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2.7* Homogeneous Linear Differential Equations with Constant Coefficients

As an introduction to this section, consider the following physical problem. A weight of mass m is attached to a vertically suspended spring that is allowed to stretch until the forces acting on the weight are in equilibrium. Suppose that the weight is now motionless and impose an xy-coordinate system with the weight at the origin and the spring lying on the positive y-axis (see Figure 2.7).

A diagram of a vertical coiled spring sits on the y axis of an x y plane. Below the spring is a solid box that is centered on the origin. — Figure 2.7

Suppose that at a certain time, say t=0 $t = 0$ , the weight is lowered a distance s along the y-axis and released. The spring then begins to oscillate.

We describe the motion of the spring. At any time t≥0 $t \geq 0$ , let F(t) denote the force acting on the weight and y(t) denote the position of the weight along the y-axis. For example, y(0)=−s $y (0) = - s$ . The second derivative of y with respect to time, y′′(t) $y^{″} (t)$ , is the acceleration of the weight at time t; hence, by Newton’s second law of motion,

F (t) = m y'' (t) . (1)

$F (t) = m y^{″} (t) . (1)$

It is reasonable to assume that the force acting on the weight is due totally to the tension of the spring, and that this force satisfies Hooke’s law: The force acting on the weight is proportional to its displacement from the equilibrium position, but acts in the opposite direction. If k>0 $k > 0$ is the proportionality constant, then Hooke’s law states that

F (t) = - k y (t) . (2)

$F (t) = - k y (t) . (2)$

Combining (1) and (2), we obtain my′′=−ky $m y^{″} = - k y$ or

y'' + k m y = 0. (3)

$y^{″} + \frac{k}{m} y = 0. (3)$

The expression (3) is an example of a differential equation. A differential equation in an unknown function y=y(t) $y = y (t)$ is an equation involving y, t, and derivatives of y. If the differential equation is of the form

a n y (n) + a n - 1 y (n - 1) + \dots + a 1 y (1) + a 0 y = f, (4)

$a_{n} y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{1} y^{(1)} + a_{0} y = f, (4)$

where a0, a1, …, an $a_{0}, a_{1}, \dots, a_{n}$ and f are functions of t and y(k) $y^{(k)}$ denotes the kth derivative of y, then the equation is said to be linear. The functions ai $a_{i}$ are called the coefficients of the differential equation (4). Thus (3) is an example of a linear differential equation in which the coefficients are constants and the function f is identically zero. When f is identically zero, (4) is called homogeneous.

In this section, we apply the linear algebra we have studied to solve homogeneous linear differential equations with constant coefficients. If an≠0 $a_{n} \neq 0$ , we say that differential equation (4) is of order n. In this case, we divide both sides by an to obtain a new, but equivalent, equation

y (n) + b n - 1 y (n - 1) + \dots + b 1 y (1) + b 0 y = 0,

$y^{(n)} + b_{n - 1} y^{(n - 1)} + \dots + b_{1} y^{(1)} + b_{0} y = 0,$

where bi=ai/an $b_{i} = a_{i} / a_{n}$ for i=0, 1, …, n−1 $i = 0, 1, \dots, n - 1$ . Because of this observation, we always assume that the coefficient an $a_{n}$ in (4) is 1.

A solution to (4) is a function that when substituted for y reduces (4) to an identity.

Example 1

The function y(t)=sink/m−−−−√ t $y (t) = sin \sqrt{k / m} t$ is a solution to (3) since

y'' (t) + k m y (t) = - k m sin k m - - - \sqrt t + k m sin k m - - - \sqrt t = 0

$y^{″} (t) + \frac{k}{m} y (t) = - \frac{k}{m} sin \sqrt{\frac{k}{m}} t + \frac{k}{m} sin \sqrt{\frac{k}{m}} t = 0$

for all t. Notice, however, that substituting y(t)=t $y (t) = t$ into (3) yields

y'' (t) + k m y (t) = k m t,

$y^{″} (t) + \frac{k}{m} y (t) = \frac{k}{m} t,$

which is not identically zero. Thus y(t)=t $y (t) = t$ is not a solution to (3).

In our study of differential equations, it is useful to regard solutions as complex-valued functions of a real variable even though the solutions that are meaningful to us in a physical sense are real-valued. The convenience of this viewpoint will become clear later. Thus we are concerned with the vector space F(R, C) (as defined in Example 3 of Section 1.2). In order to consider complex-valued functions of a real variable as solutions to differential equations, we must define what it means to differentiate such functions. Given a complex-valued function x∈F(R, C) $x \in F (R, C)$ of a real variable t, there exist unique real-valued functions x1 $x_{1}$ and x2 $x_{2}$ of t, such that

x (t) = x 1 (t) + i x 2 (t) for t \in R,

$x (t) = x_{1} (t) + i x_{2} (t) for t \in R,$

where i is the imaginary number such that i2=−1 $i^{2} = - 1$ . We call x1 $x_{1}$ the real part and x2 $x_{2}$ the imaginary part of x.

Definitions.

Given a function x∈F(R, C) $x \in F (R, C)$ with real part x1 $x_{1}$ and imaginary part x2 $x_{2}$ , we say that x is differentiable if x1 $x_{1}$ and x2 $x_{2}$ are differentiable. If x is differentiable, we define the derivative x′ $x^{'}$ of x by

x' = x' 1 + i x' 2 .

$x^{'} = {x^{'}}_{1} + i {x^{'}}_{2} .$

We illustrate some computations with complex-valued functions in the following example.

Example 2

Suppose that x(t)=cos2t+isin2t $x (t) = cos 2 t + i sin 2 t$ . Then

x' (t) = - 2 sin 2 t + 2 i cos 2 t .

$x^{'} (t) = - 2 sin 2 t + 2 i cos 2 t .$

We next find the real and imaginary parts of x2 $x^{2}$ . Since

x 2 (t) = (cos 2 t + i sin 2 t) 2 = (cos 2 2 t - sin 2 2 t) + i (2 sin 2 t cos 2 t) = cos 4 t + i sin 4 t,

$\begin{array}{l} x^{2} (t) = {(cos 2 t + i sin 2 t)}^{2} = ({cos}^{2} 2 t - {sin}^{2} 2 t) + i (2 sin 2 t cos 2 t) \\ = cos 4 t + i sin 4 t, \end{array}$

the real part of x2(t) $x^{2} (t)$ is cos4t $cos 4 t$ , and the imaginary part is sin 4t.

The next theorem indicates that we may limit our investigations to a vector space considerably smaller than F(R, C). Its proof, which is illustrated in Example 3, involves a simple induction argument, which we omit.

Theorem 2.27.

Any solution to a homogeneous linear differential equation with constant coefficients has derivatives of all orders; that is, if x is a solution to such an equation, then x(k) $x^{(k)}$ exists for every positive integer k.

Example 3

To illustrate Theorem 2.27, consider the equation

y (2) + 4 y = 0.

$y^{(2)} + 4 y = 0.$

Clearly, to qualify as a solution, a function y must have two derivatives. If y is a solution, however, then

y (2) = - 4 y .

$y^{(2)} = - 4 y .$

Thus since y(2) $y^{(2)}$ is a constant multiple of a function y that has two derivatives, y(2) $y^{(2)}$ must have two derivatives. Hence y(4) $y^{(4)}$ exists; in fact,

y (4) = - 4 y (2) .

$y^{(4)} = - 4 y^{(2)} .$

Since y(4) $y^{(4)}$ is a constant multiple of a function that we have shown has at least two derivatives, it also has at least two derivatives; hence y(6) $y^{(6)}$ exists. Continuing in this manner, we can show that any solution has derivatives of all orders.

Definition.

We use C∞ $C^{\infty}$ to denote the set of all functions in F(R, C) that have derivatives of all orders.

It is a simple exercise to show that C∞ $C^{\infty}$ is a subspace of F(R, C) and hence a vector space over C. In view of Theorem 2.27, it is this vector space that is of interest to us. For x∈C∞ $x \in C^{\infty}$ , the derivative x′ $x^{'}$ of x also lies in C∞ $C^{\infty}$ . We can use the derivative operation to define a mapping D:C∞→C∞ $D : C^{\infty} \to C^{\infty}$ by

D (x) = x' for x \in C \infty .

$D (x) = x^{'} for x \in C^{\infty} .$

It is easy to show that D is a linear operator. More generally, consider any polynomial over C of the form

p (t) = a n t n + a n - 1 t n - 1 + \dots + a 1 t + a 0 .

$p (t) = a_{n} t^{n} + a_{n - 1} t^{n - 1} + \dots + a_{1} t + a_{0} .$

If we define

p (D) = a n D n + a n - 1 D n - 1 + \dots + a 1 D + a 0 I,

$p (D) = a_{n} D^{n} + a_{n - 1} D^{n - 1} + \dots + a_{1} D + a_{0} I,$

then p(D) is a linear operator on C∞ $C^{\infty}$ . (See Appendix E.)

Definition.

For any polynomial p(t) over C of positive degree, we call p(D) a differential operator with constant coefficients, or, more simply, a differential operator. The order of the differential operator p(D) is the degree of the polynomial p(t).

Differential operators are useful since they provide us with a means of reformulating a differential equation in the context of linear algebra. Any homogeneous linear differential equation with constant coefficients,

y (n) + a n - 1 y (n - 1) + \dots + a 1 y (1) + a 0 y = 0,

$y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{1} y^{(1)} + a_{0} y = 0,$

can be rewritten using differential operators as

(D n + a n - 1 D n - 1 + \dots + a 1 D + a 0 I) (y) = 0.

$(D^{n} + a_{n - 1} D^{n - 1} + \dots + a_{1} D + a_{0} I) (y) = 0.$

Definition.

Given the differential equation above, the complex polynomial

p (t) = t n + a n - 1 t n - 1 + \dots + a 1 t + a 0

$p (t) = t^{n} + a_{n - 1} t^{n - 1} + \dots + a_{1} t + a_{0}$

is called the auxiliary polynomial associated with the equation. For example, (3) has the auxiliary polynomial

p (t) = t 2 + k m .

$p (t) = t^{2} + \frac{k}{m} .$

Any homogeneous linear differential equation with constant coefficients can be rewritten as

p (D) (y) = 0,

$p (D) (y) = 0,$

where p(t) is the auxiliary polynomial associated with the equation. Clearly, this equation implies the following theorem.

Theorem 2.28.

The set of all solutions to a homogeneous linear differential equation with constant coefficients coincides with the null space of p(D), where p(t) is the auxiliary polynomial associated with the equation.

Proof.

Exercise.

Corollary.

The set of all solutions to a homogeneous linear differential equation with constant coefficients is a subspace of C∞ $C^{\infty}$ .

In view of the preceding corollary, we call the set of solutions to a homogeneous linear differential equation with constant coefficients the solution space of the equation. A practical way of describing such a space is in terms of a basis. We now examine a certain class of functions that is of use in finding bases for these solution spaces.

For a real number s, we are familiar with the real number es $e^{s}$ , where e is the unique number whose natural logarithm is 1 (i.e., ln e=1 $ln e = 1$ ). We know, for instance, certain properties of exponentiation, namely,

e s + t = e s e t and e - t = 1 e t

$e^{s + t} = e^{s} e^{t} and e^{- t} = \frac{1}{e^{t}}$

for any real numbers s and t. We now extend the definition of powers of e to include complex numbers in such a way that these properties are preserved.

Definition.

Let c=a+ib $c = a + i b$ be a complex number with real part a and imaginary part b. Define

e c = e a (cos b + i sin b) .

$e^{c} = e^{a} (cos b + i sin b) .$

The special case

e i b = cos b + i sin b

$e^{i b} = cos b + i sin b$

is called Euler’s formula.

For example, for c=2+i(π/3), $c = 2 + i (π / 3),$

e c = e 2 (cos π 3 + i sin π 3) = e 2 (1 2 + i 3 - \sqrt 2) .

$e^{c} = e^{2} (cos \frac{π}{3} + i sin \frac{π}{3}) = e^{2} (\frac{1}{2} + i \frac{\sqrt{3}}{2}) .$

Clearly, if c is real (b=0) $(b = 0)$ , then we obtain the usual result: ec=ea $e^{c} = e^{a}$ . Using the approach of Example 2, we can show by the use of trigonometric identities that

e c + d = e c e d and e - c = 1 e c

$e^{c + d} = e^{c} e^{d} and e^{- c} = \frac{1}{e^{c}}$

for any complex numbers c and d.

Definition.

A function f:R→C $f : R \to C$ defined by f(t)=ect $f (t) = e^{c t}$ for a fixed complex number c is called an exponential function.

The derivative of an exponential function, as described in the next theorem, is consistent with the real version. The proof involves a straightforward computation, which we leave as an exercise.

Theorem 2.29.

For any exponential function f(t)=ect, f′(t)=cect $f (t) = e^{c t}, f^{'} (t) = c e^{c t}$ .

Proof.

Exercise.

We can use exponential functions to describe all solutions to a homogeneous linear differential equation of order 1. Recall that the order of such an equation is the degree of its auxiliary polynomial. Thus an equation of order 1 is of the form

y' + a 0 y = 0. (5)

$y^{'} + a_{0} y = 0. (5)$

Theorem 2.30.

The solution space for (5) is of dimension 1 and has {e−a0t} ${e^{- a_{0} t}}$ as a basis.

Proof.

Clearly (5) has e−a0t $e^{- a_{0} t}$ as a solution. Suppose that x(t) is any solution to (5). Then

x' (t) = - a 0 x (t) for all t \in R .

$x^{'} (t) = - a_{0} x (t) for all t \in R .$

Define

z (t) = e a 0 t x (t) .

$z (t) = e^{a_{0} t} x (t) .$

Differentiating z yields

z' (t) = (e a 0 t)' x (t) + e a 0 t x' (t) = a 0 e a 0 t x (t) - a 0 e a 0 t x (t) = 0.

$z^{'} (t) = (e^{a_{0} t})^{'} x (t) + e^{a_{0} t} x^{'} (t) = a_{0} e^{a_{0} t} x (t) - a_{0} e^{a_{0} t} x (t) = 0.$

(Notice that the familiar product rule for differentiation holds for complex- valued functions of a real variable. A justification of this involves a lengthy, although direct, computation.)

Since z′ $z^{'}$ is identically zero, z is a constant function. (Again, this fact, well known for real-valued functions, is also true for complex-valued functions. The proof, which relies on the real case, involves looking separately at the real and imaginary parts of z.) Thus there exists a complex number k such that

z (t) = e a 0 t x (t) = k for all t \in R .

$z (t) = e^{a_{0} t} x (t) = k for all t \in R .$

x (t) = k e - a 0 t .

$x (t) = k e^{- a_{0} t} .$

We conclude that any solution to (5) is a scalar multiple of e−a0t $e^{- a_{0} t}$ .

Another way of stating Theorem 2.30 is as follows.

Corollary.

For any complex number c, the null space of the differential operator D−cI $D - c I$ has {ect} ${{e}^{c t}}$ as a basis.

We next concern ourselves with differential equations of order greater than one. Given an nth order homogeneous linear differential equation with constant coefficients,

y (n) + a n - 1 y (n - 1) + \dots + a 1 y (1) + a 0 y = 0,

$y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{1} y^{(1)} + a_{0} y = 0,$

its auxiliary polynomial

p (t) = t n + a n - 1 t n - 1 + \dots + a 1 t + a 0

$p (t) = t^{n} + a_{n - 1} t^{n - 1} + \dots + a_{1} t + a_{0}$

factors into a product of polynomials of degree 1, that is,

p (t) = (t - c 1) (t - c 2) \dots (t - c n),

$p (t) = (t - c_{1}) (t - c_{2}) \dots (t - c_{n}),$

where c1, c2, …, cn $c_{1}, c_{2}, \dots, c_{n}$ are (not necessarily distinct) complex numbers. (This follows from the fundamental theorem of algebra in Appendix D.) Thus

p (D = (D - c 1 I) (D - c 2 I) \dots (D - c n I) .

$p (D = (D - c_{1} I) (D - c_{2} I) \dots (D - c_{n} I) .$

The operators D−ciI $D - c_{i} I$ commute, and so, by Exercise 9, we have that

N (D - c i I) \subseteq N (p (D)) for all i .

$N (D - c_{i} I) \subseteq N (p (D)) for all i .$

Since N(p(D)) coincides with the solution space of the given differential equation, we can deduce the following result from the preceding corollary.

Theorem 2.31.

Let p(t) be the auxiliary polynomial for a homogeneous linear differential equation with constant coefficients. For any complex number c, if c is a zero of p(t), then ect $e^{c t}$ is a solution to the differential equation.

Example 4

Given the differential equation

y'' - 3 y' + 2 y = 0,

$y^{″} - 3 y^{'} + 2 y = 0,$

its auxiliary polynomial is

p (t) = t 2 - 3 t + 2 = (t - 1) (t - 2) .

$p (t) = t^{2} - 3 t + 2 = (t - 1) (t - 2) .$

Hence, by Theorem 2.31, et $e^{t}$ and e2t $e^{2 t}$ are solutions to the differential equation because c=1 $c = 1$ and c=2 $c = 2$ are zeros of p(t). Since the solution space of the differential equation is a subspace of C∞ , span ({et, e2t}) $C^{\infty}, span ({e^{t}, e^{2 t}})$ lies in the solution space. It is a simple matter to show that {et, e2t} ${e^{t}, e^{2 t}}$ is linearly independent. Thus if we can show that the solution space is two-dimensional, we can conclude that {et, e2t} ${e^{t}, e^{2 t}}$ is a basis for the solution space. This result is a consequence of the next theorem.

Theorem 2.32.

For any differential operator p(D) of order n, the null space of p(D) is an n-dimensional subspace of C∞ $C^{\infty}$ .

As a preliminary to the proof of Theorem 2.32, we establish two lemmas.

Lemma 1.

The differential operator D−cI:C∞→C∞ $D - c I : C^{\infty} \to C^{\infty}$ is onto for any complex number c.

Proof.

Let v∈C∞ $v \in C^{\infty}$ . We wish to find u∈C∞ $u \in C^{\infty}$ such that (D−cI)u=v $(D - c I) u = v$ . Let w(t)=v(t)e−ct $w (t) = v (t) e^{- c t}$ for t∈R $t \in R$ . Clearly, w∈C∞ $w \in C^{\infty}$ because both v and e−ct $e^{- c t}$ lie in C∞ $C^{\infty}$ . Let w1 $w_{1}$ and w2 $w_{2}$ be the real and imaginary parts of w. Then w1 $w_{1}$ and w2 $w_{2}$ are continuous because they are differentiable. Hence they have antiderivatives, say, W1 $W_{1}$ and W2 $W_{2}$ , respectively. Let W:R→C $W : R \to C$ be defined by

W (t) = W 1 (t) + i W 2 (t) for t \in R .

$W (t) = W_{1} (t) + i W_{2} (t) for t \in R .$

Then W∈C∞ $W \in C^{\infty}$ , and the real and imaginary parts of W are W1 $W_{1}$ and W2 $W_{2}$ , respectively. Furthermore, W′=w $W^{'} = w$ . Finally, let u:R→C $u : R \to C$ be defined by u(t)=W(t)ect $u (t) = W (t) e^{c t}$ for t∈R $t \in R$ . Clearly u∈C∞ $u \in C^{\infty}$ , and since

(D - c I) u (t) = u' (t) - c u (t) = W' (t) e c t + W (t) c e c t - c W (t) e c t = w (t) e c t = v (t) e - c t e c t = v (t),

$\begin{array}{l} (D - c I) u (t) = u^{'} (t) - c u (t) \\ = W^{'} (t) e^{c t} + W (t) c e^{c t} - c W (t) e^{c t} \\ = w (t) e^{c t} \\ = v (t) e^{- c t} e^{c t} \\ = v (t), \end{array}$

we have (D−cI)u=v $(D - c I) u = v$ .

Lemma 2.

Let V be a vector space, and suppose that T and U are linear operators on V such that U is onto and the null spaces of T and U are finite-dimensional. Then the null space of TU is finite-dimensional, and

dim (N (TU)) = dim (N (T)) + dim (N (U)) .

$dim (N (TU)) = dim (N (T)) + dim (N (U)) .$

Proof.

Let p=dim(N(T)), q=dim(N(U)) $p = dim (N (T)), q = dim (N (U))$ and {u1, u2, …, up} ${u_{1}, u_{2}, \dots, u_{p}}$ and {v1, v2, …, vq} ${v_{1}, v_{2}, \dots, v_{q}}$ be bases for N(T) and N(U), respectively. Since U is onto, we can choose for each i(1≤i≤p) $i (1 \leq i \leq p)$ a vector wi∈V $w_{i} \in V$ such that U(wi)=ui $U (w_{i}) = u_{i}$ . Note that the wi $w_{i}$ ’s are distinct. Furthermore, for any i and j, wi≠vj $w_{i} \neq v_{j}$ , for otherwise ui=U(wi)=U(vj)=0 $u_{i} = U (w_{i}) = U (v_{j}) = 0$ —a contradiction. Hence the set

β = {w 1, w 2, \dots, w p, v 1, v 2, \dots, v q}

$β = {w_{1}, w_{2}, \dots, w_{p}, v_{1}, v_{2}, \dots, v_{q}}$

contains p+q $p + q$ distinct vectors. To complete the proof of the lemma, it suffices to show that ft is a basis for N(TU).

We first show that β $β$ generates N(TU). Since for any wi $w_{i}$ and vj $v_{j}$ in β, TU(wi)=T(ui)=0 $β, TU (w_{i}) = T (u_{i}) = 0$ , and TU(vj)=T(0)=0 $TU (v_{j}) = T (0) = 0$ , it follows that β⊆N(TU) $β \subseteq N (TU)$ . Now suppose that v∈N(TU) $v \in N (TU)$ . Then 0=TU(v)=T(U(v)) $0 = TU (v) = T (U (v))$ . Thus U(v)∈N(T) $U (v) \in N (T)$ . So there exist scalars a1, a2, …, ap $a_{1}, a_{2}, \dots, a_{p}$ such that

U (v) = a 1 u 1 + a 2 u 2 + \dots + a p u p = a 1 U (w 1) + a 2 U (w 2) + \dots + a p U (w p) = U (a 1 w 1 + a 2 w 2 + \dots + a p w p) .

$\begin{array}{l} U (v) = a_{1} u_{1} + a_{2} u_{2} + \dots + a_{p} u_{p} \\ = a_{1} U (w_{1}) + a_{2} U (w_{2}) + \dots + a_{p} U (w_{p}) \\ = U (a_{1} w_{1} + a_{2} w_{2} + \dots + a_{p} w_{p}) . \end{array}$

Hence

U (v - (a 1 w 1 + a 2 w 2 + \dots + a p w p)) = 0.

$U (v - (a_{1} w_{1} + a_{2} w_{2} + \dots + a_{p} w_{p})) = 0.$

Consequently, v−(a1w1+a2w2+⋯+apwp) $v - (a_{1} w_{1} + a_{2} w_{2} + \dots + a_{p} w_{p})$ lies in N(U). It follows that there exist scalars b1, b2, …, bq $b_{1}, b_{2}, \dots, b_{q}$ such that

v - (a 1 w 1 + a 2 w 2 + \dots + a p w p) = b 1 v 1 + b 2 v 2 + \dots + b q v q

$v - (a_{1} w_{1} + a_{2} w_{2} + \dots + a_{p} w_{p}) = b_{1} v_{1} + b_{2} v_{2} + \dots + b_{q} v_{q}$

v = a 1 w 1 + a 2 w 2 + \dots + a p w p + b 1 v 1 + b 2 v 2 + \dots + b q v q .

$v = a_{1} w_{1} + a_{2} w_{2} + \dots + a_{p} w_{p} + b_{1} v_{1} + b_{2} v_{2} + \dots + b_{q} v_{q} .$

Therefore β $β$ spans N(TU).

To prove that β $β$ is linearly independent, let a1, a2, …, ap, b1, b2, …, bq $a_{1}, a_{2}, \dots, a_{p}, b_{1}, b_{2}, \dots, b_{q}$ be any scalars such that

a 1 w 1 + a 2 w 2 + \dots + a p w p + b 1 v 1 + b 2 v 2 + \dots + b q v q = 0. (6)

$a_{1} w_{1} + a_{2} w_{2} + \dots + a_{p} w_{p} + b_{1} v_{1} + b_{2} v_{2} + \dots + b_{q} v_{q} = 0. (6)$

Applying U to both sides of (6), we obtain

a 1 u 1 + a 2 u 2 + \dots + a p u p = 0.

$a_{1} u_{1} + a_{2} u_{2} + \dots + a_{p} u_{p} = 0.$

Since {u1, u2, …, up} ${u_{1}, u_{2}, \dots, u_{p}}$ is linearly independent, the ai $a_{i}$ ’ s are all zero. Thus (6) reduces to

b 1 v 1 + b 2 v 2 + \dots + b q v q = 0.

$b_{1} v_{1} + b_{2} v_{2} + \dots + b_{q} v_{q} = 0.$

Again, the linear independence of {v1, v2, …, vq} ${v_{1}, v_{2}, \dots, v_{q}}$ implies that the bi $b_{i}$ ’s are all zero. We conclude that β $β$ is a basis for N(TU). Hence N(TU) is finite-dimensional, and dim(N(TU))=p+q=dim(N(T))+dim(N(U)) $dim (N (TU)) = p + q = dim (N (T)) + dim (N (U))$ .

Proof of Theorem 2.32.

The proof is by mathematical induction on the order of the differential operator p(D). The first-order case coincides with Theorem 2.30. For some integer n>1 $n > 1$ , suppose that Theorem 2.32 holds for any differential operator of order less than n, and consider a differential operator p(D) of order n. The polynomial p(t) can be factored into a product of two polynomials as follows:

p (t) = q (t) (t - c),

$p (t) = q (t) (t - c),$

where q(t) is a polynomial of degree n−1 $n - 1$ and c is a complex number. Thus the given differential operator may be rewritten as

p (D) = q (D) (D - c I) .

$p (D) = q (D) (D - c I) .$

Now, by Lemma 1, D−cI $D - c I$ is onto, and by the corollary to Theorem 2.30, dim(N(D−cI))=1 $dim (N (D - c I)) = 1$ . Also, by the induction hypothesis, dim(N(q(D))=n−1 $dim (N (q (D)) = n - 1$ . Thus, by Lemma 2, we conclude that

dim (N (p (D))) = dim (N (q (D))) + dim (N (D - c I)) = (n - 1) + 1 = n .

$\begin{array}{l} dim (N (p (D))) = dim (N (q (D))) + dim (N (D - c I)) \\ = (n - 1) + 1 = n . \end{array}$

Corollary.

The solution space of any nth-order homogeneous linear differential equation with constant coefficients is an n-dimensional subspace of C∞ $C^{\infty}$ .

The corollary to Theorem 2.32 reduces the problem of finding all solutions to an nth-order homogeneous linear differential equation with constant coefficients to finding a set of n linearly independent solutions to the equation. By the results of Chapter 1, any such set must be a basis for the solution space. The next theorem enables us to find a basis quickly for many such equations. Hints for its proof are provided in the exercises.

Theorem 2.33.

Given n distinct complex numbers c1, c2…, cn $c_{1}, c_{2} \dots, c_{n}$ , the set of exponential functions {ec1t, ec2t, …, ecnt} ${e^{c_{1} t}, e^{c_{2} t}, \dots, e^{c_{n} t}}$ is linearly independent.

Proof.

Exercise. (See Exercise 10.)

Corollary.

For any nth-order homogeneous linear differential equation with constant coefficients, if the auxiliary polynomial has n distinct zeros c1, c2…, cn $c_{1}, c_{2} \dots, c_{n}$ , then {ec1t, ec2t, …, ecnt} ${e^{c_{1} t}, e^{c_{2} t}, \dots, e^{c_{n} t}}$ is a basis for the solution space of the differential equation.

Proof.

Exercise. (See Exercise 10.)

Example 5

We find all solutions to the differential equation

y'' + 5 y' + 4 y = 0.

$y^{″} + 5 y^{'} + 4 y = 0.$

Since the auxiliary polynomial factors as (t+4)(t+1) $(t + 4) (t + 1)$ , it has two distinct zeros, −1 $- 1$ and −4 $- 4$ . Thus {e−t, e−4t} ${e^{- t}, e^{- 4 t}}$ is a basis for the solution space. So any solution to the given equation is of the form

y (t) = b 1 e - t + b 2 e - 4 t

$y (t) = b_{1} e^{- t} + b_{2} e^{- 4 t}$

for unique scalars b1 $b_{1}$ and b2 $b_{2}$ .

Example 6

We find all solutions to the differential equation

y'' + 9 y = 0.

$y^{″} + 9 y = 0.$

The auxiliary polynomial t2+9 $t^{2} + 9$ factors as (t−3i)(t+3i) $(t - 3 i) (t + 3 i)$ and hence has distinct zeros c1=3i $c_{1} = 3 i$ and c2=−3i $c_{2} = - 3 i$ . Thus {e3it, e−3it} ${e^{3 i t}, e^{- 3 i t}}$ is a basis for the solution space. Since

cos 3 t = 1 2 (e 3 i t + e - 3 i t) and sin 3 t = 1 2 i (e 3 i t - e - 3 i t),

$cos 3 t = \frac{1}{2} (e^{3 i t} + e^{- 3 i t}) and sin 3 t = \frac{1}{2 i} (e^{3 i t} - e^{- 3 i t}),$

it follows from Exercise 7 that {cos3t, sin3t} ${cos 3 t, sin 3 t}$ is also a basis for this solution space. This basis has an advantage over the original one because it consists of the familiar sine and cosine functions and makes no reference to the imaginary number i. Using this latter basis, we see that any solution to the given equation is of the form

y (t) = b 1 cos 3 t + b 2 sin 3 t

$y (t) = b_{1} cos 3 t + b_{2} sin 3 t$

for unique scalars b1 $b_{1}$ and b2 $b_{2}$ .

Next consider the differential equation

y'' + 2 y' + y = 0,

$y^{″} + 2 y^{'} + y = 0,$

for which the auxiliary polynomial is (t+1)2 ${(t + 1)}^{2}$ . By Theorem 2.31, e−t $e^{- t}$ is a solution to this equation. By the corollary to Theorem 2.32, its solution space is two-dimensional. In order to obtain a basis for the solution space, we need a solution that is linearly independent of e−t $e^{- t}$ . The reader can verify that te−t $t e^{- t}$ is a such a solution. The following lemma extends this result.

Lemma.

For a given complex number c and positive integer n, suppose that (t−c)n ${(t - c)}^{n}$ is the auxiliary polynomial of a homogeneous linear differential equation with constant coefficients. Then the set

β = {e c t, t e c t, \dots, t n - 1 e c t}

$β = {e^{c t}, t e^{c t}, \dots, t^{n - 1} e^{c t}}$

is a basis for the solution space of the equation.

Proof.

Since the solution space is n-dimensional, we need only show that ft is linearly independent and lies in the solution space. First, observe that for any positive integer k,

(D - c I) (t k e c t) = k t k - 1 e c t + c t k e c t - c t k e c t = k t k - 1 e c t .

$\begin{array}{l} (D - c I) (t^{k} e^{c t}) = k t^{k - 1} e^{c t} + c t^{k} e^{c t} - c t^{k} e^{c t} \\ = k t^{k - 1} e^{c t} . \end{array}$

Hence for k<n $k < n$ ,

(D - c I) n (t k e c t) = 0.

${(D - c I)}^{n} (t^{k} e^{c t}) = 0.$

It follows that β $β$ is a subset of the solution space.

We next show that β $β$ is linearly independent. Consider any linear combination of vectors in β $β$ such that

b 0 e c t + b 1 t e c t + \dots + b n - 1 t n - 1 e c t = 0 (7)

$b_{0} e^{c t} + b_{1} t e^{c t} + \dots + b_{n - 1} t^{n - 1} e^{c t} = 0 (7)$

for some scalars b0, b1, …, bn−1 $b_{0}, b_{1}, \dots, b_{n - 1}$ . Dividing by ect $e^{c t}$ in (7), we obtain

b 0 + b 1 t + \dots + b n - 1 t n - 1 = 0. (8)

$b_{0} + b_{1} t + \dots + b_{n - 1} t^{n - 1} = 0. (8)$

Thus the left side of (8) must be the zero polynomial function. We conclude that the coefficients b0, b1, …, bn−1 $b_{0}, b_{1}, \dots, b_{n - 1}$ are all zero. So β $β$ is linearly independent and hence is a basis for the solution space.

Example 7

We find all solutions to the differential equation

y (4) - 4 y (3) + 6 y (2) - 4 y (1) + y = 0.

$y^{(4)} - 4 y^{(3)} + 6 y^{(2)} - 4 y^{(1)} + y = 0.$

Since the auxiliary polynomial is

t 4 - 4 t 3 + 6 t 2 - 4 t + 1 = (t - 1) 4,

$t^{4} - 4 t^{3} + 6 t^{2} - 4 t + 1 = {(t - 1)}^{4},$

we can immediately conclude by the preceding lemma that {et, tet, t2et, t3et} ${e^{t}, t e^{t}, t^{2} e^{t}, t^{3} e^{t}}$ is a basis for the solution space. So any solution y to the given differential equation is of the form

y (t) = b 1 e t + b 2 t e t + b 3 t 2 e t + b 4 t 3 e t

$y (t) = b_{1} e^{t} + b_{2} t e^{t} + b_{3} t^{2} e^{t} + b_{4} t^{3} e^{t}$

for unique scalars b1, b2, b3 $b_{1}, b_{2}, b_{3}$ , and b4 $b_{4}$ .

The most general situation is stated in the following theorem.

Theorem 2.34.

Given a homogeneous linear differential equation with constant coefficients and auxiliary polynomial

(t - c 1) n 1 (t - c 2) n 2 \dots (t - c k) n k,

${(t - c_{1})}^{n_{1}} {(t - c_{2})}^{n_{2}} \dots {(t - c_{k})}^{n_{k}},$

where n1, n2, …, nk $n_{1}, n_{2}, \dots, n_{k}$ are positive integers and c1, c2, …, ck $c_{1}, c_{2}, \dots, c_{k}$ are distinct complex numbers, the following set is a basis for the solution space of the equation:

{e c 1 t, t e c 1 t, \dots, t n 1 - 1 e c 1 t, \dots e c k t, t e c k t, \dots, t n k - 1 e c k t} .

${e^{c_{1} t}, t e^{c_{1} t}, \dots, t^{n_{1} - 1} e^{c_{1} t}, \dots e^{c_{k} t}, t e^{c_{k} t}, \dots, t^{n_{k - 1}} e^{c_{k} t}} .$

Proof.

Exercise.

Example 8

The differential equation

y (3) - 4 y (2) + 5 y (1) - 2 y = 0

$y^{(3)} - 4 y^{(2)} + 5 y^{(1)} - 2 y = 0$

has the auxiliary polynomial

t 3 - 4 t 2 + 5 t - 2 = (t - 1) 2 (t - 2) .

$t^{3} - 4 t^{2} + 5 t - 2 = {(t - 1)}^{2} (t - 2) .$

By Theorem 2.34, {et, tet, e2t} ${e^{t}, t e^{t}, e^{2 t}}$ is a basis for the solution space of the differential equation. Thus any solution y has the form

y (t) = b 1 e t + b 2 t e t + b 3 e 2 t

$y (t) = b_{1} e^{t} + b_{2} t e^{t} + b_{3} e^{2 t}$

for unique scalars b1, b2 $b_{1}, b_{2}$ , and b3 $b_{3}$ .

Exercises

Label the following statements as true or false.
1. (a) The set of solutions to an nth-order homogeneous linear differential equation with constant coefficients is an n-dimensional subspace of C∞ $C^{\infty}$ .
2. (b) The solution space of a homogeneous linear differential equation with constant coefficients is the null space of a differential operator.
3. (c) The auxiliary polynomial of a homogeneous linear differential equation with constant coefficients is a solution to the differential equation.
4. (d) Any solution to a homogeneous linear differential equation with constant coefficients is of the form aect $a e^{c t}$ or atkect $a t^{k} e^{c t}$ , where a and c are complex numbers and k is a positive integer.
5. (e) Any linear combination of solutions to a given homogeneous linear differential equation with constant coefficients is also a solution to the given equation.
6. (f) For any homogeneous linear differential equation with constant coefficients having auxiliary polynomial p(t), if c1, c2, …, ck $c_{1}, c_{2}, \dots, c_{k}$ are the distinct zeros of p(t), then {ec1t, ec2t, …, eckt} ${e^{c_{1} t}, e^{c_{2} t}, \dots, e^{c_{k} t}}$ is a basis for the solution space of the given differential equation.
7. (g) Given any polynomial p(t)∈P(C) $p (t) \in P (C)$ , there exists a homogeneous linear differential equation with constant coefficients whose auxiliary polynomial is p(t).
For each of the following parts, determine whether the statement is true or false. Justify your claim with either a proof or a counterexample, whichever is appropriate.
1. (a) Any finite-dimensional subspace of C∞ $C^{\infty}$ is the solution space of a homogeneous linear differential equation with constant coefficients.
2. (b) There exists a homogeneous linear differential equation with constant coefficients whose solution space has the basis {t, t2} ${t, t^{2}}$ .
3. (c) For any homogeneous linear differential equation with constant coefficients, if x is a solution to the equation, so is its derivative x′ $x^{'}$ .
  
  Given two polynomials p(t) and q(t) in P(C), if x∈N(p(D)) $x \in N (p (D))$ and y∈N(q(D)) $y \in N (q (D))$ , then
4. (d) x+y∈N(p(D)q(D)) $x + y \in N (p (D) q (D))$ .
5. (e) xy∈N(p(D)q(D)) $x y \in N (p (D) q (D))$ .
Find a basis for the solution space of each of the following differential equations.
1. (a) y′′+2y′+y=0 $y^{″} + 2 y^{'} + y = 0$
2. (b) y′′′=y′ ${y^{'}}^{'}^{'} = y^{'}$
3. (c) y(4)−2y(2)+y=0 $y^{(4)} - 2 y^{(2)} + y = 0$
4. (d) y′′+2y′+y=0 $y^{″} + 2 y^{'} + y = 0$
5. (e) y(3)−y(2)+3y(1)+5y=0 $y^{(3)} - y^{(2)} + 3 y^{(1)} + 5 y = 0$
Find a basis for each of the following subspaces of C∞ $C^{\infty}$ .
1. (a) N(D2−D−I) $N (D^{2} - D - I)$
2. (b) N(D3−3D2+3D−I) $N (D^{3} - 3 D^{2} + 3 D - I)$
3. (c) N(D3+6D2+8D) $N (D^{3} + 6 D^{2} + 8 D)$
Show that C∞ $C^{\infty}$ is a subspace of F(R, C).
1. (a) Show that D:C∞→C∞ $D : C^{\infty} \to C^{\infty}$ is a linear operator.
2. (b) Show that any differential operator is a linear operator on C∞ $C^{\infty}$ .
Prove that if {x, y} ${x, y}$ is a basis for a vector space over C, then so is

${1 2 (x + y), 1 2 i (x - y)} .$ ${\frac{1}{2} (x + y), \frac{1}{2 i} (x - y)} .$
Consider a second-order homogeneous linear differential equation with constant coefficients in which the auxiliary polynomial has distinct conjugate complex roots a+ib $a + i b$ and a−ib $a - i b$ , where a, b∈R $a, b \in R$ . Show that {eat cos bt, eat sin bt} ${e^{a t} cos b t, e^{a t} sin b t}$ is a basis for the solution space.
Suppose that {U1, U2, …, Un} ${U_{1}, U_{2}, \dots, U_{n}}$ is a collection of pairwise commutative linear operators on a vector space V (i.e., operators such that UiUj=UjUi $U_{i} U_{j} = U_{j} U_{i}$ for all i, j). Prove that, for any i(1≤i≤n), $i (1 \leq i \leq n),$

$N (U i) \subseteq N (U 1 U 2 \dots U n) .$ $N (U_{i}) \subseteq N (U_{1} U_{2} \dots U_{n}) .$
Prove Theorem 2.33 and its corollary. Hints: For Theorem 2.33, use mathematical induction on n. In the inductive step, let a1, a2, …, an $a_{1}, a_{2}, \dots, a_{n}$ be scalars such that ∑ni=1aiecit=0 $\sum_{i = 1}^{n} a_{i} e^{c_{i} t} = 0$ . Multiply both sides of this equation by e−cnt $e^{- c_{n} t}$ , and differentiate the resulting equation with respect to t. For the corollary, use Theorems 2.31, 2.33, and 2.32. Visit goo.gl/oKTEbV for a solution.
Prove Theorem 2.34. Hint: First verify that the alleged basis lies in the solution space. Then verify that this set is linearly independent by mathematical induction on k as follows. The case k=1 $k = 1$ is the lemma to Theorem 2.34. Assuming that the theorem holds for k−1 $k - 1$ distinct ci $c_{i}$ ’s, apply the operator (D−ckI)nk ${(D - c_{k} I)}^{n_{k}}$ to any linear combination of the alleged basis that equals 0 .
Let V be the solution space of an nth-order homogeneous linear differential equation with constant coefficients having auxiliary polynomial p(t). Prove that if p(t)=g(t)h(t) $p (t) = g (t) h (t)$ , where g(t) and h(t) are polynomials of positive degree, then

$N (h (D)) = R (g (D V)) = g (D) (V),$ $N (h (D)) = R (g (D_{V})) = g (D) (V),$

where DV:V→V $D_{V} : V \to V$ is defined by DV(x)=x′ $D_{V} (x) = x^{'}$ for x∈V $x \in V$ . Hint: First prove g(D)(V)⊆N(h(D)) $g (D) (V) \subseteq N (h (D))$ . Then prove that the two spaces have the same finite dimension.
A differential equation

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

is called a nonhomogeneous linear differential equation with constant coefficients if the ai $a_{i}$ s are constant and x is a function that is not identically zero.
1. (a) Prove that for any x∈C∞ $x \in C^{\infty}$ there exists y∈C∞ $y \in C^{\infty}$ such that y is a solution to the differential equation. Hint: Use Lemma 1 to Theorem 2.32 to show that for any polynomial p(t), the linear operator p(D):C∞→C∞ $p (D) : C^{\infty} \to C^{\infty}$ .
2. (b) Let V be the solution space for the homogeneous linear equation
  
  $y (n) + a n - 1 y (n - 1) + \dots + a 1 y (1) + a 0 y = 0.$ $y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{1} y^{(1)} + a_{0} y = 0.$
  
  Prove that if z is any solution to the associated nonhomogeneous linear differential equation, then the set of all solutions to the nonhomogeneous linear differential equation is
  
  ${z + y : y \in V} .$ ${z + y : y \in V} .$
Given any nth-order homogeneous linear differential equation with constant coefficients, prove that, for any solution x and any t0∈R $t_{0} \in R$ , if x(t0)=x′(t0)=⋯=x(n−1)(t0)=0 $x (t_{0}) = x^{'} (t_{0}) = \dots = x^{(n - 1)} (t_{0}) = 0$ , then x=0 $x = 0$ (the zero function). Hint: Use mathematical induction on n as follows. First prove the conclusion for the case n=1 $n = 1$ . Next suppose that it is true for equations of order n−1 $n - 1$ , and consider an nth-order differential equation with auxiliary polynomial p(t). Factor p(t)=q(t)(t−c) $p (t) = q (t) (t - c)$ , and let z=q((D))x $z = q ((D)) x$ . Show that z(t0)=0 $z (t_{0}) = 0$ and z′−cz=0 $z^{'} - c z = 0$ to conclude that z=0 $z = 0$ . Now apply the induction hypothesis.
Let V be the solution space of an nth-order homogeneous linear differential equation with constant coefficients. Fix t0∈R $t_{0} \in R$ , and define a mapping Φ:V→Cn $Φ : V \to C^{n}$ by

$Φ (x) = ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ x (t 0) x' (t 0) ⋮ x (n - 1) (t 0) ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ for each x \in V .$ $Φ (x) = (\begin{matrix} x (t_{0}) \\ x^{'} (t_{0}) \\ ⋮ \\ x^{(n - 1)} (t_{0}) \end{matrix}) for each x \in V .$
1. (a) Prove that Φ $Φ$ is linear and its null space is the zero subspace of V. Deduce that Φ $Φ$ is an isomorphism. Hint: Use Exercise 14.
2. (b) Prove the following: For any nth-order homogeneous linear differential equation with constant coefficients, any t0∈R $t_{0} \in R$ , and any complex numbers c0, c1, …, cn−1 $c_{0}, c_{1}, \dots, c_{n - 1}$ (not necessarily distinct), there exists exactly one solution, x, to the given differential equation such that x(t0)=c0 $x (t_{0}) = c_{0}$ and x(k)(t0)=ck $x^{(k)} (t_{0}) = c_{k}$ for k=1, 2, …, n−1 $k = 1, 2, \dots, n - 1$ .
Pendular Motion. It is well known that the motion of a pendulum is approximated by the differential equation

$θ'' + g l θ = 0,$ $θ^{″} + \frac{g}{l} θ = 0,$

where θ(t) $θ (t)$ is the angle in radians that the pendulum makes with a vertical line at time t (see Figure 2.8), interpreted so that θ $θ$ is positive if the pendulum is to the right and negative if the pendulum is to the left of the vertical line as viewed by the reader. Here l is the length of the pendulum and g is the magnitude of acceleration due to gravity. The variable t and constants l and g must be in compatible units (e.g., t in seconds, l in meters, and g in meters per second per second).

Figure 2.8

Figure 2.8 Full Alternative Text
1. (a) Express an arbitrary solution to this equation as a linear combination of two real-valued solutions.
2. (b) Find the unique solution to the equation that satisfies the conditions
  
  $θ (0) = θ 0 > 0 and θ' (0) = 0.$ $θ (0) = θ_{0} > 0 and θ^{'} (0) = 0.$
  
  (The significance of these conditions is that at time t=0 $t = 0$ the pendulum is released from a position displaced from the vertical by θ0 $θ_{0}$ .)
3. (c) Prove that it takes 2πl/g−−−√ $2 π \sqrt{l / g}$ units of time for the pendulum to make one circuit back and forth. (This time is called the period of the pendulum.)
Periodic Motion of a Spring without Damping. Find the general solution to (3), which describes the periodic motion of a spring, ignoring frictional forces.
Periodic Motion of a Spring with Damping. The ideal periodic motion described by solutions to (3) is due to the ignoring of frictional forces. In reality, however, there is a frictional force acting on the motion that is proportional to the speed of motion, but that acts in the opposite direction. The modification of (3) to account for the frictional force, called the damping force, is given by

$m y'' + r y' + k y = 0,$ $m y^{″} + r y^{'} + k y = 0,$

where r>0 $r > 0$ is the proportionality constant.
1. (a) Find the general solution to this equation.
2. (b) Find the unique solution in (a) that satisfies the initial conditions y(0)=0 $y (0) = 0$ and y′(0)=v0 $y^{'} (0) = v_{0}$ , the initial velocity.
3. (c) For y(t) as in (b), show that the amplitude of the oscillation decreases to zero; that is, prove that limt→∞y(t)=0 $\lim_{t \to \infty} y (t) = 0$ .
In our study of differential equations, we have regarded solutions as complex-valued functions even though functions that are useful in describing physical motion are real-valued. Justify this approach.
The following parts, which do not involve linear algebra, are included for the sake of completeness.
1. (a) Prove Theorem 2.27. Hint: Use mathematical induction on the number of derivatives possessed by a solution.
2. (b) For any c, d∈C $c, d \in C$ , prove that
  
  $e c + d = c c e d and e - c = 1 e c .$ $e^{c + d} = c^{c} e^{d} and e^{- c} = \frac{1}{e^{c}} .$
3. (c) Prove Theorem 2.28.
4. (d) Prove Theorem 2.29.
5. (e) Prove the product rule for differentiating complex-valued functions of a real variable: For any differentiable functions x and y in F(R, C), the product xy is differentiable and
  
  $(x y)' = x' y + x y' .$ $(x y)^{'} = x^{'} y + x y^{'} .$
  
  Hint: Apply the rules of differentiation to the real and imaginary parts of xy.
6. (f) Prove that if x∈F(R, C) $x \in F (R, C)$ and x′=0 $x^{'} = 0$ , then x is a constant function.

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Table of Contents for 2.7* Homogeneous Linear Differential Equations with Constant Coefficients

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Table of Contents for
2.7* Homogeneous Linear Differential Equations with Constant Coefficients