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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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FIGURE 4.2.4 A two degree of freedom spring-mass system.
(c) Suppose that the initial conditions are
Uj (0) = 1, u\ (0) = 0, u2(0) = 2, u!2(0) = 0. (iii)
Use the first of Eqs. (i) and the initial conditions (iii) to obtain values for d{(0) and u'1"(0). Then show that the solution of Eq. (ii) that satisfies the four initial conditions on Uj is Uj (t) = cos t. Show that the corresponding solution u2 is u2(t) = 2 cos t.
(d) Now suppose that the initial conditions are
Uj (0) = -2, uj (0) = 0, u2(0) = 1, u2 (0) = 0. (iv)
Proceed as in part (c) to show that the corresponding solutions are Uj (t) = —2 cos \[61 and U2(t) = cos \[6 t.
(e) Observe that the solutions obtained in parts (c) and (d) describe two distinct modes of vibration. In the first, the frequency of the motion is 1, and the two masses move in phase, both moving up or down together. The second motion has frequency \/α, and the masses move out of phase with each other, one moving down while the other is moving up and vice versa. For other initial conditions, the motion of the masses is a combination of these two modes.
40. In this problem we outline one way to show that if rj,..., rn are all real and different, then eVernt are linearly independent on —ςξ < t < ςξ. To do this, we consider the linear relation
Cjer1t + ••• + Cnernt = 0, —ςξ < t < ςξ (i)
and show that all the constants are zero.
(a) Multiply Eq. (i) by e—ri1 and differentiate with respect to t, thereby obtaining
C2(r2 — rj )e(r2—rr) + ••• + cn (rn — r)e(rn—γ) = 0.
(b) Multiply the result of part (a) by e—(r2—ri)( and differentiate with respect to t to obtain
C3(r3 — r2)(r3 — r^-*^ + ••• + Cn (rn — r2)(rn — rj)e(rn—r2)( = 0.
(c) Continue the procedure from parts (a) and (b), eventually obtaining
c (r — r ,) ••• (r — r, )e(rn—rn—l)t = 0.
nK n n—1 K n V
222
Chapter 4. Higher Order Linear Equations
Hence c = 0 and therefore
n
c1er1t +------+ cn_ 1 ern—1‘ = 0.
(d) Repeat the preceding argument to show that cn 1 = 0. In a similar way it follows that c 2 = ••• = c1 = 0. Thus the functions er11,ernt are linearly independent.
4.3 The Method of Undetermined Coefficients
A particular solution Y of the nonhomogeneous nth order linear equation with constant coefficients
L [Σ] = a0 Σ(ο) + a, y(n—l) + ••• + a— Σ + any = g(t) (1)
can be obtained by the method of undetermined coefficients, provided that g(t) is of an appropriate form. While the method of undetermined coefficients is not as general as the method of variation of parameters described in the next section, it is usually much easier to use when applicable.
Just as for the second order linear equation, when the constant coefficient linear differential operator L is applied to a polynomial A0tm + A1tm—1 + ••• + Am, an exponential function eat, a sine function sin β t, or a cosine function cos β t, the result is a polynomial, an exponential function, or a linear combination of sine and cosine functions, respectively. Hence, if g(t) is a sum of polynomials, exponentials, sines, and cosines, or products of such functions, we can expect that it is possible to find Y(t) by choosing a suitable combination of polynomials, exponentials, and so forth, multiplied by a number of undetermined constants. The constants are then determined so that Eq. (1) is satisfied.
The main difference in using this method for higher order equations stems from the fact that roots of the characteristic polynomial equation may have multiplicity greater than 2. Consequently, terms proposed for the nonhomogeneous part of the solution may need to be multiplied by higher powers of t to make them different from terms in the solution of the corresponding homogeneous equation.
Find the general solution of
f — 3 / + 3Σ — y = 4el. (2)
The characteristic polynomial for the homogeneous equation corresponding to Eq. (2) is
r3 — 3r2 + 3r — 1 = (r — 1)3, so the general solution of the homogeneous equation is
yc (t) = c1et + c2tel + c3t2et. (3)
To find a particular solution Y(t) of Eq. (2), we start by assuming that Y(t) = Ael.
However, since el, tel, and t2el are all solutions of the homogeneous equation, we
must multiply this initial choice by t3. Thus our final assumption is that Y(t) = At3et,
4.3 The Method of Undetermined Coefficients
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