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5 Higher-Order Linear Differential Equations

5.1 Introduction: Second-Order Linear Equations

In Chapters 1 and 2 we investigated first-order differential equations. We now turn to equations of higher order $n ≧ 2,$ beginning in this chapter with equations that are linear. The general theory of linear differential equations parallels the second-order case ( $n = 2$ ), which we outline in this initial section.

Recall that a second-order differential equation in the (unknown) function y(x) is one of the form

$G (x, y, y ′, y ″) = 0.$ (1)

This differential equation is said to be linear provided that G is linear in the dependent variable y and its derivatives $y ′$ and $y ″ .$ Thus a linear second-order equation takes (or can be written in) the form

$A (x) y ″ + B (x) y ′ + C (x) y = F (x) .$ (2)

Unless otherwise noted, we will always assume that the (known) coefficient functions A(x), B(x), C(x), and F(x) are continuous on some open interval I (perhaps unbounded) on which we wish to solve this differential equation, but we do not require that they be linear functions of x. Thus the differential equation

$e^{x} y ″ + (\cos x) y ′ + (1 + \sqrt{x)} y = \tan^{- 1} x$

is linear because the dependent variable y and its derivatives $y ′$ and $y ″$ appear linearly. By contrast, the equations

$y ″ = y y ′ and y ″ + 3 {(y ′)}^{2} + 4 y^{3} = 0$

are not linear because products and powers of y or its derivatives appear.

If the function F(x) on the right-hand side of Eq. (2) vanishes identically on I, then we call Eq. (2) a homogeneous linear equation; otherwise, it is nonhomogeneous. For example, the second-order equation

$x^{2} y ″ + 2 x y ′ + 3 y = \cos x$

is nonhomogeneous; its associated homogeneous equation is

$x^{2} y ″ + 2 x y ′ + 3 y = 0.$

In general, the homogeneous linear equation associated with Eq. (2) is

$A (x) y ″ + B (x) y ′ + C (x) y = 0.$ (3)

In case the differential equation in (2) models a physical system, the nonhomogeneous term F(x) frequently corresponds to some external influence on the system.

Remark

Note that the meaning of the term “homogeneous” for a second-order linear differential equation is quite different from its meaning for a first-order differential equation (as in Section 1.6). Of course, it is not unusual—either in mathematics or in the English language more generally—for the same word to have different meanings in different contexts.

A Typical Application

Linear differential equations frequently appear as mathematical models of mechanical systems and electrical circuits. For example, suppose that a mass m is attached both to a spring that exerts on it a force $F_{S}$ and to a dashpot (shock absorber) that exerts a force $F_{R}$ on the mass (Fig. 5.1.1). Assume that the restoring force $F_{S}$ of the spring is proportional to the displacement x of the mass from its equilibrium position and acts opposite to the direction of displacement. Then

$F_{S} = - k x (with k > 0)$

FIGURE 5.1.1.

A mass–spring–dashpot system.

so $F_{S} < 0$ if $x > 0$ (spring stretched) while $F_{S} > 0$ if $x < 0$ (spring compressed). We assume that the dashpot force $F_{R}$ is proportional to the velocity $v = d x / d t$ of the mass and acts opposite to the direction of motion. Then

$F_{R} = - c v = - c \frac{d x}{d t} (with c > 0),$

so $F_{R} < 0$ if $v > 0$ (motion to the right) while $F_{R} > 0$ if $v < 0$ (motion to the left).

If $F_{R}$ and $F_{S}$ are the only forces acting on the mass m (Fig. 5.1.2) and its resulting acceleration is $a = d v / d t,$ then Newton’s law $F = m a$ gives

$m x ″ = F_{S} + F_{R};$ (4)

FIGURE 5.1.2.

Directions of the forces acting on `m`.

that is,

$\begin{array}{l} m \frac{d^{2} x}{d t^{2}} + c \frac{d x}{d t} + k x = 0. \end{array}$ (5)

Thus we have a differential equation satisfied by the position function x(t) of the mass m. This homogeneous second-order linear equation governs the free vibrations of the mass; we will return to this problem in detail in Section 5.4.

If, in addition to $F_{S}$ and $F_{R},$ the mass m is acted on by an external force F(t)—which must then be added to the right-hand side in Eq. (4)—the resulting equation is

$\begin{array}{l} m \frac{d^{2} x}{d t^{2}} + c \frac{d x}{d t} + k x = F (t) . \end{array}$ (6)

This nonhomogeneous linear differential equation governs the forced vibrations of the mass under the influence of the external force F(t).

Homogeneous Second-Order Linear Equations

Consider the general second-order linear equation

$A (x) y ″ + B (x) y ′ + C (x) y = F (x),$ (7)

where the coefficient functions A, B, C, and F are continuous on the open interval I. Here we assume in addition that $A (x) \neq 0$ at each point of I, so we can divide each term in Eq. (7)by A(x) and write it in the form

$y ″ + p (x) y ′ + q (x) y = f (x) .$ (8)

We will discuss first the associated homogeneous equation

$y ″ + p (x) y ′ + q (x) y = 0.$ (9)

A particularly useful property of this homogeneous linear equation is the fact that the sum of any two solutions of Eq. (9) is again a solution, as is any constant multiple of a solution. This is the central idea of the following theorem.

Remark

According to Theorem 1 in Section 4.2, Theorem 1 here implies that the set $S$ of all solutions of the homogeneous linear second-order equation (9) is a subspace of the vector space of all functions on the real line R—and hence is itself a vector space. We therefore call S the solution space of the differential equation.

Example 1

We can see by inspection that

$y_{1} (x) = \cos x and y_{2} (x) = \sin x$

are two solutions of the equation

$y ″ + y = 0.$

Theorem 1 tells us that any linear combination of these solutions, such as

$y (x) = 3 y_{1} (x) - 2 y_{2} (x) = 3 \cos x - 2 \sin x,$

is also a solution. We will see later that, conversely, every solution of $y ″ + y = 0$ is a linear combination of these two particular solutions $y_{1}$ and $y_{2} .$ Thus a general solution of $y ″ + y = 0$ is given by

$y (x) = c_{1} \cos x + c_{2} \sin x .$

It is important to understand that this single formula for the general solution encompasses a “twofold infinity” of particular solutions, because the two coefficients $c_{1}$ and $c_{2}$ can be selected independently. Figures 5.1.3 through 5.1.5 illustrate some of the possibilities, with either $c_{1}$ or $c_{2}$ set equal to zero, or with both nonzero.

FIGURE 5.1.3.

Solutions $y (x) = c_{1} \cos x$ of $y ″ + y = 0$ .

FIGURE 5.1.4.

Solutions $y (x) = c_{2} \cos x$ of $y ″ + y = 0$ .

FIGURE 5.1.5.

Solutions of $y ″ + y = 0$ with $c_{1}$ and $c_{2}$ both nonzero.

Earlier in this section we gave the linear equation $m x ″ + c x ′ + k x = F (t)$ as a mathematical model of the motion of the mass shown in Fig. 5.1.1. Physical considerations suggest that the motion of the mass should be determined by its initial position and initial velocity. Hence, given any preassigned values of x(0) and $x ′ (0),$ Eq. (6) ought to have a unique solution satisfying these initial conditions. More generally, in order to be a “good” mathematical model of a deterministic physical situation, a differential equation must have unique solutions satisfying any appropriate initial conditions. The following existence and uniqueness theorem (proved in the Appendix) gives us this assurance for the general second-order equation.

Remark 1

Equation (8) and the conditions in (11) constitute a second-order linear initial value problem. Theorem 2 tells us that any such initial value problem has a unique solution on the whole interval I where the coefficient functions in (8) are continuous. Recall from Section 1.3 that a nonlinear differential equation generally has a unique solution on only a smaller interval.

Remark 2

Whereas a first-order differential equation $d y / d x = F (x, y)$ generally admits only a single solution curve $y = y (x)$ passing through a given initial point (a, b), Theorem 2 implies that the second-order equation in (8) has infinitely many solution curves passing through the point $(a, b_{0})$ —namely, one for each (real number) value of the initial slope $y ′ (a) = b_{1} .$ That is, instead of there being only one line through $(a, b_{0})$ tangent to a solution curve, every nonvertical straight line through $(a, b_{0})$ is tangent to some solution curve of Eq. (8). Figure 5.1.6 shows a number of solution curves of the equation $y ″ + 3 y ′ + 2 y = 0$ all having the same initial value $y (0) = 1,$ while Fig. 5.1.7 shows a number of solution curves all having the same initial slope $y ′ (0) = 1 .$ The application at the end of this section suggests how to construct such families of solution curves for a given homogeneous second-order linear differential equation.

FIGURE 5.1.6.

Solutions of $y ″ + 3 y ′ + 2 y = 0$ with the same initial value $y (0) = 1$ but different initial slopes.

FIGURE 5.1.7.

Solutions of $y ″ + 3 y ′ + 2 y = 0$ with the same initial slope $y ′ (0) = 1$ but different initial values.

Example 1

Continued We saw in the first part of Example 1 that $y (x) = 3 \cos x - 2 \sin x$ is a solution (on the entire real line) of $y ″ + y = 0 .$ It has the initial values $y (0) = 3, y ′ (0) = - 2 .$ Theorem 2 tells us that this is the only solution with these initial values. More generally, the solution

$y (x) = b_{0} \cos x + b_{1} \sin x$

satisfies the arbitrary initial conditions $y (0) = b_{0}, y ′ (0) = b_{1};$ this illustrates the existence of such a solution, also as guaranteed by Theorem 2.

Example 1 suggests how, given a homogeneous second-order linear equation, we might actually find the solution y(x) whose existence is assured by Theorem 2. First, we find two “essentially different” solutions $y_{1}$ and $y_{2};$ second, we attempt to impose on the general solution

$y = c_{1} y_{1} + c_{2} y_{2}$ (12)

the initial conditions $y (a) = b_{0}, y ′ (a) = b_{1} .$ That is, we attempt to solve the simultaneous equations

$\begin{array}{l} c_{1} y_{1} (a) + c_{2} y_{2} (a) = b o, \\ c_{1} y_{1}^{′} (a) + c_{2} y_{2}^{′} (a) = b 1 \end{array}$ (13)

for the coefficients $c_{1}$ and $c_{2}$ .

Example 2

Verify that the functions

$y_{1} (x) = e^{x} and y_{2} (x) = x e^{x}$

are solutions of the differential equation

$y ″ - 2 y ′ + y = 0,$

and then find a solution satisfying the initial conditions $y (0) = 3, y ′ (0) = 1$ .

Solution

The verification is routine; we omit it. We impose the given initial conditions on the general solution

$y (x) = c_{1} e^{x} + c_{2} x e^{x},$

for which

$y ′ (x) = (c_{1} + c_{2}) e^{x} + c_{2} x e^{x},$

to obtain the simultaneous equations

$\begin{array}{l} y (0) & = & c_{1} & = & 3, \\ y^{′} (0) & = & c_{1} & + & c_{2} & = & 1. \end{array}$

The resulting solution is $c_{1} = 3, c_{2} = - 2 .$ Hence the solution of the original initial value problem is

$y (x) = 3 e^{x} - 2 x e^{x} .$

Figure 5.1.8 shows several additional solutions of $y ″ - 2 y ′ + y = 0,$ all having the same initial value $y (0) = 3$ .

FIGURE 5.1.8.

Different solutions $y (x) = 3 e^{x} + c_{2} x e^{x}$ of $y ″ - 2 y ′ + y = 0$ with the same initial value $y (0) = 3$ .

Linearly Independent Solutions

In order for the procedure of Example 2 to succeed, the two solutions $y_{1}$ and $y_{2}$ must have the property that the equations in (13) can always be solved for $c_{1}$ and $c_{2},$ no matter what the initial values $y (a) = b_{0}$ and $y ′ (a) = b_{1}$ might be. This is so, provided that the two functions $y_{1}$ and $y_{2}$ are linearly independent. According to the following definition, linear independence of functions defined on an interval I is analogous to linear independence of ordinary vectors.

Two functions are said to be linearly dependent on an open interval provided that they are not linearly independent there; that is, one of them is a constant multiple of the other. We can always determine whether two given functions f and g are linearly dependent on an interval I by noting at a glance whether either of the two quotients f/g or g/f is a constant-valued function on I.

Example 3

Thus it is clear that the following pairs of functions are linearly independent on the entire real line:

$\begin{array}{rcl} \sin x & and & \cos x; \\ e^{x} & and & e^{- 2 x}; \\ e^{x} & and & x e^{x}; \\ x + 1 & and & x^{2}; \\ x & and & | x | . \end{array}$

That is, neither $\sin x / \cos x = \tan x$ nor $\cos x / \sin x = \cot x$ is a constant-valued function; neither $e^{x} / e^{- 2 x} = e^{3 x}$ nor $e^{- 2 x} / e^{x}$ is a constant-valued function; and so forth. But the identically zero function $f (x) \equiv 0$ and any other function g are linearly dependent on every interval, because $0 \cdot g (x) = 0 = f (x) .$ Also, the functions

$f (x) = \sin 2 x and g (x) = \sin x \cos x$

are linearly dependent on any interval because $f (x) = 2 g (x)$ is the familiar trigonometric identity $\sin 2 x = 2 \sin x \cos x$ .

General Solutions

But does the homogeneous equation $y ″ + p y ′ + q y = 0$ always have two linearly independent solutions? Theorem 2 says yes! We need only choose $y_{1}$ and $y_{2}$ so that

$y_{1} (a) = 1, y_{1} ′ (a) = 0 and y_{2} (a) = 0, y_{2} ′ (a) = 1.$

It is then impossible that either $y_{1} = k y_{2}$ or $y_{2} = k y_{1}$ because $k \cdot 0 \neq 1$ for any constant k. Theorem 2 tells us that two such linearly independent solutions exist; actually finding them is a crucial matter that we will discuss briefly at the end of this section, and in greater detail beginning in Section 5.3.

We want to show, finally, that given any two linearly independent solutions $y_{1}$ and $y_{2}$ of the homogeneous equation

$y ″ (x) + p (x) y ′ (x) + q (x) y (x) = 0,$ (9)

every solution y of Eq. (9) can be expressed as a linear combination

$y = c_{1} y_{1} + c_{2} y_{2}$ (12)

of $y_{1}$ and $y_{2} .$ This means that the function in (12) is a general solution of Eq. (9)—it provides all possible solutions of the differential equation.

As suggested by the equations in (13), the determination of the constants $c_{1}$ and $c_{2}$ in (12) depends on a certain $2 \times 2$ determinant of values of $y_{1}, y_{2},$ and their derivatives. Given two functions f and g, the Wronskian of f and g is the determinant

$W = | \begin{array}{r} f & g \\ f^{′} & g^{′} \end{array} | = f g^{′} - f^{′} g .$

We write either W(f, g) or W(x), depending on whether we wish to emphasize the two functions or the point x at which the Wronskian is to be evaluated. For example,

$W (\cos x, \sin x) = | \begin{array}{r} \cos x & \sin x \\ - \sin x & \cos x \end{array} | = \cos^{2} x + \sin^{2} x = 1$

and

$W (e^{x}, x e^{x}) = | \begin{array}{c} e^{x} & x e^{x} \\ e^{x} & e^{x} + x e^{x} \end{array} | = e^{2 x} .$

These are examples of linearly independent pairs of solutions of differential equations (see Examples 1 and 2). Note that in both cases the Wronskian is everywhere nonzero.

On the other hand, if the functions f and g are linearly dependent, with $f = k g$ (for example), then

$W (f, g) = | \begin{array}{c} k g & g \\ k g^{′} & g^{′} \end{array} | = k g g^{′} - k g^{′} g \equiv 0.$

Thus the Wronskian of two linearly dependent functions is identically zero. In Section 5.2 we will prove that, if the two functions $y_{1}$ and $y_{2}$ are solutions of a homogeneous second-order linear equation, then the strong converse stated in part (b) of Theorem 3 holds.

Thus, given two solutions of Eq. (9), there are just two possibilities: The Wronskian W is identically zero if the solutions are linearly dependent; the Wronskian is never zero if the solutions are linearly independent. The latter fact is what we need to show that $y = c_{1} y_{1} + c_{2} y_{2}$ is the general solution of Eq. (9) if $y_{1}$ and $y_{2}$ are linearly independent solutions.

Remark

In essence, Theorem 4 tells us that, when we have found two linearly independent solutions $y_{1}$ and $y_{2}$ of the homogeneous equation in (9), we have then found all of its solutions. Specifically, ${y_{1}, y_{2}}$ is then a basis for the solution space of the differential equation. We therefore call the linear combination $Y = c_{1} y_{1} + c_{2} y_{2}$ a general solution of the differential equation.

Proof of Theorem 4:

Choose a point a of I, and consider the simultaneous equations

$\begin{array}{l} c_{1} y_{1} (a) + c_{2} y_{2} (a) = Y (a), \\ c_{1} y_{1}^{′} (a) + c_{2} y_{2}^{′} (a) = Y ′ (a) . \end{array}$ (14)

The determinant of the coefficients in this system of linear equations in the unknowns $c_{1}$ and $c_{2}$ is simply the Wronskian $W (y_{1}, y_{2})$ evaluated at $x = a .$ By Theorem 3, this determinant is nonzero, so by elementary algebra it follows that the equations in (14) can be solved for $c_{1}$ and $c_{2} .$ With these values of $c_{1}$ and $c_{2},$ we define the solution

$G (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x)$

of Eq. (9); then

$G (a) = c_{1} y_{1} (a) + c_{2} y_{2} (a) = Y (a)$

and

$G ′ (a) = c_{1} y_{1}^{′} (a) + c_{2} y_{2}^{′} (a) = Y ′ (a) .$

Thus the two solutions Y and G have the same initial values at a; likewise, so do $Y ′$ and $G ′ .$ By the uniqueness of a solution determined by such initial values (Theorem 2), it follows that Y and G agree on I. Thus we see that

$Y (x) \equiv G (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x),$

as desired.

Example 4

If $y_{1} (x) = e^{2 x}$ and $y_{2} (x) = e^{- 2 x},$ then

$y_{1}^{″} = (2) (2) e^{2 x} = 4 e^{2 x} = 4 y_{1} and y_{2}^{″} = (- 2) (- 2) e^{- 2 x} = 4 e^{- 2 x} = 4 y_{2} .$

Therefore, $y_{1}$ and $y_{2}$ are linearly independent solutions of

$y ″ - 4 y = 0.$ (15)

But $y_{3} (x) = \cosh 2 x$ and $y_{4} (x) = \sinh 2 x$ are also solutions of Eq. (15), because

$\frac{d^{2}}{d x^{2}} (\cosh 2 x) = \frac{d}{d x} (2 \sinh 2 x) = 4 \cosh 2 x$

and, similarly, $(\sinh 2 x) ″ = 4 \sinh 2 x .$ It therefore follows from Theorem 4 that the functions cosh 2 x and sinh 2 x can be expressed as linear combinations of $y_{1} (x) = e^{2 x}$ and $y_{2} (x) = e^{- 2 x} .$ Of course, this is no surprise, because

$\cosh 2 x = \frac{1}{2} e^{2 x} + \frac{1}{2} e^{- 2 x} and \sinh 2 x = \frac{1}{2} e^{2 x} - \frac{1}{2} e^{- 2 x}$

by the definitions of the hyperbolic cosine and hyperbolic sine. Thus the solution space of the differential equation $y ″ - 4 y = 0$ has the two different bases ${e^{2 x}, e^{- 2 x}}$ and {cosh 2x, sinh 2x}.

Remark

Because $e^{2 x}, e^{- 2 x}$ and cosh x, sinh x are two different pairs of linearly independent solutions of the equation $y ″ - 4 y = 0$ in (15), Theorem 4 implies that every particular solution Y(x) of this equation can be written both in the form

$Y (x) = c_{1} e^{2 x} + c_{2} e^{- 2 x}$

and in the form

$Y (x) = a \cosh 2 x + b \sinh 2 x .$

Thus these two different linear combinations (with arbitrary constant coefficients) provide two different descriptions of the set of all solutions of the same differential equation $y ″ - 4 y = 0 .$ Hence each of these two linear combinations is a general solution of the equation. Indeed, this is why it is accurate to refer to a specific such linear combination as “a general solution” rather than as “the general solution.”

Linear Second-Order Equations with Constant Coefficients

As an illustration of the general theory introduced in this section, we discuss the homogeneous second-order linear differential equation

$a y ″ + b y ′ + c y = 0$ (16)

with constant coefficients a, b, and c. We first look for a single solution of Eq. (16) and begin with the observation that

$(e^{r x})^{′} = r e^{r x} and (e^{r x})^{″} = r^{2} e^{r x},$ (17)

so any derivative of $e^{r x}$ is a constant multiple of $e^{r x} .$ Hence, if we substituted $y = e^{r x}$ in Eq. (16), then each term would be a constant multiple of $e^{r x},$ with the constant coefficients dependent on r and the coefficients a, b, and c. This suggests that we try to find a value of r so that these multiples of $e^{r x}$ will have sum zero. If we succeed, then $y = e^{r x}$ will be a solution of Eq. (16).

For example, if we substitute $y = e^{r x}$ in the equation

$y ″ - 5 y ′ + 6 y = 0,$

we obtain

$r^{2} e^{r x} - 5 r e^{r x} + 6 e^{r x} = 0.$

Thus

$(r^{2} - 5 r + 6) e^{r x} = 0; (r - 2) (r - 3) e^{r x} = 0.$

Hence $y = e^{r x}$ will be a solution if either $r = 2$ or $r = 3 .$ So, in searching for a single solution, we actually have found two solutions: $y_{1} (x) = e^{2 x}$ and $y_{2} (x) = e^{3 x}$ .

To carry out this procedure in the general case, we substitute $y = e^{r x}$ in Eq. (16). With the aid of the equations in (17), we find the result to be

$a r^{2} e^{r x} + b r e^{r x} + c e^{r x} = 0.$

Because $e^{r x}$ is never zero, we conclude that $y (x) = e^{r x}$ will satisfy the differential equation in (16) precisely when r is a root of the algebraic equation

$a r^{2} + b r + c = 0.$ (18)

This quadratic equation is called the characteristic equation of the homogeneous linear differential equation

$a y ″ + b y ′ + c y = 0.$ (16)

If Eq. (18) has two distinct (unequal) roots $r_{1}$ and $r_{2},$ then the corresponding solutions $y_{1} (x) = e^{r_{1} x}$ and $y_{2} (x) = e^{r_{2} x}$ of (16) are linearly independent. (Why?) This gives the following result.

Example 5

Find the general solution of

$2 y ″ - 7 y ′ + 3 y = 0.$

Solution

We can solve the characteristic equation

$2 r^{2} - 7 r + 3 = 0$

by factoring:

$(2 r - 1) (r - 3) = 0.$

The roots $r_{1} = \frac{1}{2}$ and $r_{2} = 3$ are real and distinct, so Theorem 5 yields the general solution

$y (x) = c_{1} e^{x / 2} + c_{2} e^{3 x} .$

Example 6

The differential equation $y ″ + 2 y ′ = 0$ has characteristic equation

$r^{2} + 2 r = r (r + 2) = 0$

with distinct real roots $r_{1} = 0$ and $r_{2} = - 2.$ Because $e^{0 \cdot x} \equiv 1,$ we get the general solution

$y (x) = c_{1} + c_{2} e^{- 2 x} .$

Figure 5.1.9 shows several different solution curves with $c_{1} = 1,$ all appearing to approach the solution curve $y (x) \equiv 1$ (with $c_{2} = 0$ ) as $x \to + \infty$ .

FIGURE 5.1.9.

Solutions $y (x) = 1 + c_{2} e^{- 2 x}$ of $y ″ + 2 y ′ = 0$ with different values of $c_{2}$ .

Remark

Note that Theorem 5 changes a problem involving a differential equation into one involving only the solution of an algebraic equation.

If the characteristic equation in (18) has equal roots $r_{1} = r_{2},$ we get (at first) only the single solution $y_{1} (x) = e^{r_{1} x}$ of Eq. (16). The problem in this case is to produce the “missing” second solution of the differential equation.

A double root $r = r_{1}$ will occur precisely when the characteristic equation is a constant multiple of the equation

${(r - r_{1})}^{2} = r^{2} - 2 r_{1} r + r_{1}^{2} = 0.$

Any differential equation with this characteristic equation is equivalent to

$y ″ - 2 r_{1} y ′ + r_{1}^{2} y = 0.$ (20)

But it is easy to verify by direct substitution that $y = {x e}^{r_{1} x}$ is a second solution of Eq. (20). It is clear (but you should verify) that

$y_{1} (x) = e^{r_{1} x} and y_{2} (x) = x e^{r_{1} x}$

are linearly independent functions, so the general solution of the differential equation in (20) is

$y (x) = c_{1} e^{r_{1} x} + c_{2} x e^{r_{1} x} .$

Example 7

To solve the initial value problem

$\begin{array}{l} y ″ + 2 y ′ + y = 0; \\ y (0) = 5, y ′ (0) = - 3, \end{array}$

we note first that the characteristic equation

$r^{2} + 2 r + 1 = {(r + 1)}^{2} = 0$

has equal roots $r_{1} = r_{2} = - 1 .$ Hence the general solution provided by Theorem 6 is

$y (x) = c_{1} e^{- x} + c_{2} x e^{- x} .$

Differentiation yields

$y (x) = - c_{1} e^{- x} + c_{2} e^{- x} - c_{2} x e^{- x},$

so the initial conditions yield the equations

$\begin{array}{llrlrlr} y (0) & = & c_{1} & = & 5, \\ y ′ (0) & = & - c_{1} & + & c_{2} & = & - 3, \end{array}$

which imply that $c_{1} = 5$ and $c_{2} = 2 .$ Thus the desired particular solution of the initial value problem is

$y (x) = 5 e^{- x} + 2 x e^{- x} .$

This particular solution, together with several others of the form $y (x) = c_{1} e^{- x} + 2 {x e}^{- x},$ is illustrated in Fig. 5.1.10.

FIGURE 5.1.10.

Solutions $y (x) = c_{1} e^{- x} + 2 x e^{- x}$ of $y ″ + 2 y ′ + y = 0$ with different values of $c_{1} .$

The characteristic equation in (18) may have either real or complex roots. The case of complex roots will be discussed in Section 5.3.

5.1 Problems

In Problems 1 through 16, a homogeneous second-order linear differential equation, two functions $y_{1}$ and $y_{2},$ and a pair of initial conditions are given. First verify that $y_{1}$ and $y_{2}$ are solutions of the differential equation. Then find a particular solution of the form $y = c_{1} y_{1} + c_{2} y_{2}$ that satisfies the given initial conditions. Primes denote derivatives with respect to x.

$y ″ - y = 0; y_{1} = e^{x}, y_{2} = e^{- x}; y (0) = 0, y ′ (0) = 5$
$y ″ - 9 y = 0; y_{1} = e^{3 x}, y_{2} = e^{- 3 x}; y (0) = - 1, y ′ (0) = 15$
$y ″ + 4 y = 0; y_{1} = \cos 2 x, y_{2} = \sin 2 x; y (0) = 3, y ′ (0) = 8$
$y ″ + 25 y = 0; y_{1} = \cos 5 x, y_{2} = \sin 5 x; y (0) = 10, y ′ (0) = - 10$
$y ″ - 3 y ′ + 2 y = 0; y_{1} = e^{x}, y_{2} = e^{2 x}; y (0) = 1, y ′ (0) = 0$
$y ″ + y ′ - 6 y = 0; y_{1} = e^{2 x}, y_{2} = e^{- 3 x}; y (0) = 7, y ′ (0) = - 1$
$y ″ + y ′ = 0; y_{1} = 1, y_{2} = e^{- x}; y (0) = - 2, y ′ (0) = 8$
$y ″ - 3 y ′ = 0; y_{1} = 1, y_{2} = e^{3 x}; y (0) = 4, y ′ (0) = - 2$
$y ″ + 2 y ′ + y = 0; y_{1} = e^{- x}, y_{2} = {x e}^{- x}; y (0) = 2, y ′ (0) = - 1$
$y ″ - 10 y ′ + 25 y = 0; y_{1} = e^{5 x}, y_{2} = {x e}^{5 x}; y (0) = 3, y ′ (0) = 13$
$y ″ - 2 y ′ + 2 y = 0; y_{1} = e^{x} \cos x, y_{2} = e^{x} \sin x; y (0) = 0, y ′ (0) = 5$
$y ″ + 6 y ′ + 13 y = 0; y_{1} = e^{- 3 x} \cos 2 x, y_{2} = e^{- 3 x} \sin 2 x; y (0) = 2, y ′ (0) = 0$
$x^{2} y ″ - 2 x y ′ + 2 y = 0; y_{1} = x, y_{2} = x^{2}; y (1) = 3, y ′ (1) = 1$
$x^{2} y ″ + 2 x y ′ - 6 y = 0; y_{1} = x^{2}, y_{2} = x^{- 3}; y (2) = 10, y ′ (2) = 15$
$x^{2} y ″ - x y ′ + y = 0; y_{1} = x, y_{2} = x \ln x; y (1) = 7, y ′ (1) = 2$
$x^{2} y ″ + x y ′ + y = 0; y_{1} = \cos (\ln x), y_{2} = \sin (\ln x); y (1) = 2, y ′ (1) = 3$

The following three problems illustrate the fact that the superposition principle does not generally hold for nonlinear equations.

Show that $y = 1 / x$ is a solution of $y ′ + y^{2} = 0,$ but that if $c \neq 0$ and $c \neq 1,$ then $y = c / x$ is not a solution.
Show that $y = x^{3}$ is a solution of $y y ″ = 6 x^{4},$ but that if $c^{2} \neq 1,$ then $y = {c x}^{3}$ is not a solution.
Show that $y_{1} \equiv 1$ and $y_{2} = \sqrt{x}$ are solutions of $y y ″ + {(y ′)}^{2} = 0,$ but that their sum $y = y_{1} + y_{2}$ is not a solution.

Determine whether the pairs of functions in Problems 20 through 26 are linearly independent or linearly dependent on the real line.

$f (x) = π, g (x) = \cos^{2} x + \sin^{2} x$
$f (x) = x^{3}, g (x) = x^{2} | x |$
$f (x) = 1 + x, g (x) = 1 + | x |$
$f (x) = {x e}^{x}, g (x) = | x | e^{x}$
$f (x) = \sin^{2} x, g (x) = 1 - \cos 2 x$
$f (x) = e^{x} \sin x, g (x) = e^{x} \cos x$
$f (x) = 2 \cos x + 3 \sin x, g (x) = 3 \cos x - 2 \sin x$
Let $y_{p}$ be a particular solution of the nonhomogeneous equation $y ″ + p y ′ + q y = f (x)$ and let $y_{c}$ be a solution of its associated homogeneous equation. Show that $y = y_{c} + y_{p}$ is a solution of the given nonhomogeneous equation.
With $y_{p} = 1$ and $y_{c} = c_{1} \cos x + c_{2} \sin x$ in the notation of Problem 27, find a solution of $y ″ + y = 1$ satisfying the initial conditions $y (0) = - 1 = y ′ (0)$ .

Problems 29 through 32 explore the propeties of the Wronskian.

Show that $y_{1} = x^{2}$ and $y_{2} = x^{3}$ are two different solutions of $x^{2} y ″ - 4 x y ′ + 6 y = 0,$ both satisfying the initial conditions $y (0) = 0 = y ′ (0) .$ Explain why these facts do not contradict Theorem 2 (with respect to the guaranteed uniqueness).
(a) Show that $y_{1} = x^{3}$ and $y_{2} = | x^{3} |$ are linearly independent solutions on the real line of the equation $x^{2} y ″ - 3 x y ′ + 3 y = 0$ . (b) Verify that $W (y_{1}, y_{2})$ is identically zero. Why do these facts not contradict Theorem 3?
Show that $y_{1} = \sin x^{2}$ and $y_{2} = \cos x^{2}$ are linearly independent functions, but that their Wronskian vanishes at $x = 0 .$ Why does this imply that there is no differential equation of the form $y ″ + p (x) y ′ + q (x) y = 0,$ with both p and q continuous everywhere, having both $y_{1}$ and $y_{2}$ as solutions?
Let $y_{1}$ and $y_{2}$ be two solutions of $A (x) y ″ + B (x) y ′ + C (x) y = 0$ on an open interval I where A, B, and C are continuous and A(x) is never zero. (a) Let $W = W (y_{1}, y_{2}) .$ Show that

$A (x) \frac{d w}{d x} = (y_{1}) (A {y^{″}}_{2}) - (y_{2}) (A {y^{″}}_{1}) .$

Then substitute for $A {y^{″}}_{2} and A {y^{″}}_{2}$ from the original differential equation to show that

$A (x) \frac{d w}{d x} = - B (x) W (x) .$

(b) Solve this first-order equation to deduce Abel’s formula

$W (x) = K \exp (- \int \frac{B (x)}{A (x)} d x),$

where K is a constant. (c) Why does Abel’s formula imply that the Wronskian $W (y_{1}, y_{2})$ is either zero everywhere or nonzero everywhere (as stated in Theorem 3)?

Apply Theorems 5 and 6 to find general solutions of the differential equations given in Problems 33 through 42. Primes denote derivatives with respect to x.

$y ″ - 3 y ′ + 2 y = 0$
$y ″ + 2 y ′ - 15 y = 0$
$y ″ + 5 y ′ = 0$
$2 y ″ + 3 y ′ = 0$
$2 y ″ - y ′ - y = 0$
$4 y ″ + 8 y ′ + 3 y = 0$
$4 y ″ + 4 y ′ + y = 0$
$9 y ″ - 12 y ′ + 4 y = 0$
$6 y ″ - 7 y ′ - 20 y = 0$
$35 y ″ - y ′ - 12 y = 0$

Each of Problems 43 through 48 gives a general solution y(x) of a homogeneous second-order differential equation $a y ″ + b y ′ + c y = 0$ with constant coefficients. Find such an equation.

$y (x) = c_{1} + c_{2} e^{- 10 x}$
$y (x) = c_{1} e^{10 x} + c_{2} e^{- 10 x}$
$y (x) = c_{1} e^{- 10 x} + c_{2} {x e}^{- 10 x}$
$y (x) = c_{1} e^{10 x} + c_{2} e^{100 x}$
$y (x) = c_{1} + c_{2} x$
$y (x) = e^{x} (c_{1} e^{x \sqrt{2}} + c_{2} e^{- x \sqrt{2}})$

Problems 49 and 50 deal with the solution curves of $y ″ + 3 y ′ + 2 y = 0$ shown in Figs. 5.1.6 and 5.1.7.

Find the highest point on the solution curve with $y (0) = 1$ and $y ′ (0) = 6$ in Fig. 5.1.6.
Figure 5.1.7 suggests that the solution curves shown all meet at a common point in the third quadrant. Assuming that this is indeed the case, find the coordinates of that point.
A second-order Euler equation is one of the form

$a x^{2} y ″ + b x y ′ + c y = 0$ (22)

where a, b, c are constants. (a) Show that if $x > 0,$ then the substitution $v = \ln x$ transforms Eq. (22) into the constant-coefficient linear equation

$a \frac{d^{2} y}{d v^{2}} + (b - a) \frac{d y}{d v} + c y = 0$ (23)

with independent variable v. (b) If the roots $r_{1}$ and $r_{2}$ of the characteristic equation of Eq. (23) are real and distinct, conclude that a general solution of the Euler equation in (22) is $y (x) = c_{1} x^{r_{1}} + c_{2} x^{r_{2}}$ .

Make the substitution $v = \ln x$ of Problem 51 to find general solutions (for $x > 0$ ) of the Euler equations in Problems 52–56.

$x^{2} y ″ + x y ′ - y = 0$
$x^{2} y ″ + 2 x y ′ - 12 y = 0$
$4 x^{2} y ″ + 8 x y ′ - 3 y = 0$
$x^{2} y ″ + x y ′ = 0$
$x^{2} y ″ - 3 x y ′ + 4 y = 0$

5.1 Application Plotting Second-Order Solution Families

This application deals with the plotting by computer of families of solutions such as those illustrated in Figs. 5.1.6 and 5.1.7. Show first that the general solution of the differential equation

$y ″ + 3 y ′ + 2 y = 0$ (1)

$y (x) = c_{1} e^{- x} + c_{2} e^{- 2 x} .$ (2)

This is equivalent to the graphing calculator result shown in Figure 5.1.11, and to the $Wolfram | Alpha$ output generated by the simple query

y″ + 3y′ + 2y = 0

FIGURE 5.1.11.

TI-Nspire CX CAS screen showing the general solution of $y ″ + 3 y ′ + 2 y = 0$ .

Next show that the particular solution of Eq. (1) satisfying $y (0) = a, y ′ (0) = b$ corresponds to $c_{1} = 2 a + b$ and $c_{2} = - a - b,$ that is

$y (x) = (2 a + b) e^{- x} - (a + b) e^{- 2 x} .$ (3)

For Fig. 5.1.6, we fix $a = 1,$ leading to the particular solution

$y (x) = (b + 2) e^{- x} - (b + 1) e^{- 2 x} .$ (4)

The Matlab loop

x = - 1 :  0.02 :  5 % x-vector from x = - 1 to x = 5
for b = -6 :  2 :  6 % for b = -6 to 6 with db = 2 do
   y = (b + 2)*exp(-x) - (b + 1)*exp(-2*x);
   plot(x,y)
end

was used to generate Fig. 5.1.6.

For Fig. 5.1.7, we instead fix $b = 1,$ leading to the particular solution

$y (x) = (2 a + 1) e^{- x} - (a + 1) e^{- 2 x} .$ (5)

The Matlab loop

x = -2 :  0.02 :  4   % x-vector from x = -2 to x = 4
for a = -3 :  1 :  3  % for a = -3 to 3 with da = 1 do
   y = (2*a + 1)*exp(-x) - (a + 1)*exp(-2*x);
   plot(x,y)
end

was used to generate Fig. 5.1.7.

Computer systems, such as Maple and Mathematica, as well as graphing calculators, have commands to carry out for-loops such as the two shown here. Moreover, such systems often allow for interactive investigation, in which the solution curve is immediately redrawn in response to on-screen input. For example, Fig. 5.1.12 was generated using Matlab’s uicontrol command; moving the sliders allows the user to experiment with various combinations of the initial conditions $y (0) = a$ and $y ′ (0) = b$ .

FIGURE 5.1.12.

Matlab graph of Eq. (3) with $a = 2$ and $b = 1 .$ Using the sliders, `a` and `b` can be changed interactively.

The Mathematica command

Manipulate[
    Plot[(2*a+b)*Exp[-x]+(-b-a)*Exp[-2*x],
    {x,-1,5}, PlotRange -> {-5,5}],
    {a,-3,3}, {b,-6,6}]

produces a similar display, as does the Maple command

Explore(plot((2*a+b)*exp(-x)+(-b-a)*exp(-2*x),
    x = -1..5, y=-5..5))

(after first bringing up a dialog box in which the ranges of values for a and b are specified). Likewise some graphing calculators feature a touchpad or other screen navigation method allowing the user to vary specified parameters in real time (see Fig. 5.1.13).

FIGURE 5.1.13.

TI-Nspire CX CAS graph of Eq. (3) with $a = 2$ and $b = 1.$ The arrows allow a and b to be changed interactively.

Begin by either reproducing Figs. 5.1.6 and 5.1.7 or by creating an interactive display that shows the graph of (3) for any desired combination of a and b. Then, for each of the following differential equations, modify your commands to examine the family of solution curves satisfying $y (0) = 1,$ as well as the family of solution curves satisfying the initial condition $y ′ (0) = 1$ .

$y ″ - y = 0$
$y ″ - 3 y ′ + 2 y = 0$
$2 y ″ + 3 y ′ + y = 0$
$y ″ + y = 0$ (see Example 1)
$y ″ + 2 y ′ + 2 y = 0,$ which has general solution $y (x) = e^{- x} (c_{1} \cos x + c_{2} \sin x)$

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Table of Contents for 5 Higher-Order Linear Differential Equations

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Table of Contents for
5 Higher-Order Linear Differential Equations