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5.2 General Solutions of Linear Equations

We now show that our discussion in Section 5.1 of second-order linear equations generalizes in a very natural way to the general nth-order linear differential equation of the form

P 0 (x) y (n) + P 1 (x) y (n - 1) + \dots + P n - 1 (x) y' + P n (x) y = F (x) .

$P_{0} (x) y^{(n)} + P_{1} (x) y^{(n - 1)} + \dots + P_{n - 1} (x) y ′ + P_{n} (x) y = F (x) .$ (1)

Unless otherwise noted, we will always assume that the coefficient functions Pi(x) $P_{i} (x)$ and F(x)are continuous on some open interval I (perhaps unbounded) where we wish to solve the equation. Under the additional assumption that P0(x)≠0 $P_{0} (x) \neq 0$ at each point of I, we can divide each term in Eq. (1) by P0(x) $P_{0} (x)$ to obtain an equation with leading coefficient 1, of the form

y (n) + p 1 (x) y (n - 1) + \dots + p n - 1 (x) y' + p n (x) y = f (x) .

$y^{(n)} + p_{1} (x) y^{(n - 1)} + \dots + p_{n - 1} (x) y ′ + p_{n} (x) y = f (x) .$ (2)

The homogeneous linear equation associated with Eq. (2) is

y (n) + p 1 (x) y (n - 1) + \dots + p n - 1 (x) y' + p n (x) y = 0.

$y^{(n)} + p_{1} (x) y^{(n - 1)} + \dots + p_{n - 1} (x) y ′ + p_{n} (x) y = 0.$ (3)

Just as in the second-order case, a homogeneous nth-order linear differential equation has the valuable property that any superposition, or linear combination, of solutions of the equation is again a solution. The proof of the following theorem is essentially the same—a routine verification—as that of Theorem 1 of Section 5.1.

Example 1

It is easy to verify that the three functions

y 1 (x) = e - 3 x, y 2 (x) = cos 2 x, and y 3 (x) = sin 2 x

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

are all solutions of the homogeneous third-order equation

y (3) + 3 y'' + 4 y' + 12 y = 0

$y^{(3)} + 3 y ″ + 4 y ′ + 12 y = 0$

on the entire real line. Theorem 1 tells us that any linear combination of these solutions, such as

y (x) = - 3 y 1 (x) + 3 y 2 (x) - 2 y 3 (x) = - 3 e - 3 x + 3 cos 2 x - 2 sin 2 x,

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

is also a solution on the entire real line. We will see that, conversely, every solution of the differential equation of this example is a linear combination of the three particular solutions y1, y2, $y_{1}, y_{2},$ and y3. $y_{3} .$ Thus a general solution is given by

y (x) = c 1 e - 3 x + c 2 cos 2 x + c 3 sin 2 x .

$y (x) = c_{1} e^{- 3 x} + c_{2} \cos 2 x + c_{3} \sin 2 x .$

Existence and Uniqueness of Solutions

We saw in Section 5.1 that a particular solution of a second-order linear differential equation is determined by two initial conditions. Similarly, a particular solution of an nth-order linear differential equation is determined by n initial conditions. The following theorem, proved in the Appendix, is the natural generalization of Theorem 2 of Section 5.1.

Equation (2) and the conditions in (5) constitute an nth-order 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 (2) are continuous. It tells us nothing, however, about how to find this solution. In Section 5.3 we will see how to construct explicit solutions of initial value problems in the constant-coefficient case that occurs often in applications.

Example 1

Continued We saw earlier that

y (x) = - 3 e - 3 x + 3 cos 2 x - 2 sin 2 x

$y (x) = - 3 e^{- 3 x} + 3 \cos 2 x - 2 \sin 2 x$

is a solution of

y (3) + 3 y'' + 4 y' + 12 y = 0

$y^{(3)} + 3 y ″ + 4 y ′ + 12 y = 0$

on the entire real line. This particular solution has initial values y(0)=0, y'(0)=5 $y (0) = 0, y ′ (0) = 5$ , and y''(0)=−39, $y ″ (0) = - 39,$ and Theorem 2 implies that there is no other solution with these same initial values. Note that its graph (in Fig. 5.2.1) looks periodic on the right. Indeed, because of the negative exponent, we see that y(x)≈3 cos 2x−2 sin 2x $y (x) \approx 3 \cos 2 x - 2 \sin 2 x$ for large positive x.

FIGURE 5.2.1.

The particular solution y(x)=−3e−3x+3 cos 2x−2 sin 2x $y (x) = - 3 e^{- 3 x} + 3 \cos 2 x - 2 \sin 2 x$ .

Remark

Because its general solution involves the three arbitrary constants c1, c2, $c_{1}, c_{2},$ and c3, $c_{3},$ the third-order equation in Example 1 has a “threefold infinity” of solutions, including three families of especially simple solutions:

y(x)=c1e−3x $y (x) = c_{1} e^{- 3 x}$ (obtained from the general solution with c2=c3=0 $c_{2} = c_{3} = 0$ ),
y(x)=c2cos 2x $y (x) = c_{2} \cos 2 x$ (with c1=c3=0 $c_{1} = c_{3} = 0$ ), and
y(x)=c3sin 2x $y (x) = c_{3} \sin 2 x$ (with c1=c2=0 $c_{1} = c_{2} = 0$ ).

Alternatively, Theorem 2 suggests a threefold infinity of particular solutions corresponding to independent choices of the three initial values y(0)=b0, y'(0)=b1, $y (0) = b_{0}, y ′ (0) = b_{1},$ and y''(0)=b2. $y ″ (0) = b_{2} .$ Figures 5.2.2 through 5.2.4 illustrate three corresponding families of solutions—for each of which, two of these three initial values are zero.

FIGURE 5.2.2.

Solutions of y(3)+3y''+4y'+12y=0 $y^{(3)} + 3 y ″ + 4 y ′ + 12 y = 0$ with y'(0)=y''(0)=0 $y ′ (0) = y ″ (0) = 0$ but with different values for y(0).

FIGURE 5.2.3.

Solutions of y(3)+3y''+4y'+12y=0 $y^{(3)} + 3 y ″ + 4 y ′ + 12 y = 0$ with y(0)=y''(0)=0 $y (0) = y ″ (0) = 0$ but with different values for y'(0) $y ′ (0)$ .

FIGURE 5.2.4.

Solutions of y(3)+3y''+4y'+12y=0 $y^{(3)} + 3 y ″ + 4 y ′ + 12 y = 0$ with y(0)=y'(0)=0 $y (0) = y ′ (0) = 0$ but with different values for y''(0) $y ″ (0)$ .

Note that Theorem 2 implies that the trivial solution y(x)≡0 $y (x) \equiv 0$ is the only solution of the homogeneous equation

y (n) + p 1 (x) y (n - 1) + \dots + p n - 1 (x) y' + p n (x) y = 0

$y^{(n)} + p_{1} (x) y^{(n - 1)} + \dots + p_{n - 1} (x) y ′ + p_{n} (x) y = 0$ (3)

that satisfies the trivial initial conditions

y (a) = y' (a) = \dots = y (n - 1) (a) = 0.

$y (a) = y ′ (a) = \dots = y^{(n - 1)} (a) = 0.$

Example 2

It is easy to verify that

y 1 (x) = x 2 and y 2 (x) = x 3

$y_{1} (x) = x^{2} and y_{2} (x) = x^{3}$

are two different solutions of

x 2 y'' - 4 x y' + 6 y = 0,

$x^{2} y ″ - 4 x y ′ + 6 y = 0,$

and that both satisfy the initial conditions y(0)=y'(0)=0. $y (0) = y ′ (0) = 0 .$ Why does this not contradict the uniqueness part of Theorem 2? It is because the leading coefficient in this differential equation vanishes at x=0, $x = 0,$ so this equation cannot be written in the form of Eq. (3) with coefficient functions continuous on an open interval containing the point x=0 $x = 0$ .

Linearly Independent Solutions

On the basis of our knowledge of general solutions of second-order linear equations, we anticipate that a general solution of the homogeneous nth-order linear equation

y (n) + p 1 (x) y (n - 1) + \dots + p n - 1 (x) y' + p n (x) y = 0

$y^{(n)} + p_{1} (x) y^{(n - 1)} + \dots + p_{n - 1} (x) y ′ + p_{n} (x) y = 0$ (3)

will be a linear combination

y = c 1 y 1 + c 2 y 2 + \dots + c n y n,

$y = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n},$ (4)

where y1, y2, …, $y_{1}, y_{2}, \dots,$ yn $y_{n}$ are particular solutions of Eq. (3). But these n particular solutions must be “sufficiently independent” that we can always choose the coefficients c1, c2, …, $c_{1}, c_{2}, \dots,$ cn $c_{n}$ in (4) to satisfy arbitrary initial conditions of the form in (5). The question is this: What should be meant by independence of three or more functions?

Recall that two functions f1 $f_{1}$ and f2 $f_{2}$ are linearly dependent if one is a constant multiple of the other; that is, if either f1=kf2 $f_{1} = k f_{2}$ or f2=kf1 $f_{2} = k f_{1}$ for some constant k. If we write these equations as

(1) f 1 + (- k) f 2 = 0 or (k) f 1 + (- 1) f 2 = 0,

$(1) f_{1} + (- k) f_{2} = 0 or (k) f_{1} + (- 1) f_{2} = 0,$

we see that the linear dependence of f1 $f_{1}$ and f2 $f_{2}$ implies that there exist two constants c1 $c_{1}$ and c2 $c_{2}$ not both zero such that

c 1 f 1 + c 2 f 2 = 0.

$c_{1} f_{1} + c_{2} f_{2} = 0.$ (6)

Conversely, if c1 $c_{1}$ and c2 $c_{2}$ are not both zero, then Eq. (6) certainly implies that f1 $f_{1}$ and f2 $f_{2}$ are linearly dependent.

In analogy with Eq. (6), we say that n functions f1, f2, …, $f_{1}, f_{2}, \dots,$ fn $f_{n}$ are linearly dependent provided that some nontrivial linear combination

c 1 f 1 + c 2 f 2 + \dots + c n f n

$c_{1} f_{1} + c_{2} f_{2} + \dots + c_{n} f_{n}$

of them vanishes identically; nontrivial means that not all of the coefficients c1 $c_{1}$ , c2, …, $c_{2}, \dots,$ cn $c_{n}$ are zero (although some of them may be zero).

Thus linear dependence of functions on an interval I is precisely analogous to linear dependence of ordinary vectors (Section 4.3).

If not all the coefficients in Eq. (7) are zero, then clearly we can solve for at least one of the functions as a linear combination of the others, and conversely. Thus the functions f1, f2, …, $f_{1}, f_{2}, \dots,$ fn $f_{n}$ are linearly dependent if and only if at least one of them is a linear combination of the others.

Example 3

The functions

f 1 (x) = sin 2 x, f 2 (x) = sin x cos x, and f 3 (x) = e x

$f_{1} (x) = \sin 2 x, f_{2} (x) = \sin x \cos x, and f_{3} (x) = e^{x}$

are linearly dependent on the real line because

(1) f 1 + (- 2) f 2 + (0) f 3 = 0

$(1) f_{1} + (- 2) f_{2} + (0) f_{3} = 0$

(by the familiar trigonometric identity sin 2x=2 sin x cos x $\sin 2 x = 2 \sin x \cos x$ ).

The n functions f1, f2, …, $f_{1}, f_{2}, \dots,$ fn $f_{n}$ are called linearly independent on the interval I provided that they are not linearly dependent there. Equivalently, they are linearly independent on I provided that the identity

c 1 f 1 + c 2 f 2 + \dots + c n f n = 0

$c_{1} f_{1} + c_{2} f_{2} + \dots + c_{n} f_{n} = 0$ (7)

holds on I only in the trivial case

c 1 = c 2 = \dots = c n = 0;

$c_{1} = c_{2} = \dots = c_{n} = 0;$

that is, no nontrivial linear combination of these functions vanishes on I. Put yet another way, the functions f1, f2, …, $f_{1}, f_{2}, \dots,$ fn $f_{n}$ are linearly independent if no one of them is a linear combination of the others. (Why?)

Sometimes one can show that n given functions are linearly dependent by finding, as in Example 3, nontrivial values of the coefficients so that Eq. (7) holds. But in order to show that n given functions are linearly independent, we must prove that nontrivial values of the coefficients cannot be found, and this is seldom easy to do in any direct or obvious manner.

Fortunately, in the case of n solutions of a homogeneous nth-order linear equation, there is a tool that makes the determination of their linear dependence or independence a routine matter in many examples. This tool is the Wronskian determinant, which we introduced (for the case n=2 $n = 2$ ) in Section 5.1. Suppose that the n functions f1, f2, …, $f_{1}, f_{2}, \dots,$ fn $f_{n}$ are each n−1 $n - 1$ times differentiable. Then their Wronskian is the n×n $n \times n$ determinant

W = ∣ ∣ ∣ ∣ ∣ ∣ ∣ f 1 f' 1 ⋮ f (n - 1) 1 f 2 f' 2 ⋮ f (n - 1) 2 \dots \dots \dots f n f' n ⋮ f (n - 1) n ∣ ∣ ∣ ∣ ∣ ∣ ∣ .

$W = | \begin{array}{c} f_{1} & f_{2} & \dots & f_{n} \\ {f^{′}}_{1} & {f^{′}}_{2} & \dots & {f^{′}}_{n} \\ ⋮ & ⋮ & ⋮ \\ f_{1}^{(n - 1)} & f_{2}^{(n - 1)} & \dots & f_{n}^{(n - 1)} \end{array} | .$ (8)

We write W(f1,f2,…,fn) $W (f_{1}, f_{2}, \dots, f_{n})$ or W(x), depending on whether we wish to emphasize the functions or the point x at which their Wronskian is to be evaluated. The Wronskian is named after the Polish mathematician J. M. H. Wronski (1778–1853).

We saw in Section 5.1 that the Wronskian of two linearly dependent functions vanishes identically. More generally, the Wronskian of n linearly dependent functions f1, f2, …, $f_{1}, f_{2}, \dots,$ fn $f_{n}$ is identically zero. To prove this, assume that Eq. (7) holds on the interval I for some choice of the constants c1, c2, …, $c_{1}, c_{2}, \dots,$ cn $c_{n}$ not all zero. We then differentiate this equation n−1 $n - 1$ times in succession, obtaining the n equations

c 1 f 1 (x) c 1 f' 1 (x) c 1 f (n - 1) 1 (x) + + + c 2 f 2 (x) c 2 f' 2 (x) c 2 f (n - 1) 2 (x) + + + \dots \dots \dots + + + c n f n (x) c n f' n (x) c n f (n - 1) n (x) = = ⋮ = 0, 0, 0,

$\begin{aligned} c_{1} f_{1} (x) & + & c_{2} f_{2} (x) & + & \dots & + & c_{n} f_{n} (x) & = & 0, \\ c_{1} f_{1}^{′} (x) & + & c_{2} f_{2}^{′} (x) & + & \dots & + & c_{n} f_{n}^{′} (x) & = & 0, \\ ⋮ \\ c_{1} f_{1}^{(n - 1)} (x) & + & c_{2} f_{2}^{(n - 1)} (x) & + & \dots & + & c_{n} f_{n}^{(n - 1)} (x) & = & 0, \end{aligned}$ (9)

which hold for all x in I. We recall from Theorem 7 in Section 3.5 that a homogeneous n×n $n \times n$ linear system of equations has a nontrivial solution if and only if its coefficient matrix is not invertible, which by Theorem 2 in Section 3.6 is so if and only if the coefficient determinant vanishes. In Eq. (9) the unknowns are the constants c1, c2, …, cn $c_{1}, c_{2}, \dots, c_{n}$ and the determinant of coefficients is simply the Wronskian W(f1,f2,…,fn) $W (f_{1}, f_{2}, \dots, f_{n})$ evaluated at the typical point x of I. Because we know that the ci $c_{i}$ are not all zero, it follows that W(x)≡0, $W (x) \equiv 0,$ as we wanted to prove.

Therefore, to show that the functions f1, f2, …, $f_{1}, f_{2}, \dots,$ fn $f_{n}$ are linearly independent on the interval I, it suffices to show that their Wronskian is nonzero at just one point of I.

Example 4

Show that the functions y1(x)=e−3x, y2(x)=cos 2x, $y_{1} (x) = e^{- 3 x}, y_{2} (x) = \cos 2 x,$ and y3(x)=sin 2x $y_{3} (x) = \sin 2 x$ (of Example 1) are linearly independent.

Solution

Their Wronskian is

W = = ∣ ∣ ∣ ∣ ∣ e - 3 x - 3 e - 3 x 9 e - 3 x cos 2 x - 2 sin 2 x - 4 cos 2 x sin 2 x 2 cos 2 x - 4 sin 2 x ∣ ∣ ∣ ∣ ∣ e - 3 x ∣ ∣ ∣ - 2 sin 2 x - 4 cos 2 x 2 cos 2 x - 4 sin 2 x ∣ ∣ ∣ + 3 e - 3 x ∣ ∣ ∣ cos 2 x - 4 cos 2 x sin 2 x - 4 sin 2 x ∣ ∣ ∣ + 9 e - 3 x ∣ ∣ ∣ cos 2 x - 2 sin 2 x sin 2 x 2 cos 2 x ∣ ∣ ∣ = 26 e - 3 x \neq 0.

$\begin{aligned} W & = & | \begin{array}{r} e^{- 3 x} & \cos 2 x & \sin 2 x \\ - 3 e^{- 3 x} & - 2 \sin 2 x & 2 \cos 2 x \\ 9 e^{- 3 x} & - 4 \cos 2 x & - 4 \sin 2 x \end{array} | \\ = & e^{- 3 x} | \begin{array}{r} - 2 \sin 2 x & 2 \cos 2 x \\ - 4 \cos 2 x & - 4 \sin 2 x \end{array} | + 3 e^{- 3 x} | \begin{array}{r} \cos 2 x & \sin 2 x \\ - 4 \cos 2 x & - 4 \sin 2 x \end{array} | \\ + 9 e^{- 3 x} | \begin{array}{r} \cos 2 x & \sin 2 x \\ - 2 \sin 2 x & 2 \cos 2 x \end{array} | = 26 e^{- 3 x} \neq 0. \end{aligned}$

Because W≠0 $W \neq 0$ everywhere, it follows that y1, y2, $y_{1}, y_{2},$ and y3 $y_{3}$ are linearly independent on any open interval (including the entire real line).

Example 5

Show first that the three solutions

y 1 (x) = x, y 2 (x) = x ln x, and y 3 (x) = x 2

$y_{1} (x) = x, y_{2} (x) = x \ln x, and y_{3} (x) = x 2$

of the third-order equation

x 3 y (3) - x 2 y'' + 2 x y' - 2 y = 0

$x^{3} y^{(3)} - x^{2} y ″ + 2 x y ′ - 2 y = 0$ (10)

are linearly independent on the open interval x>0. $x > 0 .$ Then find a particular solution of Eq. (10) that satisfies the initial conditions

y (1) = 3, y' (1) = 2, y'' (1) = 1.

$y (1) = 3, y ′ (1) = 2, y ″ (1) = 1.$ (11)

Solution

Note that for x>0, $x > 0,$ we could divide each term in (10) by x3 $x^{3}$ to obtain a homogeneous linear equation of the standard form in (3). When we compute the Wronskian of the three given solutions, we find that

W = ∣ ∣ ∣ ∣ ∣ x 10 x ln x 1 + ln x 1 x x 2 2 x 2 ∣ ∣ ∣ ∣ ∣ = x .

$W = | \begin{array}{c} x & x \ln x & x^{2} \\ 1 & 1 + \ln x & 2 x \\ 0 & \frac{1}{x} & 2 \end{array} | = x .$

Thus W(x)≠0 $W (x) \neq 0$ for x>0, $x > 0,$ so y1, y2, $y_{1}, y_{2},$ and y3 $y_{3}$ are linearly independent on the interval x>0. $x > 0 .$ To find the desired particular solution, we impose the initial conditions in (11) on

y (x) y' (x) y'' (x) = = = c 1 x c 1 0 + + + c 2 x ln x c 2 (1 + ln x) c 2 x + + + c 3 x 2, 2 c 3 x, 2 c 3 .

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

This yields the simultaneous equations

y (1) y' (1) y'' (1) = = = c 1 c 1 + c 2 c 2 + + + c 3 2 c 3 2 c 3 = = = 3, 2, 1;

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

we solve to find c1=1, c2=−3, $c_{1} = 1, c_{2} = - 3,$ and c3=2. $c_{3} = 2 .$ Thus the particular solution in question is

y (x) = x - 3 x ln x + 2 x 2 .

$y (x) = x - 3 x \ln x + 2 x^{2} .$

Provided that W(y1,y2,…,yn)≠0, $W (y_{1}, y_{2}, \dots, y_{n}) \neq 0,$ it turns out (Theorem 4) that we can always find values of the coefficients in the linear combination

y = c 1 y 1 + c 2 y 2 + \dots + c n y n

$y = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n}$

that satisfy any given initial conditions of the form in (5). Theorem 3 provides the necessary nonvanishing of W in the case of linearly independent solutions.

Proof:

We have already proven part (a). To prove part (b), it is sufficient to assume that W(a)=0 $W (a) = 0$ at some point of I, and show this implies that the solutions y1, y2, …, $y_{1}, y_{2}, \dots,$ yn $y_{n}$ are linearly dependent. But W(a) is simply the determinant of coefficients of the system of n homogeneous linear equations

c 1 y 1 (a) c 1 y' 1 (a) c 1 y (n - 1) 1 (a) + + + c 2 y 2 (a) c 2 y' 2 (a) c 2 y (n - 1) 2 (a) + + + \dots \dots \dots + + + c n y n (a) c n y' n (a) c n (n - 1) n (a) = = ⋮ = 0, 0, 0

$\begin{aligned} c_{1} y_{1} (a) & + & c_{2} y_{2} (a) & + & \dots & + & c_{n} y_{n} (a) & = & 0, \\ c_{1} y_{1}^{′} (a) & + & c_{2} y_{2}^{′} (a) & + & \dots & + & c_{n} y_{n}^{′} (a) & = & 0, \\ ⋮ \\ c_{1} y_{1}^{(n - 1)} (a) & + & c_{2} y_{2}^{(n - 1)} (a) & + & \dots & + & c n_{n}^{(n - 1)} (a) & = & 0 \end{aligned}$ (12)

in the n unknowns c1, c2, …, $c_{1}, c_{2}, \dots,$ cn. $c_{n} .$ Because W(a)=0, $W (a) = 0,$ the basic fact from linear algebra quoted just after (9) implies that the equations in (12) have a nontrivial solution. That is, the numbers c1, c2, …, $c_{1}, c_{2}, \dots,$ cn $c_{n}$ are not all zero.

We now use these values to define the particular solution

Y (x) = c 1 y 1 (x) + c 2 y 2 (x) + \dots + c n y n (x)

$Y (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x) + \dots + c_{n} y_{n} (x)$ (13)

of Eq. (3). The equations in (12) then imply that Y satisfies the trivial initial conditions

Y (a) = y' (a) = \dots = Y (n - 1) (a) = 0.

$Y (a) = y ′ (a) = \dots = Y^{(n - 1)} (a) = 0.$

Theorem 2 (uniqueness) therefore implies that Y(x)≡0 $Y (x) \equiv 0$ on I. In view of (13) and the fact that c1, c2, …, $c_{1}, c_{2}, \dots,$ cn $c_{n}$ are not all zero, this is the desired conclusion that the solutions y1, y2, …, $y_{1}, y_{2}, \dots,$ yn $y_{n}$ are linearly dependent. This completes the proof of Theorem 3.

Remark

According to Problem 35, the Wronskian of Theorem 3 satisfies the first-order equation W'=−p1(x)W. $W ′ = - p_{1} (x) W .$ Solution of this equation yields Abel’s formula

W (x) = K exp (- \int p 1 (x) d x)

$W (x) = K \exp (- \int p_{1} (x) d x)$

for the homogeneous linear equation in (3). According as the constant K=0 $K = 0$ or K≠0, $K \neq 0,$ it follows immediately from Abel’s formula that either W vanishes everywhere or W is nonzero everywhere.

General Solutions

We can now show that, given any fixed set of n linearly independent solutions of a homogeneous nth-order equation, every (other) solution of the equation can be expressed as a linear combination of those n particular solutions. Using the fact from Theorem 3 that the Wronskian of n linearly independent solutions is nonzero, the proof of the following theorem is essentially the same as the proof of Theorem 4 of Section 5.1 (the case n=2 $n = 2$ ).

Remark

Theorem 4 tells us that, once we have found n linearly independent solutions y1,y2,…,yn $y_{1}, y_{2}, \dots, y_{n}$ of the nth-order homogeneous linear equation in (3), we really have found all of its solutions. For then the solution space S $S$ of the equation is an n-dimensional vector space with basis {y1,y2,…,yn}. ${y_{1}, y_{2}, \dots, y_{n}} .$ Because every solution of (3) can be expressed as a linear combination of the form

y = c 1 y 1 + c 2 y 2 + \dots + c n y n,

$y = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n},$ (14)

we call such a linear combination of n linearly independent particular solutions a general solution of the homogeneous linear differential equation.

Example 6

According to Example 4, the particular solutions y1(x)=e−3x, y2(x)=cos 2x, $y_{1} (x) = e^{- 3 x}, y_{2} (x) = \cos 2 x,$ and y3(x)=sin 2x $y_{3} (x) = \sin 2 x$ of the linear differential equation y(3)+3y''+4y'+12y=0 $y^{(3)} + 3 y ″ + 4 y ′ + 12 y = 0$ are linearly independent. Now Theorem 2 says that—given b0, b1, $b_{0}, b_{1},$ and $b_{2}$ —there exists a particular solution y(x) satisfying the initial conditions $y (0) = b_{0}, y ′ (0) = b_{1},$ and $y ″ (0) = b_{2} .$ Hence Theorem 4 implies that this particular solution is a linear combination of $y_{1}, y_{2},$ and $y_{3} .$ That is, there exist coefficients $c_{1}, c_{2},$ and $c_{3}$ such that

$y (x) = c_{1} e^{- 3 x} + c_{2} \cos 2 x + c_{3} \sin 2 x .$

Upon successive differentiation and substitution of $x = 0,$ we discover that to find these coefficients, we need only solve the three linear equations

$\begin{array}{r} c_{1} & + & c_{2} & = & b_{0}, \\ - 3 c_{1} & + & 2 c_{3} & = & b_{1}, \\ 9 c_{1} & - & 4 c_{2} & = & b_{2}, \end{array}$

(See the application for this section.)

Nonhomogeneous Equations

We now consider the nonhomogeneous nth-order linear differential equation

$y^{(n)} + p_{1} (x) y^{(n - 1)} + \dots + p_{n - 1} (x) y ′ + p_{n} (x) y = f (x)$ (2)

with associated homogeneous equation

$y^{(n)} + p_{1} (x) y^{(n - 1)} + \dots + p_{n - 1} (x) y ′ + p_{n} (x) y = 0.$ (3)

Suppose that a single fixed particular solution $y_{p}$ of the nonhomogeneous equation in (2) is known, and that Y is any other solution of Eq. (2). If $y_{c} = Y - y_{p},$ then subsitution of $y_{c}$ in the differential equation gives (using the linearity of differentiation)

$\begin{array}{l} y_{c}^{(n)} + p_{1} y_{c}^{(n - 1)} + \dots + p_{n - 1} y_{c}^{′} + p_{n} y_{c} \\ \begin{array}{l} = & [(Y^{(n)} + p_{1} Y^{(n - 1)} + \dots + p_{n - 1} Y ′ + p_{n} Y] \\ - [(y_{p}^{(n)} + p_{1} y_{p}^{(n - 1)} + \dots + p_{n - 1} y_{p}^{′} + p_{n} y_{p}] \\ = & f (x) - f (x) = 0. \end{array} \end{array}$

Thus $y_{c} = Y - y_{p}$ is a solution of the associated homogeneous equation in (3). Then

$Y = y_{c} + y_{p},$ (14)

and it follows from Theorem 4 that

$y_{c} = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n},$ (15)

where $y_{1}, y_{2}, \dots,$ $y_{n}$ are linearly independent solutions of the associated homogeneous equation. We call $y_{c}$ a complementary function of the nonhomogeneous equation and have thus proved that a general solution of the nonhomogeneous equation in (2) is the sum of its complementary function $y_{c}$ and a single particular solution $y_{p}$ of Eq. (2).

Example 7

It is evident that $y_{p} = 3 x$ is a particular solution of the equation

$y ″ + 4 y = 12 x,$ (17)

and that $y_{c} (x) = c_{1} \cos 2 x + c_{2} \sin 2 x$ is its complementary solution. Find a solution of Eq. (17) that satisfies the initial conditions $y (0) = 5, y ′ (0) = 7$ .

Solution

The general solution of Eq. (17) is

$y (x) = c_{1} \cos 2 x + c_{2} \sin 2 x + 3 x .$

Now

$y ′ (x) = - 2 c_{1} \sin 2 x + 2 c_{2} \cos 2 x + 3.$

Hence the initial conditions give

$\begin{array}{llrlrll} y (0) & = & c_{1} & = & 5, \\ y^{′} (0) & = & 2 c_{2} & + & 3 & = & 7. \end{array}$

We find that $c_{1} = 5$ and $c_{2} = 2 .$ Thus the desired solution is

$y (x) = 5 \cos 2 x + 2 \sin 2 x + 3 x .$

5.2 Problems

In Problems 1 through 6, show directly that the given functions are linearly dependent on the real line. That is, find a nontrivial linear combination of the given functions that vanishes identically.

$f (x) = 2 x, g (x) = 3 x^{2}, h (x) = 5 x - 8 x^{2}$
$f (x) = 5, g (x) = 2 - 3 x^{2}, h (x) = 10 + 15 x^{2}$
$f (x) = 0, g (x) = \sin x, h (x) = e^{x}$
$f (x) = 17, g (x) = 2 \sin^{2} x, h (x) = 3 \cos^{2} x$
$f (x) = 17, g (x) = \cos^{2} x, h (x) = \cos 2 x$
$f (x) = e^{x}, g (x) = \cosh x, h (x) = \sinh x$

In Problems 7 through 12, use the Wronskian to prove that the given functions are linearly independent on the indicated interval.

$f (x) = 1, g (x) = x, h (x) = x^{2};$ the real line
$f (x) = e^{x}, g (x) = e^{2 x}, h (x) = e^{3 x};$ the real line
$f (x) = e^{x}, g (x) = \cos x, h (x) = \sin x;$ the real line
$f (x) = e^{x}, g (x) = x^{- 2}, h (x) = x^{- 2} \ln x; x > 0$
$f (x) = x, g (x) = {x e}^{x}, h (x) = x^{2} e^{x};$ the real line
$f (x) = x, g (x) = \cos (\ln x), h (x) = \sin (\ln x); x > 0$

In Problems 13 through 20, a third-order homogeneous linear equation and three linearly independent solutions are given. Find a particular solution satisfying the given initial conditions.

$y^{(3)} + 2 y ″ - y ′ - 2 y = 0; y (0) = 1, y ′ (0) = 2, y ″ (0) = 0; y_{1} = e^{x}, y_{2} = e^{- x}, y_{3} = e^{- 2 x}$
$y^{(3)} - 6 y ″ + 11 y ′ - 6 y = 0; y (0) = 0, y ′ (0) = 0, y ″ (0) = 3; y_{1} = e^{x}, y_{2} = e^{2 x}, y_{3} = e^{3 x}$
$y^{(3)} - 3 y ″ + 3 y ′ - y = 0; y (0) = 2, y ′ (0) = 0, y ″ (0) = 0; y_{1} = e^{x}, y_{2} = {x e}^{x}, y_{3} = x^{2} e^{x}$
$y^{(3)} - 5 y ″ + 8 y ′ - 4 y = 0; y (0) = 1, y ′ (0) = 4, y ″ (0) = 0; y_{1} = e^{x}, y_{2} = e^{2 x}, y_{3} = {x e}^{2 x}$
$y^{(3)} + 9 y ′ = 0; y (0) = 3, y ′ (0) = - 1, y ″ (0) = 2; y_{1} = 1, y_{2} = \cos 3 x, y_{3} = \sin 3 x$
$y^{(3)} - 3 y ″ + 4 y ′ - 2 y = 0; y (0) = 1, y ′ (0) = 0, y ″ (0) = 0; y_{1} = e^{x}, y_{2} = e^{x} \cos x, y_{3} = e^{x} \sin x$ .
$x^{3} y^{(3)} - 3 x^{2} y ″ + 6 x y ′ - 6 y = 0; y (1) = 6, y ′ (1) = 14, y ″ (1) = 22; y_{1} = x, y_{2} = x^{2}, y_{3} = x^{3}$
$x^{3} y^{(3)} + 6 x^{2} y ″ + 4 x y ′ - 4 y = 0; y (1) = 1, y ′ (1) = 5, y ″ (1) = - 11; y_{1} = x, y_{2} = x^{- 2}, y_{3} = x^{- 2} \ln x$

In Problems 21 through 24, a nonhomogeneous differential equation, a complementary solution $y_{c},$ and a particular solution $y_{p}$ are given. Find a solution satisfying the given initial conditions.

$y ″ + y = 3 x; y (0) = 2, y ′ (0) = - 2; y_{c} = c_{1} \cos x + c_{2} \sin x; y_{p} = 3 x$
$y ″ - 4 y = 12; y (0) = 0, y ′ (0) = 10; y_{c} = c_{1} e^{2 x} + c_{2} e^{- 2 x}; y_{p} = - 3$
$y ″ - 2 y ′ - 3 y = 6; y (0) = 3, y ′ (0) = 11; y_{c} = c_{1} e^{- x} + c_{2} e^{3 x}; y_{p} = - 2$
$y ″ - 2 y ′ + 2 y = 2 x; y (0) = 4, y ′ (0) = 8; y_{c} = c_{1} e^{x} \cos x + c_{2} e^{x} \sin x; y_{p} = x + 1$
Let $L y = y ″ + p y ′ + q y .$ Suppose that $y_{1}$ and $y_{2}$ are two functions such that

$L y_{1} = f (x) and L y_{2} = g (x) .$

Show that their sum $y = y_{1} + y_{2}$ satisfies the nonhomogeneous equation $L y = f (x) + g (x)$ .
(a) Find by inspection particular solutions of the two nonhomogeneous equations

$y ″ + 2 y = 4 and y ″ + 2 y = 6 x .$

(b) Use the method of Problem 25 to find a particular solution of the differential equation $y ″ + 2 y = 6 x + 4$ .
Prove directly that the functions

$f_{1} (x) \equiv 1, f_{2} (x) = x, and f_{3} (x) = x^{2}$

are linearly independent on the whole real line. (Suggestion: Assume that $c_{1} + c_{2} x + c_{3} x^{2} = 0 .$ Differentiate this equation twice, and conclude from the equations you get that $c_{1} = c_{2} = c_{3} = 0$ .)
Generalize the method of Problem 27 to prove directly that the functions

$f_{0} (x) \equiv 1, f_{1} (x) = x, f_{2} (x) = x^{2}, \dots, f_{n} (x) = x^{n}$

are linearly independent on the real line.
Use the result of Problem 28 and the definition of linear independence to prove directly that, for any constant r, the functions

$f_{0} (x) = e^{r x}, f_{1} (x) = x e^{r x}, \dots, f_{n} (x) = x^{n} e^{r x}$

are linearly independent on the whole real line.
Verify that $y_{1} = x$ and $y_{2} = x^{2}$ are linearly independent solutions on the entire real line of the equation

$x^{2} y ″ - 2 x y ′ + 2 y = 0,$

but that $W (x, x^{2})$ vanishes at $x = 0 .$ Why do these observations not contradict part (b) of Theorem 3?
This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation

$y ″ + p y ′ + q y = 0,$

explain why the value of $y ″ (a)$ is determined by the values of y(a) and $y ′ (a)$ . (b) Prove that the equation

$y ″ - 2 y ′ - 5 y = 0$

has a solution satisfying the conditions

$y (0) = 1, y ′ (0) = 0 and y ″ (0) = c$

if and only if $C = 5$ .
Prove that an nth-order homogeneous linear differential equation satisfying the hypotheses of Theorem 2 has n linearly independent solutions $y_{1}, y_{2}, \dots, y_{n} .$ (Suggestion: Let $y_{i}$ be the unique solution such that

$y_{i}^{(i - 1)} (a) = 1 and y_{i}^{(k)} (a) = 0 if k \neq i - 1.)$
Suppose that the three numbers $r_{1}, r_{2},$ and $r_{3}$ are distinct. Show that the three functions $\exp (r_{1} x), \exp (r_{2} x),$ and $\exp (r_{3} x)$ are linearly independent by showing that their Wronskian

$W = \exp [(r_{1} + r_{2} + r_{3}) x] \cdot | \begin{array}{c} 1 & 1 & 1 \\ r_{1} & r_{2} & r_{3} \\ r_{1}^{2} & r_{2}^{2} & r_{3}^{3} \end{array} |$

is nonzero for all x.
Assume as known that the Vandermonde determinant

$V = | \begin{array}{c} 1 & 1 & \dots & 1 \\ r_{1} & r_{2} & \dots & r_{n} \\ r_{1}^{2} & r_{2}^{2} & \dots & r_{n}^{2} \\ ⋮ & ⋮ & ⋮ \\ r_{1}^{n - 1} & r_{2}^{n - 1} & \dots & r_{n}^{n - 1} \end{array} |$

is nonzero if the numbers $r_{1}, r_{2}, \dots, r_{n}$ are distinct. Prove by the method of Problem 33 that the functions

$f_{i} (x) = \exp (r_{i} x), 1 ≦ i ≦ n$

are linearly independent.
According to Problem 32 of Section 5.1, the Wronskian $W (y_{1}, y_{2})$ of two solutions of the second-order equation

$y ″ + p_{1} (x) y ′ + p_{2} (x) y = 0$

is given by Abel’s formula

$W (x) = K \exp (- \int p_{1} (x) d x)$

for some constant K. It can be shown that the Wronskian of n solutions $y_{1}, y_{2}, \dots, y_{n}$ of the nth-order equation

$y^{(n)} + p_{1} (x) y^{(n - 1)} + \dots + p_{n - 1} (x) y ′ + p_{n} (x) y = 0$

satisfies the same identity. Prove this for the case $n = 3$ as follows: (a) The derivative of a determinant of functions is the sum of the determinants obtained by separately differentiating the rows of the original determinant. Conclude that

$W^{′} = | \begin{array}{l} y_{1} & y_{2} & y_{3} \\ y_{1}^{′} & y_{2}^{′} & y_{3}^{′} \\ y_{1}^{(3)} & y_{2}^{(3)} & y_{3}^{(3)} \end{array} | .$

(b) Substitute for $y_{1}^{(3)}, y_{2}^{(3)},$ and $y_{3}^{(3)}$ from the equation

$y^{(3)} + p_{1} y ″ + p_{2} y ′ + p_{3} y = 0,$

and then show that $W ′ = - p_{1} W .$ Integration now gives Abel’s formula.
Suppose that one solution $y_{1} (x)$ of the homogeneous second-order linear differential equation

$y ″ + p (x) y ′ + q (x) y = 0$ (18)

is known (on an interval I where p and q are continuous functions). The method of reduction of order consists of substituting $y_{2} (x) = v (x) y_{1} (x)$ in (18) and attempting to determine the function v(x) so that $y_{2} (x)$ is a second linearly independent solution of (18). After substituting $y = v (x) y_{1} (x)$ in Eq. (18), use the fact that $y_{1} (x)$ is a solution to deduce that

$y_{1} v^{″} + (2 y_{1}^{′} + p y_{1}) v^{′} = 0.$ (19)

If $y_{1} (x)$ is known, then (19) is a separable equation that is readily solved for the derivative $v ′ (x)$ of v(x). Integration of $v ′ (x)$ then gives the desired (nonconstant) function v(x).
Before applying Eq. (19) with a given homogeneous second-order linear differential equation and a known solution $y_{1} (x),$ the equation must first be written in the form of (18) with leading coefficient 1 in order to correctly determine the coefficient function p(x). Frequently it is more convenient to simply substitute $y = v (x) y_{1} (x)$ in the given differential equation and then proceed directly to find v(x). Thus, starting with the readily verified solution $y_{1} (x) = x^{3}$ of the equation

$x^{2} y ″ - 5 x y ′ + 9 y = 0 (x > 0),$

substitute $y = {v x}^{3}$ and deduce that $x v ″ + v ′ = 0 .$ Thence solve for $v (x) = C \ln x,$ and thereby obtain (with $C = 1$ ) the second solution $y_{2} (x) = x^{3} \ln x$ .

In each of Problems 38 through 42, a differential equation and one solution $y_{1}$ are given. Use the method of reduction of order as in Problem 37 to find a second linearly independent solution $y_{2}$ .

$x^{2} y ″ + x y ′ - 9 y = 0$ $(x > 0)$ ; $y_{1} (x) = x^{3}$
$4 y ″ - 4 y ′ + y = 0$ ; $y_{1} (x) = e^{x / 2}$
$x^{2} y ″ - x (x + 2) y ′ + (x + 2) y = 0$ $(x > 0)$ ; $y_{1} (x) = x$
$(x + 1) y ″ - (x + 2) y ′ + y = 0$ $(x > - 1)$ ; $y_{1} (x) = e^{x}$
$(1 - x^{2}) y ″ + 2 x y ′ - 2 y = 0$ $(- 1 < x < 1)$ ; $y_{1} (x) = x$
First note that $y_{1} (x) = x$ is one solution of Legendre’s equation of order 1,

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

Then use the method of reduction of order to derive the second solution

$y_{2} (x) = 1 - \frac{x}{2} \ln \frac{1 + x}{1 - x} (for - 1 < x < 1) .$
First verify by substitution that $y_{1} (x) = x^{- 1/2} \cos x$ is one solution (for $x > 0$ ) of Bessel’s equation of order $\frac{1}{2}$ ,

$x^{2} y ″ + x y ′ + (x^{2} - \frac{1}{4}) y = 0.$

Then derive by reduction of order the second solution $y_{2} (x) = x^{- 1/2} \sin x$ .

5.2 Application Plotting Third-Order Solution Families

This application deals with the plotting by computer of families of solutions such as those illustrated in Figs. 5.2.2 through 5.2.4. We know from Example 6 that the general solution of

$y^{(3)} + 3 y ″ + 4 y ′ + 12 y = 0$ (1)

$y (x) = c_{1} e^{- 3 x} + c_{2} \cos 2 x + c_{3} \sin 2 x .$ (2)

For Fig. 5.2.2, use the method of Example 6 to show that the particular solution of Eq. (1)satisfying the initial conditions $y (0) = a, y ′ (0) = 0,$ and $y ″ (0) = 0$ is given by

$y (x) = \frac{a}{13} (4 e^{- 3 x} + 9 \cos 2 x + 6 \sin 2 x) .$ (3)

The Matlab loop

x = -1.5 :  0.02 :  5 % x-vector from x = -1.5 to x = 5
for a = -3 :  1 :  3  % for a = -3 to 3 with da = 1 do
   c1 = 4*a/13;
   c2 = 9*a/13;
   c3 = 6*a/13;
   y = c1*exp(-3*x) + c2*cos(2*x) + c3*sin(2*x);
   plot(x,y)
end

was used to generate Fig. 5.2.2.

For Fig. 5.2.3, show that the particular solution of Eq. (1) satisfying the initial conditions $y (0) = 0, y ′ (0) = b,$ and $y ″ (0) = 0$ is given by

$y (x) = \frac{b}{2} \sin 2 x,$ (4)

and alter the preceding for-loop accordingly.

For Fig. 5.2.4, show that the particular solution of Eq. (1) satisfying the initial conditions $y (0) = 0, y ′ (0) = 0,$ and $y ″ (0) = c$ is given by

$y (x) = \frac{c}{26} (2 e^{- 3 x} - 2 \cos 2 x + 3 \sin 2 x) .$ (5)

Computer algebra systems such as Maple and Mathematica, as well as graphing calculators, have commands to carry out for-loops such as the one shown here. Begin by reproducing Figs. 5.2.2 through 5.2.4. Then plot similar families of solution curves for the differential equations in Problems 13 through 20.

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Table of Contents for 5.2 General Solutions of Linear Equations

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Table of Contents for
5.2 General Solutions of Linear Equations