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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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30. xy" - (cos x)y'+ (sinx)y = 0, x > 0 31. x2y" + xy' - y = 0, x > 0
32. The Adjoint Equation. If a second order linear homogeneous equation is not exact, it can be made exact by multiplying by an appropriate integrating factor /x(x). Thus we require that i^(x) be such that p,(x) P (x)y" + /x(x) Q (x)y' + p,(x) R(x )y = 0 can be written in the form [^(x)P (x)y']' + [ f (x)y]' = 0. By equating coefficients in these two equations and eliminating f (x), show that the function ä must satisfy
Pä" + (2P' - Q)ä + (P" - Q' + ß)ä = 0.
This equation is known as the adjoint of the original equation and is important in the advanced theory of differential equations. In general, the problem of solving the adjoint differential equation is as difficult as that of solving the original equation, so only occasionally is it possible to find an integrating factor for a second order equation.
3.3 Linear Independence and the Wronskian
147
In each of Problems 33 through 35 use the result of Problem 32 to find the adjoint of the given
differential equation.
33. x2y" + xy' + (x2 - v2)y = 0, Bessel’s equation
34. (1 - x2)y" - 2xy' + a(a + 1)y = 0, Legendre’s equation
35. y" - xy = 0, Airy’s equation
36. For the second order linear equation P (x )y" + Q (x )y' + R(x )y = 0, show that the adjoint of the adjoint equation is the original equation.
37. A second order linear equation P(x)y" + Q(x)y' + R(x)y = 0 is said to be self-adjoint if its adjoint is the same as the original equation. Show that a necessary condition for this equation to be self-adjoint is that P' (x) = Q (x). Determine whether each of the equations in Problems 33 through 35 is self-adjoint.
3.3 Linear Independence and the Wronskian
I
The representation of the general solution of a second order linear homogeneous differential equation as a linear combination of two solutions whose Wronskian is not zero is intimately related to the concept of linear independence of two functions. This is a very important algebraic idea and has significance far beyond the present context; we briefly discuss it in this section.
We will refer to the following basic property of systems of linear homogeneous algebraic equations. Consider the two-by-two system
ai1 xi + ai2x2 = 0 (2)
a21 xi + a22x2 = 0
and let Ä = a21a22 - a12a21 be the corresponding determinant of coefficients. Then x = 0, y = 0 is the only solution of the system (1) if and only if Ä = 0. Further, the system (1) has nonzero solutions if and only if Ä = 0.
Two functions f and g are said to be linearly dependent on an interval I if there exist two constants k2 and k2, not both zero, such that
k2 f (1) + k2 g(1) = 0 (2)
for all 1 in I. The functions f and g are said to be linearly independent on an interval I if they are not linearly dependent; that is, Eq. (2) holds for all 1 in I only if k2 = k2 = 0. In Section 4.1 these definitions are extended to an arbitrary number of
functions. Although it may be difficult to determine whether a large set of functions is
linearly independent or linearly dependent, it is usually easy to answer this question for a set of only two functions: they are linearly dependent if they are proportional to each other, and linearly independent otherwise. The following examples illustrate these definitions.
EXAMPLE Determine whether the functions sin 1 and cos(1 - n/2) are linearly independent or
1 linearly dependent on an arbitrary interval.
148
Chapter 3. Second Order Linear Equations
The given functions are linearly dependent on any interval since k2 sin 1 + k2cos(t - n/2) = 0 for all 1 if we choose k2 = 1 and k2 = -1.
Show that the functions el and e21 are linearly independent on any interval.
To establish this result we suppose that
k2 ef + k2e21 = 0 (3)
for all 1 in the interval; we must then show that k2 = k2 = 0. Choose two points t0 and 12 in the interval, where 12 = t0. Evaluating Eq. (3) at these points, we obtain
k1et0 + k2e210 = 0,
1 2 (4)
k1e1i + k2e21i = 0.
The determinant of coefficients is
el0e2t1 - e210e11 = e*0et1 (e11 - ef0).
Since this determinant is not zero, it follows that the only solution of Eq. (4) is k\ = k2 = 0. Hence el and e21 are linearly independent.
The following theorem relates linear independence and dependence to the Wronskian.
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