Initial-Value Problem

In Section 1.2 we defined an initial-value problem for a general nth-order differential equation. For a linear differential equation, an nth-order initial-value problem (IVP) is

(1)

Recall that for a problem such as this, we seek a function defined on some interval I containing x₀ that satisfies the differential equation and the n initial conditions specified at x₀: y(x₀) = y₀, y′(x₀) = y₁, …, y⁽ⁿ⁻¹⁾(x₀) = y_n−1. We have already seen that in the case of a second-order initial-value problem, a solution curve must pass through the point (x₀, y₀) and have slope y₁ at this point.

Existence and Uniqueness

In Section 1.2 we stated a theorem that gave conditions under which the existence and uniqueness of a solution of a first-order initial-value problem were guaranteed. The theorem that follows gives sufficient conditions for the existence of a unique solution of the problem in (1).

EXAMPLE 1 Unique Solution of an IVP

EXAMPLE 2 Unique Solution of an IVP

The requirements in Theorem 3.1.1 that a_i(x), i = 0, 1, 2, …, n be continuous and a_n(x) ≠ 0 for every x in I are both important. Specifically, if a_n(x) = 0 for some x in the interval, then the solution of a linear initial-value problem may not be unique or even exist. For example, you should verify that the function y = cx² + x + 3 is a solution of the initial-value problem

x²y″ − 2xy′ + 2y = 6, y(0) = 3, y′(0) = 1

on the interval (−∞, ∞) for any choice of the parameter c. In other words, there is no unique solution of the problem. Although most of the conditions of Theorem 3.1.1 are satisfied, the obvious difficulties are that a₂(x) = x² is zero at x = 0 and that the initial conditions are also imposed at x = 0.

Boundary-Value Problem

Another type of problem consists of solving a linear differential equation of order two or greater in which the dependent variable y or its derivatives are specified at different points. A problem such as

is called a two-point boundary-value problem, or simply a boundary-value problem (BVP). The prescribed values y(a) = y₀ and y(b) = y₁ are called boundary conditions (BC). A solution of the foregoing problem is a function satisfying the differential equation on some interval I, containing a and b, whose graph passes through the two points (a, y₀) and (b, y₁). See FIGURE 3.1.1.

For a second-order differential equation, other pairs of boundary conditions could be

y′(a) = y₀, y(b) = y₁

y(a) = y₀, y′(b) = y₁

y′(a) = y₀, y′(b) = y₁,

where y₀ and y₁ denote arbitrary constants. These three pairs of conditions are just special cases of the general boundary conditions

A₁y(a) + B₁y′(a) = C₁

A₂y(b) + B₂y′(b) = C₂.

The next example shows that even when the conditions of Theorem 3.1.1 are fulfilled, a boundary-value problem may have several solutions (as suggested in Figure 3.1.1), a unique solution, or no solution at all.

EXAMPLE 3 A BVP Can Have Many, One, or No Solutions

In Example 10 of Section 1.1 we saw that the two-parameter family of solutions of the differential equation x″ + 16x = 0 is

x = c₁ cos 4t +c₂ sin 4t. (2)

(a) Suppose we now wish to determine that solution of the equation that further satisfies the boundary conditions x(0) = 0, x(π/2) = 0. Observe that the first condition 0 = c₁ cos 0 + c₂ sin 0 implies c₁ = 0, so that x = c₂ sin 4t. But when t = π/2, 0 = c₂ sin 2π is satisfied for any choice of c₂ since sin 2π = 0. Hence the boundary-value problem

(3)

has infinitely many solutions. FIGURE 3.1.2 shows the graphs of some of the members of the one-parameter family x = c₂ sin 4t that pass through the two points (0, 0) and (π/2, 0).

(b) If the boundary-value problem in (3) is changed to

, (4)

then x(0) = 0 still requires c₁ = 0 in the solution (2). But applying x(π/8) = 0 to x = c₂ sin 4t demands that 0 = c₂ sin(π/2) = c₂ · 1. Hence x = 0 is a solution of this new boundary-value problem. Indeed, it can be proved that x = 0 is the only solution of (4).

(c) Finally, if we change the problem to

x″ + 16x = 0, x(0) = 0, x(π/2) = 1, (5)

we find again that c₁ = 0 from x(0) = 0, but that applying x(π/2) = 1 to x = c₂ sin 4t leads to the contradiction 1 = c₂ sin 2π = c₂ · 0 = 0. Hence the boundary-value problem (5) has no solution. ≡

A linear nth-order differential equation of the form

(6)

is said to be homogeneous, whereas an equation

(7)

with g(x) not identically zero is said to be nonhomogeneous. For example, 2y″ + 3y′ − 5y = 0 is a homogeneous linear second-order differential equation, whereas x²y‴ + 6y′ + 10y = e^x is a nonhomogeneous linear third-order differential equation. The word homogeneous in this context does not refer to coefficients that are homogeneous functions as in Section 2.5; rather, the word has exactly the same meaning as in Section 2.3.

We shall see that in order to solve a nonhomogeneous linear equation (7), we must first be able to solve the associated homogeneous equation (6).

To avoid needless repetition throughout the remainder of this section, we shall, as a matter of course, make the following important assumptions when stating definitions and theorems about the linear equations (6) and (7). On some common interval I,

• the coefficients a_i(x), i = 0, 1, 2, …, n, are continuous;
• the right-hand member g(x) is continuous; and
• a_n(x) ≠ 0 for every x in the interval.

Differential Operators

In calculus, differentiation is often denoted by the capital letter D; that is, dy/dx = Dy. The symbol D is called a differential operator because it transforms a differentiable function into another function. For example, D(cos 4x) = −4 sin 4x, and D(5x³ − 6x²) = 15x² − 12x. Higher-order derivatives can be expressed in terms of D in a natural manner:

where y represents a sufficiently differentiable function. Polynomial expressions involving D, such as D + 3, D² + 3D − 4, and 5x³D³ − 6x²D² + 4xD + 9, are also differential operators. In general, we define an nth-order differential operator to be

L = a_n(x)Dⁿ + a_n−1(x)Dⁿ⁻¹ + … + a₁(x)D +a₀(x). (8)

As a consequence of two basic properties of differentiation, D(cf(x)) = cDf(x), c a constant, and D{f(x) + g(x)} = Df(x) + Dg(x), the differential operator L possesses a linearity property; that is, L operating on a linear combination of two differentiable functions is the same as the linear combination of L operating on the individual functions. In symbols, this means

L{αf(x) + βg(x)} = αL(f(x)) + βL(g(x)), (9)

where α and β are constants. Because of (9) we say that the nth-order differential operator L is a linear operator.

Differential Equations

Any linear differential equation can be expressed in terms of the D notation. For example, the differential equation y″ + 5y′ + 6y = 5x − 3 can be written as D²y + 5Dy + 6y = 5x − 3 or (D² + 5D + 6)y = 5x − 3. Using (8), the linear nth-order differential equations (6) and (7) can be written compactly as

L(y) = 0 and L(y) = g(x),

respectively.

Superposition Principle

In the next theorem we see that the sum, or superposition, of two or more solutions of a homogeneous linear differential equation is also a solution.

PROOF:

We prove the case k = 2. Let L be the differential operator defined in (8), and let y₁(x) and y₂(x) be solutions of the homogeneous equation L(y) = 0. If we define y = c₁y₁(x) + c₂y₂(x), then by linearity of L we have

L(y) = L{c₁y₁(x) + c₂y₂(x)} = c₁L(y₁) + c₂L(y₂) = c₁ · 0 + c₂ · 0 = 0. ≡

EXAMPLE 4 Superposition—Homogeneous DE

The function y = e^7x is a solution of y″ − 9y′ + 14y = 0. Since the differential equation is linear and homogeneous, the constant multiple y = ce^7x is also a solution. For various values of c we see that y = 9e^7x, y = 0, , …, are all solutions of the equation.

Linear Dependence and Linear Independence

The next two concepts are basic to the study of linear differential equations.

In other words, a set of functions is linearly independent on an interval if the only constants for which

c₁f₁(x) + c₂f₂(x) + … + c_nf_n(x) = 0

for every x in the interval are c₁ = c₂ = … = c_n = 0.

It is easy to understand these definitions in the case of two functions f₁(x) and f₂(x). If the functions are linearly dependent on an interval, then there exist constants c₁ and c₂ that are not both zero such that for every x in the interval c₁f₁(x) + c₂f₂(x) = 0. Therefore, if we assume that c₁ ≠ 0, it follows that f₁(x) = (−c₂/c₁)f₂(x); that is,

Conversely, if f₁(x) = c₂f₂(x) for some constant c₂, then (−1) · f₁(x) + c₂f₂(x) = 0 for every x on some interval. Hence the functions are linearly dependent, since at least one of the constants (namely, c₁ = −1) is not zero. We conclude that:

For example, the functions f₁(x) = sin 2x and f₂(x) = sin x cos x are linearly dependent on (−∞, ∞) because f₁(x) is a constant multiple of f₂(x). Recall from the double angle formula for the sine that sin 2x = 2 sin x cos x. On the other hand, the functions f₁(x) = x and f₂(x) = | x | are linearly independent on (−∞, ∞). Inspection of FIGURE 3.1.3 should convince you that neither function is a constant multiple of the other on the interval.

It follows from the preceding discussion that the ratio f₂(x)/f₁(x) is not a constant on an interval on which f₁(x) and f₂(x) are linearly independent. This little fact will be used in the next section.

EXAMPLE 5 Linearly Dependent Functions

A set of n functions f₁(x), f₂(x), …, f_n(x) is linearly dependent on an interval I if at least one of the functions can be expressed as a linear combination of the remaining functions. For example, three functions f₁(x), f₂(x), and f₃(x) are linearly dependent on I if at least one of these functions is a linear combination of the other two, say,

f₃(x) = c₁f₁(x) + c₂f₂(x)

for all x in I. A set of n functions is linearly independent on I if no one function is a linear combination of the other functions.

EXAMPLE 6 Linearly Dependent Functions

Solutions of Differential Equations

We are primarily interested in linearly independent functions or, more to the point, linearly independent solutions of a linear differential equation. Although we could always appeal directly to Definition 3.1.1, it turns out that the question of whether n solutions y₁, y₂, …, y_n of a homogeneous linear nth-order differential equation (6) are linearly independent can be settled somewhat mechanically using a determinant.

The Wronskian determinant is named after the Polish philosopher, inventor, lawyer, physicist, and mathematician Józef Maria Hoëné-Wronski (1776–1853).

It follows from Theorem 3.1.3 that when y₁, y₂, …, y_n are n solutions of (6) on an interval I, the Wronskian W(y₁, y₂, …,y_n) is either identically zero or never zero on the interval. Thus, if we can show that W(y₁, y₂, …, y_n) ≠ 0 for some x₀ in I, then the solutions y₁, y₂, …, y_n are linearly independent on I. For example, the functions

are solutions of the differential equation

x²y″ + 7xy′ + 13y = 0

on the interval (0, ∞). Note that the coefficient functions a₂(x) = x², a₁(x) = 7x, and a₀(x) = 13 are continuous on (0, ∞) and that a₂(x) ≠ 0 for every value of x in the interval. The Wronskian is

Rather than expanding this unwieldy determinant, we choose x = 1 in the interval (0, ∞) and find

The fact that W(y₁(1), y₂(1)) = 2 ≠ 0 is sufficient to conclude that y₁(x) and y₂(x) are linearly independent on (0, ∞).

A set of n linearly independent solutions of a homogeneous linear nth-order differential equation is given a special name.

The basic question of whether a fundamental set of solutions exists for a linear equation is answered in the next theorem.

Analogous to the fact that any vector in three dimensions can be expressed uniquely as a linear combination of the linearly independent vectors i, j, k, any solution of an nth-order homogeneous linear differential equation on an interval I can be expressed uniquely as a linear combination of n linearly independent solutions on I. In other words, n linearly independent solutions y₁, y₂, …, y_n are the basic building blocks for the general solution of the equation.

Theorem 3.1.5 states that if Y(x) is any solution of (6) on the interval, then constants C₁, C₂, …, C_n can always be found so that

Y(x) = C₁y₁(x) + C₂y₂(x) + … + C_n y_n(x).

We will prove the case when n = 2.

PROOF:

Let Y be a solution and y₁ and y₂ be linearly independent solutions of a₂y″ + a₁y′ + a₀y = 0 on an interval I. Suppose x = t is a point in I for which W(y₁(t), y₂(t)) ≠ 0. Suppose also that Y(t) = k₁ and Y′ (t) = k₂. If we now examine the equations

it follows that we can determine C₁ and C₂ uniquely, provided that the determinant of the coefficients satisfies

But this determinant is simply the Wronskian evaluated at x = t, and, by assumption, W ≠ 0. If we define G(x) = C₁y₁(x) + C₂y₂(x), we observe that G(x) satisfies the differential equation, since it is a superposition of two known solutions; G(x) satisfies the initial conditions

Y(x) satisfies the same linear equation and the same initial conditions. Since the solution of this linear initial-value problem is unique (Theorem 3.1.1), we have Y(x) = G(x) or Y(x) = C₁y₁(x) + C₂y₂(x). ≡

EXAMPLE 7 General Solution of a Homogeneous DE

EXAMPLE 8 A Solution Obtained from a General Solution

EXAMPLE 9 General Solution of a Homogeneous DE

Any function y_p free of arbitrary parameters that satisfies (7) is said to be a particular solution of the equation. For example, it is a straightforward task to show that the constant function y_p = 3 is a particular solution of the nonhomogeneous equation y″ + 9y = 27.

Now if y₁, y₂, …, y_k are solutions of (6) on an interval I and y_p is any particular solution of (7) on I, then the linear combination

y = c₁y₁(x) + c₂y₂(x) + … + c_ky_k(x) + y_p(x) (10)

is also a solution of the nonhomogeneous equation (7). If you think about it, this makes sense, because the linear combination c₁y₁(x) + c₂y₂(x) + … + c_ky_k(x) is mapped into 0 by the operator L = a_nDⁿ + a_n−1Dⁿ⁻¹ + … + a₁D + a₀, whereas y_p is mapped into g(x). If we use k = n linearly independent solutions of the nth-order equation (6), then the expression in (10) becomes the general solution of (7).

PROOF:

Let L be the differential operator defined in (8), and let Y(x) and y_p(x) be particular solutions of the nonhomogeneous equation L(y) = g(x). If we define u(x) = Y(x) − y_p(x), then by linearity of L we have

L(u) = L{Y(x) − y_p(x)} = L(Y(x)) − L(y_p(x)) = g(x) − g(x) = 0.

This shows that u(x) is a solution of the homogeneous equation L(y) = 0. Hence, by Theorem 3.1.5, u(x) = c₁y₁(x) + c₂y₂(x) + … + c_ny_n(x), and so

Y(x) − y_p(x) = c₁y₁(x) + c₂y₂(x) + … + c_ny_n(x)

or

Y(x) = c₁y₁(x) + c₂y₂(x) + … + c_ny_n(x) + y_p(x). ≡

Complementary Function

We see in Theorem 3.1.6 that the general solution of a nonhomogeneous linear equation consists of the sum of two functions:

y = c₁y₁(x) + c₂y₂(x) + … + c_ny_n(x) + y_p(x) = y_c(x) + y_p(x).

The linear combination y_c(x) = c₁y₁(x) + c₂y₂(x) + … + c_ny_n(x), which is the general solution of (6), is called the complementary function for equation (7). In other words, to solve a nonhomogeneous linear differential equation we first solve the associated homogeneous equation and then find any particular solution of the nonhomogeneous equation. The general solution of the nonhomogeneous equation is then

EXAMPLE 10 General Solution of a Nonhomogeneous DE

Another Superposition Principle

The last theorem of this discussion will be useful in Section 3.4, when we consider a method for finding particular solutions of nonhomogeneous equations.

PROOF:

We prove the case k = 2. Let L be the differential operator defined in (8), and let y_p1(x) and y_p2(x) be particular solutions of the nonhomogeneous equations L(y) = g₁(x) and L(y) = g₂(x), respectively. If we define y_p(x) = (x) + (x), we want to show that y_p is a particular solution of L(y) = g₁(x) + g₂(x). The result follows again by the linearity of the operator L:

L(y_p) = L{(x) + (x)} = L((x)) + L((x)) = g₁(x) + g₂(x). ≡

EXAMPLE 11 Superposition—Nonhomogeneous DE

If the are particular solutions of (12) for i = 1, 2, …, k, then the linear combination

where the c_i are constants, is also a particular solution of (14) when the right-hand member of the equation is the linear combination

c₁g₁(x) + c₂g₂(x) + … + c_kg_k(x).

Before we actually start solving homogeneous and nonhomogeneous linear differential equations, we need one additional bit of theory presented in the next section.

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

3.1.1 Initial-Value and Boundary-Value Problems

In Problems 1–4, the given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem.

1. y = c₁e^x + c₂e^−x, (−∞, ∞); y″ − y = 0, y(0) = 0, y′(0) = 1
2. y = c₁e^4x + c₂e^−x, (−∞, ∞); y″ − 3y′ − 4y = 0, y(0) = 1, y′(0) = 2
3. y = c₁x + c₂x ln x, (0, ∞); x²y″ − xy′ + y = 0, y(1) = 3, y′(1) = −1
4. y = c₁ + c₂ cos x + c₃ sin x, (−∞, ∞); y‴ + y′ = 0, y(π) = 0, y′(π) = 2, y″(π) = −1
5. Given that y = c₁ + c₂x² is a two-parameter family of solutions of xy″ − y′ = 0 on the interval (−∞, ∞), show that constants c₁ and c₂ cannot be found so that a member of the family satisfies the initial conditions y(0) = 0, y′(0) = 1. Explain why this does not violate Theorem 3.1.1.
6. Find two members of the family of solutions in Problem 5 that satisfy the initial conditions y(0) = 0, y′(0) = 0.
7. Given that x(t) = c₁ cos ωt + c₂ sin ωt is the general solution of x″ + ω²x = 0 on the interval (−∞, ∞), show that a solution satisfying the initial conditions x(0) = x₀, x′(0) = x₁, is given by
   

   x(t) = x₀ cos ωt + sin ωt.

8. Use the general solution of x″ + ω²x = 0 given in Problem 7 to show that a solution satisfying the initial conditions x(t₀) = x₀, x′(t₀) = x₁, is the solution given in Problem 7 shifted by an amount t₀:
   

   x(t) = x₀ cos ω (t − t₀) + sin ω (t − t₀).

In Problems 9 and 10, find an interval centered about x = 0 for which the given initial-value problem has a unique solution.

9. (x − 2)y″ + 3y = x,y(0) = 0, y′(0) = 1
10. y″ + (tan x)y = e^x, y(0) = 1, y′(0) = 0
11. 1. Use the family in Problem 1 to find a solution of y″ − y = 0 that satisfies the boundary conditions y(0) = 0, y(1) = 1.
    2. The DE in part (a) has the alternative general solution y = c₃ cosh x + c₄ sinh x on (−∞, ∞). Use this family to find a solution that satisfies the boundary conditions in part (a).
    3. Show that the solutions in parts (a) and (b) are equivalent.
12. Use the family in Problem 5 to find a solution of xy″ − y′ = 0 that satisfies the boundary conditions y(0) = 1, y′(1) = 6.

In Problems 13 and 14, the given two-parameter family is a solution of the indicated differential equation on the interval (−∞, ∞). Determine whether a member of the family can be found that satisfies the boundary conditions.

13. y = c₁e^x cos x + c₂e^x sin x; y″ −2y′ + 2y = 0
    1. y(0) = 1, y′(π) = 0
    2. y(0) = 1, y(π) = −1
    3. y(0) = 1, y = 1
    4. y(0) = 0, y(π) = 0
14. y = c₁x² + c₂x⁴ + 3; x²y″ − 5xy′ + 8y = 24
    1. y(−1) = 0, y(1) = 4
    2. y(0) = 1, y(1) = 2
    3. y(0) = 3, y(1) = 0
    4. y(1) = 3, y(2) = 15

3.1.2 Homogeneous Equations

In Problems 15–22, determine whether the given set of functions is linearly dependent or linearly independent on the interval (−∞, ∞).

15. f₁(x) = x, f₂(x) = x², f₃(x) = 4x −3x²
16. f₁(x) = 0, f₂(x) = x, f₃(x) = e^x
17. f₁(x) = 5, f₂(x) = cos²x, f₃(x) = sin²x
18. f₁(x) = cos 2x, f₂(x) = 1, f₃(x) = cos²x
19. f₁(x) = x, f₂(x) = x − 1, f₃(x) = x + 3
20. f₁(x) = 2 + x, f₂(x) = 2 + | x |
21. f₁(x) = 1 + x, f₂(x) = x, f₃(x) = x²
22. f₁(x) = e^x, f₂(x) = e^−x, f₃(x) = sinh x

In Problems 23–30, verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. Form the general solution of the equation.

23. y″ − y′ − 12y = 0; e^−3x, e^4x, (−∞, ∞)
24. y″ − 4y = 0; cosh 2x, sinh 2x, (−∞, ∞)
25. y″ − 2y′ + 5y = 0; e^x cos 2x,e^x sin 2x, (−∞, ∞)
26. 4y″ − 4y′ + y = 0; e^x/2, xe^x/2, (−∞, ∞)
27. x²y″ − 6xy′ + 12y = 0; x³, x⁴, (0, ∞)
28. x²y″ + xy′ + y = 0; cos(ln x), sin(ln x), (0, ∞)
29. x³y‴ + 6x²y″ + 4xy′ − 4y = 0; x, x⁻², x⁻² ln x, (0, ∞)
30. y⁽⁴⁾ + y″ = 0; 1, x, cos x, sin x, (−∞, ∞)

3.1.3 Nonhomogeneous Equations

In Problems 31–34, verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval.

31. y″ − 7y′ + 10y = 24e^x;
    y = c₁e^2x + c₂e^5x + 6e^x, (−∞, ∞)
32. y″ + y = sec x;
    y = c₁ cos x + c₂ sin x + x sin x + (cos x) ln(cos x),
    (− π/2,π/2)
33. y″ − 4y′ + 4y = 2e^2x + 4x − 12;
    y = c₁e^2x + c₂xe^2x + x²e^2x + x − 2, (−∞, ∞)
34. 2x²y″ + 5xy′ + y = x² − x;
    y = c₁ + c₂x⁻¹ + x² − x, (0, ∞)
35. (a) Verify that = 3e^2x and = x² + 3x are, respectively, particular solutions of
    

    y″ − 6y′ + 5y = −9e^2x

    and

    y″ − 6y′ + 5y = 5x² + 3x −16.

    (b) Use part (a) to find particular solutions of

    y″ − 6y′ + 5y = 5x² + 3x − 16 − 9e^2x

    and

    y″ − 6y′ + 5y = −10x² − 6x + 32 + e^2x.

36. (a) By inspection, find a particular solution of
    

    y″ + 2y = 10.

    (b) By inspection, find a particular solution of

    y″ + 2y = −4x.

    (c) Find a particular solution of y″ + 2y = −4x + 10.

    (d) Find a particular solution of y″ + 2y = 8x + 5.

Discussion Problems

37. Let n = 1, 2, 3, …. Discuss how the observations Dⁿxⁿ⁻¹ = 0 and Dⁿxⁿ = n! can be used to find the general solutions of the given differential equations.
    1. y″ = 0
    2. y‴ = 0
    3. y⁽⁴⁾ = 0
    4. y″ = 2
    5. y‴ = 6
    6. y⁽⁴⁾ = 24
38. Suppose that y₁ = e^x and y₂ = e^−x are two solutions of a homogeneous linear differential equation. Explain why y₃ = cosh x and y₄ = sinh x are also solutions of the equation.
39. 1. Verify that y₁ = x³ and y₂ = | x |³ are linearly independent solutions of the differential equation x²y″ −4xy′ + 6y = 0 on the interval (−∞, ∞).
    2. For the functions y₁ and y₂ in part (a), show that W(y₁, y₂) = 0 for every real number x. Does this result violate Theorem 3.1.3? Explain.
    3. Verify that Y₁ = x³ and Y₂ = x² are also linearly independent solutions of the differential equation in part (a) on the interval (−∞, ∞).
    4. Besides the functions y₁, y₂, Y₁, and Y₂ in parts (a) and (c), find a solution of the differential equation that satisfies y(0) = 0, y′(0) = 0.
    5. By the superposition principle, Theorem 3.1.2, both linear combinations y = c₁y₁ + c₂y₂ and Y = c₁Y₁ + c₂Y₂ are solutions of the differential equation. Discuss whether one, both, or neither of the linear combinations is a general solution of the differential equation on the interval (−∞, ∞).
40. Is the set of functions f₁(x) = e^x+2, f₂(x) = e^x−3 linearly dependent or linearly independent on the interval (−∞, ∞)? Discuss.
41. By substituting into the differential equation
    

    

    find four linearly independent solutions of the equation. Give the general solution of the equation on (−∞, ∞).

42. Suppose y₁, y₂, …, y_k are k linearly independent solutions on (−∞, ∞) of a homogeneous linear nth-order differential equation with constant coefficients. By Theorem 3.1.2 it follows that y_k+1 = 0 is also a solution of the differential equation. Is the set of solutions y₁, y₂, …, y_k, y_k+1 linearly dependent or linearly independent on (−∞, ∞)? Discuss.
43. Suppose that y₁, y₂, …, y_k are k nontrivial solutions of a homogeneous linear nth-order differential equation with constant coefficients and that k = n + 1. Is the set of solutions y₁, y₂, …, y_k linearly dependent or linearly independent on (−∞, ∞)? Discuss.
44. If and are particular solutions of the nonhomogeneous linear differential equation (7), then show that the difference of these particular solutions is a solution of the associated homogeneous differential equation (6).