is released at t = 0 from 1 meter above the equilibrium position with a downward velocity of 3 m/s, the amplitude of vibrations is meters.
8. If simple harmonic motion is described by x(t) = (/2) sin (2t + φ), the phase angle φ is when x(0) = − and x′(0) = 1.
9. Give an interval over which f₁(x) = x² and f₂(x) = x |x| are linearly independent. Then give an interval on which f₁ and f₂ are linearly dependent.
10. Without the aid of the Wronskian determine whether the given set of functions is linearly independent or linearly dependent on the indicated interval.
    

    (a) f₁(x) = ln x, f₂(x) = ln x², (0, )

    (b) f₁(x) = xⁿ, f₂(x) = xⁿ⁺¹, n = 1, 2, …, (−, )

    (c) f₁(x) = x, f₂(x) = x + 1, (−, )

    (d) f₁(x) = cos (x + ), f₂(x) = sin x, (−, )

    (e) f₁(x) = 0, f₂(x) = x, (−5, 5)

    (f) f₁(x) = 2, f₂(x) = 2x, (−, )

    (g) f₁(x) = x², f₂(x) = 1 − x², f₃(x) = 2 + x², (−, )

    (h) f₁(x) = xe^x+1, f₂(x) = (4x − 5)e^x, f₃(x) = xe^x, (−, )

11. Suppose m₁ = 3, m₂ = −5, and m₃ = 1 are roots of multiplicity one, two, and three, respectively, of an auxiliary equation. Write down the general solution of the corresponding homogeneous linear DE if it is
    

    (a) an equation with constant coefficients,

    (b) a Cauchy–Euler equation.

12. Find a Cauchy–Euler differential equation ax²y″ + bxy′ + cy = 0, where a,b, and c are real constants, if it is known that
    

    (a) m₁ = 3 and m₂ = −1 are roots of its auxiliary equation,

    (b) m₁ = i is a complex root of its auxiliary equation.

In Problems 13 and 14, verify that the indicated function is a solution of the given differential equation. Use formula (5) of Section 3.2 to find the general solution of the equation on the interval

In Problems 15−30, use the procedures developed in this chapter to find the general solution of each differential equation.

15. y″ − 2y′ − 2y = 0
16. 2y″ + 2y′ + 3y = 0
17. y‴ + 10y″ + 25y′ = 0
18. 2y‴ + 9y″ + 12y′ + 5y = 0
19. 3y‴ + 10y″ + 15y′ + 4y = 0
20. 2y⁽⁴⁾ + 3y‴ + 2y″ + 6y′ − 4y = 0
21. y″ − 3y′ + 5y = 4x³ − 2x
22. y″ − 2y′ + y = x²e^x
23. y‴ − 5y″ + 6y′ = 8 + 2 sin x
24. y‴ − y″ = 6
25. y″ − 2y′ + 2y = e^x tan x
26. y″ − y =
27. 6x²y″ + 5xy′ − y = 0
28. 2x³y‴ + 19x²y″ + 39xy′ + 9y = 0
29. x²y″ − 4xy′ + 6y = 2x⁴ + x²
30. x²y″ − xy′ + y = x³
31. Write down the form of the general solution y = y_c + y_p of the given differential equation in the two cases ω ≠ α and ω =α. Do not determine the coefficients in y_p.
    1. y″ + ω²y = sin α x
    2. y″ − ω²y = e^αx

32. (a) Given that y = sin x is a solution of y⁽⁴⁾ + 2y‴ + 11y″ + 2y′ + 10y = 0, find the general solution of the DE without the aid of a calculator or a computer.

    (b) Find a linear second-order differential equation with constant coefficients for which y₁ = 1 and y₂ = e^−x are solutions of the associated homogeneous equation and y_p = x² − x is a particular solution of the nonhomogeneous equation.

33. 1. Write the general solution of the fourth-order DE y⁽⁴⁾ − 2y″ + y = 0 entirely in terms of hyperbolic functions.
    2. Write down the form of a particular solution of y⁽⁴⁾ − 2y″ + y = sinh x.
34. Consider the differential equation x²y″ − (x² + 2x)y′ + (x + 2)y = x³. Verify that y₁ = x is one solution of the associated homogeneous equation. Then show that the method of reduction of order discussed in Section 3.2 leads both to a second solution y₂ of the homogeneous equation and to a particular solution y_p of the nonhomogeneous equation. Form the general solution of the DE on the interval (0, ).

In Problems 35−40, solve the given differential equation subject to the indicated conditions.

35. y″ − 2y′ + 2y = 0, y = 0, y(π) = −1
36. y″ + 2y′ + y = 0, y(−1) = 0, y′(0) = 0
37. y″ − y = x + sin x, y(0) = 2, y′(0) = 3
38. y″ + y = sec³x, y(0) = 1, y′(0) =
39. y′ y″ = 4x, y(1) = 5, y′(1) = 2
40. 2y″ = 3y², y(0) = 1, y′(0) = 1
41. 1. Use a CAS as an aid in finding the roots of the auxiliary equation for 12y⁽⁴⁾ + 64y ‴ + 59y″ − 23y′ − 12y = 0. Give the general solution of the equation.
    2. Solve the DE in part (a) subject to the initial conditions y(0) = −1, y′(0) = 2, y″(0) = 5, y ‴(0) = 0. Use a CAS as an aid in solving the resulting systems of four equations in four unknowns.
42. Find a member of the family of solutions of
    

    xy″ + y′ + = 0

    whose graph is tangent to the x-axis at x = 1. Use a graphing utility to obtain the solution curve.

In Problems 43−46, use systematic elimination to solve the given system of differential equations.

43. 
44. 
45. 
46. 
47. A free undamped spring/mass system oscillates with a period of 3 s. When 8 lb is removed from the spring, the system then has a period of 2 s. What was the weight of the original mass on the spring?
48. A 12-pound weight stretches a spring 2 feet. The weight is released from a point 1 foot below the equilibrium position with an upward velocity of 4 ft/s.
    
    1. Find the equation describing the resulting simple harmonic motion.
    2. What are the amplitude, period, and frequency of motion?
    3. At what times does the weight return to the point 1 foot below the equilibrium position?
    4. At what times does the weight pass through the equilibrium position moving upward? Moving downward?
    5. What is the velocity of the weight at t = 3π/16 s?
    6. At what times is the velocity zero?
49. A spring with constant k = 2 is suspended in a liquid that offers a damping force numerically equal to four times the instantaneous velocity. If a mass m is suspended from the spring, determine the values of m for which the subsequent free motion is nonoscillatory.
50. A 32-pound weight stretches a spring 6 inches. The weight moves through a medium offering a damping force numerically equal to β times the instantaneous velocity. Determine the values of β for which the system will exhibit oscillatory motion.
51. A series circuit contains an inductance of L = 1 h, a capacitance of C = 10⁻⁴ F, and an electromotive force of E(t) = 100 sin 50t V. Initially the charge q and current i are zero.
    
    1. Find the equation for the charge at time t.
    2. Find the equation for the current at time t.
    3. Find the times for which the charge on the capacitor is zero.
52. Show that the current i(t) in an LRC-series circuit satisfies the differential equation
    

    

    where E′(t) denotes the derivative of E(t).

53. Consider the boundary-value problem
    

    y″ + λy = 0, y(0) = y(2π), y′(0) = y′(2π).

    Show that except for the case λ = 0, there are two independent eigenfunctions corresponding to each eigenvalue.

54. Sliding Bead A bead is constrained to slide along a frictionless rod of length L. The rod is rotating in a vertical plane with a constant angular velocity ω about a pivot P fixed at the midpoint of the rod, but the design of the pivot allows the bead to move along the entire length of the rod. Let r(t) denote the position of the bead relative to this rotating coordinate system, as shown in FIGURE 3.R.1. In order to apply Newton’s second law of motion to this rotating frame of reference it is necessary to use the fact that the net force acting on the bead is the sum of the real forces (in this case, the force due to gravity) and the inertial forces (coriolis, transverse, and centrifugal). The mathematics is a little complicated, so we give just the resulting differential equation for r,
    

    = mω²r − mg sin (ωt).

    1. Solve the foregoing DE subject to the initial conditions r(0) = r₀, r′(0) = v₀.
    2. Determine initial conditions for which the bead exhibits simple harmonic motion. What is the minimum length L of the rod for which it can accommodate simple harmonic motion of the bead?
    3. For initial conditions other than those obtained in part (b), the bead must eventually fly off the rod. Explain using the solution r(t) in part (a).
    4. Suppose ω = 1 rad/s. Use a graphing utility to plot the graph of the solution r(t) for the initial conditions r(0) = 0, r′(0) = v₀, where v₀ is 0, 10, 15, 16, 16.1, and 17.
    5. Suppose the length of the rod is L = 40 ft. For each pair of initial conditions in part (d), use a root-finding application to find the total time that the bead stays on the rod.
       

       

55. Suppose a mass m lying on a flat, dry, frictionless surface is attached to the free end of a spring whose constant is k. In FIGURE 3.R.2(a) the mass is shown at the equilibrium position x = 0; that is, the spring is neither stretched nor compressed. As shown in Figure 3.R.2(b), the displacement x(t) of the mass to the right of the equilibrium position is positive and negative to the left. Derive a differential equation for the free horizontal (sliding) motion of the mass. Discuss the difference between the derivation of this DE and the analysis leading to (1) of Section 3.8.
    

    

56. What is the differential equation of motion in Problem 55 if kinetic friction (but no other damping forces) acts on the sliding mass? [Hint: Assume that the magnitude of the force of kinetic friction is f_k = µmg, where mg is the weight of the mass and the constant µ > 0 is the coefficient of kinetic friction. Then consider two cases: x′ > 0 and x′ < 0. Interpret these cases physically.]

In Problems 57 and 58, use a Green’s function to solve the given initial-value problem.

57. 
58. 
59. The Caught Pendulum Suppose the massless rod in the discussion of the nonlinear pendulum in Section 3.11 is actually a string of length l. A mass m is attached to the end of the string and the pendulum is released from rest at a small displacement angle When the pendulum reaches the equilibrium position, the dashed red line in FIGURE 3.R.3, the string hits a nail and gets caught at this point above the mass. The mass oscillates from this new pivot point as shown in the figure.
    
    1. Construct and solve a linear initial-value problem that gives the displacement angle, denote it for where T represents the time when the string first hits the nail.
    2. Find the time T in part (a).
    3. Construct and solve a linear initial-value problem that gives the displacement angle, denote it for where T is the time in part (a). Compare the amplitude and period of oscillations in this case with that predicted by the initial-value problem in part (a).
       

       

60. The Caught Pendulum—Continued
    1. Use a graphing utility to obtain the graphs of the displacement angles and in parts (a) and (c) of Problem 59. Use the same coordinate axes. Take and .
    2. Then use a numerical solver or a CAS to obtain the graphs of the solutions and of the corresponding nonlinear initial-value problems and compare with part (a). Take and .
    3. Experiment with values of until you discern a noticeable difference between the solutions of the linear and nonlinear initial-value problems.
61. Use a Maclaurin series to show that a power series solution of the initial-value problem
    

    

    is given by

    .

    [Hint: See Example 3 in Section 3.7.]

62. Spring Pendulum The rotational form of Newton’s second law of motion is:
    

    The time rate of change of angular momentum about a point is equal to the moment of the resultant force (torque).

In the absence of damping or other external forces, an analogue of (14) in Section 3.11 for the pendulum shown in Figure 3.11.3 is then

. (1)

1. When m and l are constant show that (1) reduces to (6) of Section 3.11.
2. Now suppose the rod in Figure 3.11.3 is replaced with a spring of negligible mass. When a mass m is attached to its free end the spring hangs in the vertical equilibrium position shown in FIGURE 3.R.4 and has length . When the spring pendulum is set in motion we assume that the motion takes place in a vertical plane and the spring is stiff enough not to bend. For the length of the spring is then , where x is the displacement from the equilibrium position. Find the differential equation for the displacement angle defined by (1).
   

   

Contributed Problem

Jeff Dodd, Professor
Department of Mathematical Sciences, Jacksonville State University

63. The Paris Guns The first mathematically correct theory of projectile motion was originally formulated by the Italian astronomer/mathematician Galileo Galilei (1564–1642). Galileo’s younger Italian collaborators Bonaventura Cavalieri (1598–1647) and Evangelista Torricelli (1608–1647) then clarified and extended his theory. Galileo’s theory was based on two simple hypotheses suggested by experimental observations: that a projectile moves with constant horizontal velocity and with constant downward vertical acceleration. Galileo, Cavalieri, and Torricelli did not have calculus at their disposal, so their arguments were largely geometric, but we can reproduce their results using a system of differential equations. Suppose that a projectile is launched from ground level at an angle θ with respect to the horizontal and with an initial velocity of magnitude m/s. Let the projectile’s height above the ground be y meters and its horizontal distance from the launch site be x meters, and for convenience take the launch site to be the origin in the xy-plane. Then Galileo’s hypotheses can be represented by the following initial-value problem:
    

    (2)

    where g = 9.8 m/s², x(0) = 0, y(0) = 0, x′(0) = v₀ cos θ (the x-component of the initial velocity), and y′(0) = v₀ sin θ (the y-component of the initial velocity). See FIGURE 3.R.5 and Problem 23 in Exercises 3.12.

    1. Note that the system of equations in (2) is decoupled; that is, it consists of separate differential equations for x(t) and y(t). Moreover, each of these differential equations can be solved simply by antidifferentiating twice. Solve (2) to obtain explicit formulas for x(t) and y(t) in terms of v₀ and θ. Then algebraically eliminate t to show that the trajectory of the projectile in the xy-plane is a parabola.
    2. A central question throughout the history of ballistics has been this: Given a gun that fires a projectile with a certain initial speed v₀, at what angle with respect to the horizontal should the gun be fired to maximize its range? The range is the horizontal distance traversed by a projectile before it hits the ground. Show that according to (2), the range of the projectile is so that a maximum range is achieved for the launch angle
    3. Show that the maximum height attained by the projectile if launched with for maximum range is
       

       

64. The Paris Guns—Continued Mathematically, Galileo’s model in Problem 63 is perfect. But in practice it is only as accurate as the hypotheses upon which it is based. The motion of a real projectile is resisted to some extent by the air, and the stronger this effect, the less realistic are the hypotheses of constant horizontal velocity and constant vertical acceleration, as well as the resulting independence of the projectile’s motion in the x andy directions.
    

    The first successful model of air resistance was formulated by the Dutch scientist Christiaan Huygens (1629–1695) and by the English polymath Sir Isaac Newton (1643–1727). It was based not so much on a detailed mathematical formulation of the underlying physics involved, which was beyond what anyone could manage at the time, but on physical intuition and groundbreaking experimental work. Newton’s version, which is known to this day as Newton’s law of air resistance, states that the resisting force or drag force on an object moving through a resisting medium acts in the direction opposite to the direction of the object’s motion with a magnitude proportional to the density ρ of the medium, the cross sectional area A of the object taken perpendicular to the direction of motion, and the square of the speed of the object. In modern vector notation, the drag force is a vector f_D given in terms of the velocity v of the object:

    

    where in the coordinate system of Problem 63 the velocity is the vector When the force of gravity and this drag force are combined according to Newton’s second law of motion we get:

    

    The initial-value problem (2) is then modified to read:

    

    (3)

    where x(0) = 0, y(0) = 0, x′(0) = v₀ cos θ, y′(0) = v₀ sin θ.

    Huygens seemed to believe that the proportionality of drag force to the square of the speed was universal, but Newton suspected that multiple different physical effects contributed to drag force, not all of which behaved that way. He turned out to be correct. For example, when the speed of an object is low enough compared to the viscosity (internal resistance to flow) of the medium, drag force ends up being approximately proportional to its speed (not the square of its speed), a relationship known as Stokes’ law of air resistance. Even today, there is no quick recipe for predicting the drag force for all objects under all conditions. The modeling of air resistance is complicated and is done in practice by a combination of theoretical and empirical methods. The coefficient C in (3) is called the drag coefficient. It is dimensionless (that is, it is the same no matter what units are used to measure mass, distance, and time) and it can usually be regarded as depending only on the shape of a projectile and not on its size. The drag coefficient is such a convenient index for measuring how much air resistance is felt by a projectile of a given shape that it is now defined in terms of the drag force to be even when this ratio cannot be regarded as constant. For example, under Stokes’ law of air resistance, C would be proportional to the reciprocal of the speed. Of greater concern to us is the fact that the drag coefficient of a projectile in air increases sharply as its speed approaches the speed of sound (approximately 340 m/s in air), then decreases gradually for even higher speeds, becoming nearly constant again for speeds several times the speed of sound. This was first discovered by the military engineer Benjamin Robins, whose book Principles of Gunnery is generally regarded as inaugurating the modern age of artillery and of the science of ballistics in general. As guns were used to shoot projectiles further and further with greater and greater initial speeds throughout the eighteenth and nineteenth centuries, the dependence of the drag coefficient on speed took on greater and greater practical importance. Moreover, as these projectiles went higher and higher, the fact that the density of the air decreases with increasing altitude also became important. By World War I, the density of the air as a function of altitude y above sea level in meters was commonly modeled by the function:

    

    and military engineers were accustomed to incorporating into (3) the dependence of C on velocity and of ρ on y. But one last major surprise was stumbled upon by German engineers during World War I. Our version of this story is based on the book Paris Kanonen—the Paris Guns (Wilhelmgeschütze) and Project HARP by Gerald V. Bull (Verlag E. S. Mittler & Sohn GmbH, Herford, 1988).

    In the fall of 1914, the German Navy charged the Friedrich Krupp engineering firm with designing a system (gun and shells) capable of bombing the English port of Dover from the French coast. This would require firing a shell approximately 37 kilometers, a range some 16 kilometers greater than had ever been achieved before. Krupp was ready for this challenge because it had already succeeded in designing and building shells with innovative shapes that had lower drag coefficients than any prewar shells. The drag coefficient for one of these shells can be well approximated by the following piecewise linear function, where the speed v is in m/s:

    

    In addition, Krupp’s engineers already had built an experimental gun having a 35.5 cm diameter barrel that could fire 535 kg shells with an initial speed of 940 m/s. They calculated that if they built one of their new low-drag shells with that diameter and mass and used this gun to launch it at a 43° angle to the horizontal, the shell should have a range of about 39 km. The shell was built, and a test firing was conducted on October 21, 1914. For the results, we quote a first-hand account by Professor Fritz Rausenberger, managing director of the Krupp firm at the time (from pages 24–25 in Bull’s book):

    After the firing of the first shot, with a top zone propelling charge and at 43°elevation, we all waited with anxiety for the spotting report to be telephoned back to us giving the location of the inert shell impact. The anxiety was that normally associated when trying to reach a range never before achieved. But the spotter’s report on impact never came. None of the observers located along the full length of the range had observed impact

    Since no observation posts had been established beyond the 40km mark, any impact outside of the area would have to be located and reported by local inhabitants using normal telephone communication between the neighbouring farms and villages. Thus it took several hours before the range staff received notification that the shell had impacted in a garden (without causing damage) some 49km down-range from the battery. This was an unexpectedly favorable result but raised the question of how the range increase of 25% over that predicted using standard exterior ballistic techniques occurred After careful study of the method of calculating range, it was clear that in the computations an average, constant air density was used which was larger than the average along the trajectory. The method of calculating trajectory was therefore changed to allow for variation of density along the trajectory. This was done by dividing the atmosphere into 3km bands from the Earth’s surface upwards. For each band an average air density value was determined and applied over that portion of the trajectory falling in this band. This step-by-step calculation technique was carried out from the muzzle until impact. The resultant calculated trajectory, using the drag coefficient as determined from small calibre firings, matched closely the experimental results from the 21st of October Meppen firing.

    Note that Rausenberger does not tell us what “average, constant” air density was used in the faulty calculation, or how it was determined. Actually, there is a logical problem here in that it is not possible to know how low the air density will become along the path of a trajectory without already knowing how high the trajectory will go. Nevertheless the engineers were confident of their calculations, so it seems likely that they did not regard the air density as a critical parameter when their concern was only to find an approximate range. (After all, they had never shot anything so high before.)

    1. Use a computer algebra system to write a routine that can numerically solve (3) with the piecewise defined C and exponentially decaying ρ given above and graph the resulting trajectory in the xy-plane. (You may need to rewrite (3) as a first-order system. See Section 6.4.) The area A is that of a circle with the diameter of the shell. Test your routine on the case ρ = 0, which was solved analytically in Problem 63.
    2. Suppose that, as Krupp engineers, we use the results of part (c) in Problem 63 to calculate the maximum height M that would have been attained by the test shell had it been launched in a vacuum (ρ = 0), then figure that the real test shell might reach about half that height, and finally settle on a “constant, average” value for ρ of (ρ(M/2) + ρ(0))/2. Plot the resulting trajectory, and show that the resulting range is uncannily close to that predicted by the Krupp engineers for the October 14, 1914, test.
    3. Note that the launch angle for this test was not the angle that yields maximum range in a vacuum, but a smaller angle. Does this smaller launch angle lead to greater range under the constant air density assumption that you used in part (b)? To the nearest degree, what launch angle yields maximum range according to this model?
    4. Now plot the trajectory of the test shell using the proper exponentially decaying ρ. What happens? (In evaluating the results, it should make you feel better that we are not pretending to take everything into account in this model. For example, a missile moving at an angle relative to its axis of symmetry can experience a substantial lift force of the same sort that makes airplanes fly. Our model does not account for the possibility of lift, the curvature and rotation of the Earth, or numerous other effects.)

    Rausenberger notes that once his engineers realized how important the exponentially decaying ρ was in calculating range, they did a series of calculations and found that the maximum range for their test would actually have been achieved with a launch angle of to . In hindsight, they realized that this was because a larger launch angle would result in the shell traveling higher and therefore in less dense air.

    5. Check this by plotting the trajectory of the test shell with exponentially decaying ρ every two degrees from to . What do you find?

    After these surprising results, engineers at Krupp became interested in the challenge of attaining even larger ranges. The only way they could think of to talk the German High Command into committing to the trouble and expense of pursuing this goal was to sell them on the possibility of bombing Paris from behind the German front line, which would require a range of some 120 km. The German High Command quickly approved this idea, and after several years of work Krupp produced what are now known as the Paris Guns. These guns were designed to launch a 106 kg shell having a diameter of 216 mm with an initial velocity of 1646 m/s. At a launch angle, such a shell was predicted to travel over 120 km.

    

    6. Simulate the trajectory of a shell from a Paris Gun using (2) with exponentially decaying ρ. Evaluate the results keeping in mind the caveats about our modeling noted in part (d). How high does the shell go? Now change the launch angle from to . What happens?

Seven Paris Guns were built, but only three were used. They fired a total of 351 shells towards Paris between March 23 and August 9 of 1918. The damage and casualties that they caused were not tactically significant. They were never intended to be so; there was no control over where the shells would fall in Paris, and the amount of explosive carried by each shell was quite small. Instead, they were intended as a form of intimidation, a “scare tactic.” However, military historians agree that they were not effective in that sense either. Their significance turned out to be more scientific than military. The shells that they launched were the first man-made objects to reach the stratosphere, initiating the space age in the science of ballistics.