INTRODUCTION

We have seen that the linear first-order DE y′ + ay = 0, where a is a constant, possesses the exponential solution y = c₁e^−ax on the interval (−∞, ∞). Therefore, it is natural to ask whether exponential solutions exist for homogeneous linear higher-order DEs

a_ny⁽ⁿ⁾ + a_n−1y⁽ⁿ⁻¹⁾ + … + a₁y′ + a₀y = 0, (1)

where the coefficients a_i, i = 0, 1, …, n are real constants and a_n ≠ 0. The surprising fact is that all solutions of these higher-order equations are either exponential functions or are constructed out of exponential functions.

Auxiliary Equation

We begin by considering the special case of a homogeneous linear second-order equation

ay″ + by′ + cy = 0. (2)

If we try a solution of the form y = e^mx, then after substituting y′ = me^mx and y″ = m²e^mx equation (2) becomes

am²e^mx + bme^mx + ce^mx = 0 or e^mx(am² + bm + c) = 0.

Since e^mx is never zero for real values of x, it is apparent that the only way that this exponential function can satisfy the differential equation (2) is to choose m as a root of the quadratic equation

am² + bm + c = 0. (3)

This last equation is called the auxiliary equation of the differential equation (2). Since the two roots of (3) are m₁ = (− b + and m₂ = (− b −, there will be three forms of the general solution of (1) corresponding to the three cases:

• m₁ and m₂ are real and distinct (b² − 4ac > 0),
• m₁ and m₂ are real and equal (b² − 4ac = 0), and
• m₁ and m₂ are conjugate complex numbers (b² − 4ac < 0).

We discuss each of these cases in turn.

Case I: Distinct Real Roots Under the assumption that the auxiliary equation (3) has two unequal real roots m₁ and m₂, we find two solutions, y₁ = and y₂ = , respectively. We see that these functions are linearly independent on (−∞, ∞) and hence form a fundamental set. It follows that the general solution of (2) on this interval is

y = c₁ + c₂. (4)

Case II: Repeated Real Roots When m₁ = m₂ we necessarily obtain only one exponential solution, y₁ = . From the quadratic formula we find that m₁ = −b/2a since the only way to have m₁ = m₂ is to have b² − 4ac = 0. It follows from the discussion in Section 3.2 that a second solution of the equation is

(5)

In (5) we have used the fact that − b/a = 2m₁. The general solution is then

y = c₁ + c₂x. (6)

Case III: Conjugate Complex Roots If m₁ and m₂ are complex, then we can write m₁ = α + iβ and m₂ = α − iβ, where α and β > 0 are real and i² = −1. Formally, there is no difference between this case and Case I, hence

y = C₁e^{(α + iβ)x} + C₂e^{(α − iβ)x}.

However, in practice we prefer to work with real functions instead of complex exponentials. To this end we use Euler’s formula:

e^iθ = cos θ + i sin θ,

where θ is any real number.^* It follows from this formula that

e^iβx = cos βx + i sin βx and e^−iβx = cos βx − i sin βx, (7)

where we have used cos(−βx) = cos βx and sin(−βx) = −sin βx. Note that by first adding and then subtracting the two equations in (7), we obtain, respectively,

e^iβx + e^−iβx = 2 cos βx and e^iβx − e^−iβx = 2i sin βx.

Since y = C₁e^(α+iβ)x + C₂e^(α−iβ)x is a solution of (2) for any choice of the constants C₁ and C₂, the choices C₁ = C₂ = 1 and C₁ = 1, C₂ = −1 give, in turn, two solutions:

Hence from Corollary (a) of Theorem 3.1.2 the last two results show that e^αx cos βx and e^αx sin βx are real solutions of (2). Moreover, these solutions form a fundamental set on (−∞, ∞). Consequently, the general solution of (2) on (−∞, ∞) is

y = c₁e^αx cos βx + c₂e^αx sin βx = e^αx(c₁ cos βx + c₂ sin βx). (8)

EXAMPLE 1 Second-Order DEs

EXAMPLE 2 An Initial-Value Problem

Two Equations Worth Knowing

The two linear differential equations

y″ + k²y = 0 and y″ − k²y = 0,

k real, are important in applied mathematics. For y″ + k²y = 0, the auxiliary equation m² + k² = 0 has imaginary roots m₁ = ki and m₂ = − ki. With α = 0 and β = k in (8), the general solution of the DE is seen to be

y = c₁ cos kx + c₂ sin kx. (9)

On the other hand, the auxiliary equation m² − k² = 0 for y″ − k²y = 0 has distinct real roots m₁ = k and m₂ = − k and so by (4) the general solution of the DE is

y = c₁e^kx + c₂e^−kx. (10)

Notice that if we choose c₁ = c₂ = and c₁ = , c₂ = − in (10), we get the particular solutions y = (e^kx + e^−kx) = cosh kx and y = (e^kx − e^−kx) = sinh kx. Since cosh kx and sinh kx are linearly independent on any interval of the x-axis, an alternative form for the general solution of y″ − k²y = 0 is

y = c₁ cosh kx + c₂ sinh kx. (11)

See Problems 43, 44, and 66 in Exercises 3.3.

Higher-Order Equations

In general, to solve an nth-order differential equation

a_n y⁽ⁿ⁾ + a_n−1 y⁽ⁿ⁻¹⁾ + … + a₂ y″ + a₁ y′ + a₀ y = 0, (12)

where the a_i, i = 0, 1, …, n are real constants, we must solve an nth-degree polynomial equation

a_nmⁿ + a_n−1mⁿ⁻¹ + … + a₂m² + a₁m + a₀ = 0. (13)

If all the roots of (13) are real and distinct, then the general solution of (12) is

It is somewhat harder to summarize the analogues of Cases II and III because the roots of an auxiliary equation of degree greater than two can occur in many combinations. For example, a fifth-degree equation could have five distinct real roots, or three distinct real and two complex roots, or one real and four complex roots, or five real but equal roots, or five real roots but with two of them equal, and so on. When m₁ is a root of multiplicity k of an nth-degree auxiliary equation (that is, k roots are equal to m₁), it can be shown that the linearly independent solutions are

, …,

and the general solution must contain the linear combination

Lastly, it should be remembered that when the coefficients are real, complex roots of an auxiliary equation always appear in conjugate pairs. Thus, for example, a cubic polynomial equation can have at most two complex roots.

EXAMPLE 3 Third-Order DE

EXAMPLE 4 Fourth-Order DE

Repeated Complex Roots

Example 4 illustrates a special case when the auxiliary equation has repeated complex roots. In general, if m₁ = α + iβ, β > 0, is a complex root of multiplicity k of an auxiliary equation with real coefficients, then its conjugate m₂ = α − iβ is also a root of multiplicity k. From the 2k complex-valued solutions

we conclude, with the aid of Euler’s formula, that the general solution of the corresponding differential equation must then contain a linear combination of the 2k real linearly independent solutions

In Example 4 we identify k = 2, α = 0, and β = 1.

Rational Roots

Of course the most difficult aspect of solving constant-coefficient differential equations is finding roots of auxiliary equations of degree greater than two. Something we can try is to test the auxiliary equation for rational roots. Recall from precalculus mathematics, if m₁ = p/q is a rational root (expressed in lowest terms) of a polynomial equation a_nmⁿ + . . . + a₁m + a₀ = 0 with integer coefficients, then the integer p is a factor of the constant term a₀ and the integer q is a factor of the leading coefficient a_n.

EXAMPLE 5 Finding Rational Roots

Use of Computers

Finding roots or approximations of roots of polynomial equations is a routine problem with an appropriate calculator or computer software. The computer algebra systems Mathematica and Maple can solve polynomial equations (in one variable) of degree less than five in terms of algebraic formulas. For the auxiliary equation in the preceding paragraph, the commands

yield immediately their representations of the roots . For auxiliary equations of higher degree it may be necessary to resort to numerical commands such as NSolve and FindRoot in Mathematica. Because of their capability of solving polynomial equations, it is not surprising that some computer algebra systems are also able to give explicit solutions of homogeneous linear constant-coefficient differential equations. For example, the inputs

give, respectively,

(14)

and

Translated, this means y = c₂e^−x cos x + c₁e^−x sin x is a solution of y″ + 2y′ + 2y = 0.

In the classic text Differential Equations by Ralph Palmer Agnew^* (used by the author as a student), the following statement is made:

Although it is debatable whether computing skills have improved in the intervening years, it is a certainty that technology has. If one has access to a computer algebra system, equation (15) could be considered reasonable. After simplification and some relabeling of the output, Mathematica yields the (approximate) general solution

We note in passing that the DSolve and dsolve commands in Mathematica and Maple, like most aspects of any CAS, have their limitations.

Finally, if we are faced with an initial-value problem consisting of, say, a fourth-order differential equation, then to fit the general solution of the DE to the four initial conditions we must solve a system of four linear equations in four unknowns (the c₁, c₂, c₃, c₄ in the general solution). Using a CAS to solve the system can save lots of time. See Problems 37, 38, 75, and 76 in Exercises 3.3.

3.3 Exercises Answers to selected odd-numbered problems begin on page ANS-5.

In Problems 1–14, find the general solution of the given second-order differential equation.

1. 4y″ + y′ = 0
2. y″ − 36y = 0
3. y″ − y′ − 6y = 0
4. y″ − 3y′ + 2y = 0
5. y″ + 8y′ + 16y = 0
6. y″ − 10y′ + 25y = 0
7. 12y″ − 5y′ − 2y = 0
8. y″ + 4y′ − y = 0
9. y″ + 9y = 0
10. 3y″ + y = 0
11. y″ − 4y′ + 5y = 0
12. 2y″ + 2y′ + y = 0
13. 3y″ + 2y′ + y = 0
14. 2y″ − 3y′ + 4y = 0

In Problems 15–30, find the general solution of the given higher-order differential equation.

15. y‴ − 4y″ − 5y′ = 0
16. y‴ − y = 0
17. y‴ − 5y″ + 3y′ + 9y = 0
18. y‴ + 3y″ − 4y′ − 12y = 0
19. 
20. 
21. y‴ + 3y″ + 3y′ + y = 0
22. y‴ − 6y″ + 12y′ − 8y = 0
23. y⁽⁴⁾ + y‴ + y″ = 0
24. y⁽⁴⁾ − 2y″ + y = 0
25. 
26. 
27. 
28. 
29. 
30. 

In Problems 31–38, solve the given initial-value problem.

31. y″ + 16y = 0, y(0) = 2, y′(0) = −2
32. ,
33. ,
34. 4y″ − 4y′ − 3y = 0, y(0) = 1, y′(0) = 5
35. y″ + y′ + 2y = 0, y(0) = y′(0) = 0
36. y″ − 2y′ + y = 0, y(0) = 5, y′(0) = 10
37. y‴ + 12y″ + 36y′ = 0, y(0) = 0, y′(0) = 1, y″(0) = −7
38. y‴ + 2y″ − 5y′ − 6y = 0, y(0) = y′(0) = 0, y″(0) = 1

In Problems 39−42, solve the given boundary-value problem.

39. y″ − 10y′ + 25y = 0, y(0) = 1, y(1) = 0
40. y″ + 4y = 0, y(0) = 0, y(π) = 0
41. y″ + y = 0, y′(0) = 0, y′(π/2) = 0
42. y″ − 2y′ + 2y = 0, y(0) = 1, y(π) = 1

In Problems 43 and 44, solve the given problem first using the form of the general solution given in (10). Solve again, this time using the form given in (11).

43. y″ − 3y = 0, y(0) = 1, y′(0) = 5
44. y″ − y = 0, y(0) = 1, y′(1) = 0

In Problems 45 and 46, find all numbers α > 0 so that the given boundary-value problem possesses nontrivial (that is, nonzero) solutions.

45. y″ + α²y = 0, y(0) = 0, y(2) = 0
46. y″ + α²y = 0, y′(0) = 0, y(4) = 0

In Problems 47–52, each figure represents the graph of a particular solution of one of the following differential equations:

1. y″ − 3y′ − 4y = 0
2. y″ + 4y = 0
3. y″ + 2y′ + y = 0
4. y″ + y = 0
5. y″ + 2y′ + 2y = 0
6. y″ − 3y′ + 2y = 0

Match a solution curve with one of the differential equations. Explain your reasoning.

In Problems 53–62, find a homogeneous linear differential equation with constant coefficients whose general solution is given.

Discussion Problems

63. Two roots of a cubic auxiliary equation with real coefficients are m₁ = − and m₂ = 3 + i. What is the corresponding homogeneous linear differential equation?
64. Find the general solution of y‴ + 6y″ + y′ − 34y = 0 if it is known that y₁ = e^−4x cos x is one solution.
65. To solve y⁽⁴⁾ + y = 0 we must find the roots of m⁴ + 1 = 0. This is a trivial problem using a CAS, but it can also be done by hand working with complex numbers. Observe that m⁴ + 1 = (m² + 1)² − 2m². How does this help? Solve the differential equation.
66. Verify that y = sinh x − 2 cos (x + π/6) is a particular solution of y⁽⁴⁾ − y = 0. Reconcile this particular solution with the general solution of the DE.
67. The complex number 3 + 2i is a root of multiplicity 3 of the auxiliary equation of a homogeneous linear sixth-order differential equation with real constant coefficients. Find the general solution of the differential equation.
68. Find the general solution of the differential equation
    

    

    if the auxiliary equation is

    

69. Consider the boundary-value problem y″ + λy = 0, y(0) = 0, y(π/2) = 0. Discuss: Is it possible to determine real values of λ so that the problem possesses (a) trivial solutions? (b) nontrivial solutions?
70. In the study of techniques of integration in calculus, certain indefinite integrals of the form ∫ e^ax f(x) dx could be evaluated by applying integration by parts twice, recovering the original integral on the right-hand side, solving for the original integral, and obtaining a constant multiple k ∫ e^ax f(x) dx on the left-hand side. Then the value of the integral is found by dividing by k. Discuss: For what kinds of functions f does the described procedure work? Your solution should lead to a differential equation. Carefully analyze this equation and solve for f.

Computer Lab Assignments

In Problems 71–74, use a computer either as an aid in solving the auxiliary equation or as a means of directly obtaining the general solution of the given differential equation. If you use a CAS to obtain the general solution, simplify the output and, if necessary, write the solution in terms of real functions.

71. y‴ − 6y″ + 2y′ + y = 0
72. 6.11y‴ + 8.59y″ + 7.93y′ + 0.778y = 0
73. 3.15y⁽⁴⁾ − 5.34y″ + 6.33y′ − 2.03y = 0
74. y⁽⁴⁾ + 2y″ − y′ + 2y = 0

In Problems 75 and 76, use a CAS as an aid in solving the auxiliary equation. Form the general solution of the differential equation. Then use a CAS as an aid in solving the system of equations for the coefficients c_i, i = 1, 2, 3, 4 that result when the initial conditions are applied to the general solution.

75. 2y⁽⁴⁾ + 3y‴ − 16y″ + 15y′ − 4y = 0,
    

    y(0) = −2, y′(0) = 6, y″(0) = 3, y‴(0) =

76. y⁽⁴⁾ − 3y‴ + 3y″ − y′ = 0,
    

    y(0) = y′(0) = 0, y″(0) = y‴(0) = 1

^* A formal derivation of Euler’s formula can be obtained from the Maclaurin series by substituting x = iθ, using i² = −1, i³ = − i, . . ., and then separating the series into real and imaginary parts. The plausibility thus established, we can adopt cos θ + i sin θ as the definition of e^iθ.

^*McGraw-Hill, New York, 1960.