# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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e—3+6i = e-3 cos 6 + ie-3 sin 6 = 0.0478041 - 0.0139113i.

With the definitions (9) and (13) it is straightforward to show that the usual laws of exponents are valid for the complex exponential function. It is also easy to verify that the differentiation formula

d (ert) = rert (14)

d1

also holds for complex values ofr.

Real-Valued Solutions. The functions y2(1) and y2(1), given by Eqs. (5) and with the meaning expressed by Eq. (13), are solutions of Eq. (1) when the roots of the characteristic equation (2) are complex numbers X ± i ë. Unfortunately, the solutions y2 and y2 are complex-valued functions, whereas in general we would prefer to have realvalued solutions, if possible, because the differential equation itself has real coefficients. Such solutions can be found as a consequence of Theorem 3.2.2, which states that if y1 and y2 are solutions of Eq. (1), then any linear combination of y1 and y2 is also a solution. In particular, let us form the sum and then the difference of y2 and y2. We have

y2(1) + y2(1) = eXt (cos i1 + i sin i1) + eXt (cos i1 — i sin i1)

= 2eXt cos i1

and

y2(1) — y2(1) = eXt (cos i1 + i sin i1) — eXt (cos i1 — i sin i1)

= 2ieXt sin /ë1.

Hence, neglecting the constant multipliers 2 and 2i, respectively, we have obtained a pair of real-valued solutions

u(1) = eXt cos i1, v(1) = eXt sin i1. (15)

Observe that u and v are simply the real and imaginary parts, respectively, of y2.

By direct computation you can show that the Wronskian of u and v is

W(u, v)(t) = ie2Xt. (16)

Thus, as long as ë = 0, the Wronskian W is not zero, so u and v form a fundamental set of solutions. (Of course, if ë = 0, then the roots are real and the discussion in this section is not applicable.) Consequently, if the roots of the characteristic equation are complex numbers X ± i ë, with ë = 0, then the general solution ofEq. (1) is

y = c1eXt cos ë1 + c2eXt sin fit, (17)

where c2 and c2 are arbitrary constants. Note that the solution (17) can be written down as soon as the values of X and ë are known.

156

Chapter 3. Second Order Linear Equations

Find the general solution of

y" + y' + y = 0. (18)

The characteristic equation is

r2 + r + 1 = 0,

and its roots are

_ -1 ± (1 - 4)1/2 1 , _.V3

r = 2 = _2 i T".

Thus X = —1/2 and ë = ë/3/2, so the general solution ofEq. (18) is

y = c1e-1/2 cos(V31 /2) + c2e~t/2 sin(V31 /2). (19)

EXAMPLE

2

Find the general solution of

y" + 9 y = 0. (20)

The characteristic equation is r2 + 9 = 0 with the roots r = ±3i; thus X = 0 and ë = 3. The general solution is

y = c2cos3t + c2sin3t; (21)

note that if the real part of the roots is zero, as in this example, then there is no exponential factor in the solution.

Find the solution of the initial value problem

16/'- 8y' + 145y = 0, y(0) = -2, /(0) = 1. (22)

The characteristic equation is 16r2 — 8r + 145 = 0 and its roots are r = 1/4 ± 3i. Thus the general solution of the differential equation is

y = c1et/4 cos 31 + c2et/4 sin 31.

(23)

To apply the first initial condition we set 1 = 0 in Eq. (23); this gives

y(0) = ci = -2.

For the second initial condition we must differentiate Eq. (23) and then set 1 = 0. In this way we find that

/(0) = 2 ci + 3c2 = 1, from which c2 = 1/2. Using these values of c2 and c2 in Eq. (23), we obtain

y = —2e1/4 cos 31 + 2 e1 /4 sin 31

(24)

as the solution of the initial value problem (22).

We will discuss the geometrical properties of solutions such as these more fully in Section 3.8, so we will be very brief here. Each of the solutions u and v in Eqs. (15) represents an oscillation, because of the trigonometric factors, and also either grows or

3.4 Complex Roots ofthe Characteristic Equation

157

decays exponentially, depending on the sign of X (unless X = 0). In Example 1 we have X = —1/2 < 0, so solutions are decaying oscillations. The graph of a typical solution of Eq. (18) is shown in Figure 3.4.1. On the other hand, X = 1/4 > 0 in Example

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