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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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? 36. / + 6/" + 17/ + 22/ + 14y = 0; y(0) = 1, /(0) = -2, /(0) = 0,
y" (0) = 3
37. Show that the general solution of / - y = 0 can be written as
y = c1 cos t + c2 sin t + c3 cosh t + c4 sinh t.
Determine the solution satisfying the initial conditions y(0) = 0, / (0) = 0, /(0) = 1, y" (0) = 1. Why is it convenient to use the solutions cosh t and sinh t rather than et and
e-t ?
38. Consider the equation / - y = 0.
(a) Use Abel’s formula [Problem 20(d) of Section 4.1] to find the Wronskian of a fundamental set of solutions of the given equation.
(b) Determine the Wronskian of the solutions et, e-t, cos t, and sin t.
(c) Determine the Wronskian of the solutions cosh t, sinh t, cos t, and sin t.
39. Consider the spring-mass system, shown in Figure 4.2.4, consisting of two unit masses
suspended from springs with spring constants 3 and 2, respectively. Assume that there is
no damping in the system.
(a) Show that the displacements u1 and u2 of the masses from their respective equilibrium positions satisfy the equations
/ + 5u1 = 2u2, u'2 + 2u2 = 2u1. (i)
(b) Solve the first of Eqs. (i) for u2 and substitute into the second equation, thereby
obtaining the following fourth order equation for u1:
u)v + 7/ + 6u1 = 0. Find the general solution of Eq. (ii).
(ii)
4.2 Homogeneous Equations with Constant Coefficients
221
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, u 2 (0) = 2, u'2(0) = 0. (iii)
Use the first of Eqs. (i) and the initial conditions (iii) to obtain values for U{(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, u'j(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 \/61.
(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 \/6, 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 er1t, ernt are linearly independent on -to < t < to. To do this, we consider the linear relation
cjer1t + ••• + cnernt = 0, -to < t < to (i)
and show that all the constants are zero.
(a) Multiply Eq. (i) by e-rj1 and differentiate with respect to t, thereby obtaining
c2(r2 - rj)e(r2-r1)t + ••• + cn(rn - r)e(rn-r1)l = 0.
(b) Multiply the result of part (a) by e-(r2-rj)t and differentiate with respect to t to obtain
c3(r3 - r2)(r3 - rj)e(r3-r2) + ••• + 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-j)t = 0.
nK n n-1 K n V
222
Chapter 4. Higher Order Linear Equations
Hence c = 0 and therefore
n
Clerit +------+ Cn_l ern-lt = 0.
(d) Repeat the preceding argument to show that c 1 = 0. In a similar way it follows that c 2 = ••• = Cj = 0. Thus the functions efit,..., 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 [ y] = a* y(n) + a! yn-1) + ••• + an_j / + 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 31, or a cosine function cos 31, 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.
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