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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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21. y + 2y" y- 3y = 0 23. tf + 2f y+ ty = 0
22. yiv + y = 0
24. t2 ylv + tf + y 4y = 0
L [ y] = y(n) + p (t) y(n-V) + ??? + Pn (t) y = 0 on I, then W(y1,..., y ) is nowhere zero in I.
(i)
214
Chapter 4. Higher Order Linear Equations
(a) Suppose that W(y.j,..., yn)(t0) = 0, and suppose that
Cj y.(t) + -^ + cnyn (f) = 0 (ii)
for all t in I. By writing the equations corresponding to the first n 1 derivatives of Eq. (ii) at t0, show that c. = = cn = 0. Therefore, yj,..., yn are linearly independent.
(b) Suppose that y.,..., yn are linearly independent solutions of Eq. (i). If W (y.,..., yn )(t0) = 0forsome t0, show that there is a nonzero solution of Eq. (i) satisfying the initial conditions
y(t0) = ?(t0) = = y(n1)(t0) = 0.
Since y = 0 is a solution of this initial value problem, the uniqueness part of Theorem 4.1.1 yields a contradiction. Thus W is never zero.
26. Show that if yj is a solution of
/" + Pj (t)f + P2(t)y + P3(t)y = 0,
then the substitution y = y. (t)v(t) leads to the following second order equation for v':
yjl)'" + (3/j + Pj yj)v" + (3/{ + 2 Pj yj + P2 y.)v' = .
In each of Problems 27 and 28 use the method of reduction of order (Problem 26) to solve the given differential equation.
27. (2 t) f+ (2t 3)y" ty + y = 0, t < 2; yj(t) = ef
28. t2(t + 3)/" 3t(t + 2)y + 6(j + t)y 6y = 0, t > 0; yj(t) = t2, y2(t) = t3
4.2 Homogeneous Equations with Constant Coefficients
Consider the nth order linear homogeneous differential equation
L [y] = 30 y(n) + 3l y(n-1) + ? ? ? + an-1 y7 + any = 0, (1)
where a0, a17, an are real constants. From our knowledge of second order linear equations with constant coefficients it is natural to anticipate that y = eft is a solution of Eq. (1) for suitable values of r. Indeed,
L [ert] = ert(a0rn + a^-1 + ? ? ? + a^r + an) = ert Z(r) (2)
for all r, where
Z(r) = a0rn + a1rn-1 + ? ? ? + an-1r + an. (3)
For those values of r for which Z(r) = 0, it follows that L[ert] = 0 and y = ert is a
solution of Eq. (1). The polynomial Z(r) is called the characteristic polynomial, and
the equation Z(r) = 0 is the characteristic equation of the differential equation (1).
4.2 Homogeneous Equations with Constant Coefficients
215
A polynomial of degree n has n zeros,1 say r1, r2,..., rn, some of which may be equal; hence we can write the characteristic polynomial in the form
Z(r) = a0(r - r1)(r - r2) (r - rn). (4)
Real and Unequal Roots. If the roots of the characteristic equation are real and no
two are equal, then we have n distinct solutions er1t, er2,..., ernt of Eq. (1). If these functions are linearly independent, then the general solution of Eq. (1) is
y = c1er1t + c2eV + ??? + cnernl. (5)
One way to establish the linear independence of er11, e^,..., erZ is to evaluate their
Wronskian determinant. Another way is outlined in Problem 40.
EXAMPLE 1
Find the general solution of
/" + /'- 7/- / + 6y = 0. (6)
Also find the solution that satisfies the initial conditions
y(0) = 1, y (0) = 0, /(0) = -2, /'(0) = -1 (7)
and plot its graph.
Assuming that y = ert, we must determine r by solving the polynomial equation
r4 + r3 - 7r2 - r + 6 = 0. (8)
The roots of this equation are r1 = 1, r2 = -1, r3 = 2, and r4 = 3. Therefore the
general solution of Eq. (6) is
y = C1et + c2e~t + c3e2t + c4e-3t. (9)
The initial conditions (7) require that c1t, c4 satisfy the four equations
c1 + c2 + c3 + c4 = 1 ,
c1 - c2 + 2c3 - 3c4 = 0,
c1 + c2 + 4c3 + 9c4 = -2, c1 - c2 + 8c3 - 27c4 = -1.
(10)
By solving this system of four linear algebraic equations, we find that
c1 = 11/8, c2 = 5/12, c3 = -2/3, c4 = -1/8.
Therefore the solution of the initial value problem is
y = - e + -2 e-t - f e2t - 8 e-3t. (11)
The graph of the solution is shown in Figure 4.2.1.
*An important question in mathematics for more than 200 years was whether every polynomial equation has at least one root. The affirmative answer to this question, the fundamental theorem of algebra, was given by Carl Friedrich Gauss in his doctoral dissertation in 1799, although his proof does not meet modern standards of rigor. Several other proofs have been discovered since, including three by Gauss himself. Today, students often meet the fundamental theorem of algebra in a first course on complex variables, where it can be established as a consequence of some of the basic properties of complex analytic functions.
216
Chapter 4. Higher Order Linear Equations
FIGURE 4.2.1 Solution of the initial value problem of Example 1.
As Example 1 illustrates, the procedure for solving an nth order linear differential equation with constant coefficients depends on finding the roots of a corresponding nth degree polynomial equation. If initial conditions are prescribed, then a system of n linear algebraic equations must be solved to determine the proper values of the constants cv ..., cn. While each of these tasks becomes much more complicated as n increases, they can often be handled without difficulty with a calculator or computer.
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