# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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2P' - Q = -4x + 2x = -2x = Q and

P" - Q' + R = -2 + 2 + a(a+1) = a(a+1) = R.

Write the adjoint D.E. given in Problem 32 as

AAA A A

Pp" + Qm' + Rm = 0 where P = P, Q = 2P' - Q, and

A

R = P" - Q' + R. The adjoint of this equation, namely the adjoint of the adjoint, is

A A A AAA

Py" + (2P' - Q)y' + (P" - Q' + R)y = 0. After

A A A

substituting for P, Q, and R and simplifying, we obtain

Py" + Qy' + Ry = 0. This is the same as the original

equation.

Section 3.3

41

37. From Problem 32 the adjoint of Py" + Qy' + Ry = 0 is

Pm" + (2P' - Q)m' + (P" - Q' + R)m = 0. The two equations

are the same if 2P' - Q = Q and P" - Q' + R = R. This

will be true if P' = Q. Hence the original D.E. is self-

2

adjoint if P' = Q. For Problem 33, P(x) = x so P'(x) = 2x and Q(x) = x. Hence the Bessel equation of order v is not self-adjoint. In a similar manner we find that Problems 34 and 35 are self-adjoint.

Section 3.3, Page 152

2. Since cos30 = 4cos30 - 3cos0 we have

cos30 - (4cos30-3cos0) = 0 for all 0. From Eq.(1) we have k1 = 1 and k2 = -1 and thus cos30 and 4cos30 - 3cos0 are linearly dependent.

= -2/t n 0.

6. W(t,t-1) =

t t-1

1 -t-2

7. For t>0 g(t) = t and hence f(t) - 3g(t) = 0 for all t.

Therefore f and g are linearly dependent on 0<t. For t<0 g(t) = -t and f(t) + 3g(t) = 0, so again f and g are linearly dependent on t<0. For any interval that includes the origin, such as -1<t<2, there is no c for which f(t) +cg(t) = 0 for all t, and hence f and g are linearly independent on this interval.

12. The D.E. is linear and homogeneous. Hence, if y1 and y2 are solutions, then y3 = y1 + y2 and y4 = y1 - y2 are

solutions. W(y3,y4) = y3y4 - y'3y4 = (yj_ + y2 )(y 1 - y2) -

(y! + y^)(yi - y2) = -2(yiy2 - y1y2) = -2W(y1,y2), is not zero since y1 and y2 are linearly independent solutions. Hence y3 and y4 form a fundamental set of solutions. Conversely, solving the first two equations for y1 and y2, we have y1 = (y3+y4)/2 and y2 = (y3-y4)/2, so y1 and y2 are solutions. Finally, from above we have W(y1y2) = -W(y3,y4)/2.

15. Writing the D.E. in the form of Eq.(7), we have p(t) = -(t+2)/t. Thus Eq.(8) yields

42

Section 3 . 3

r -(t+2) 2 t

W(t) = cexp[-J ----------- dt] = ct e .

t

20. From Eq.(8) we have W(y1(y2) = cexp[-| p(t)dt], where

2

p(t) = 2/t from the D.E. Thus W(y1(y2) = c/t . Since W(y1,y2)(1) = 2 we find c = 2 and thus W(y1fy2)(5) = 2/25.

24. Let c be the point in I at which both y1 and y2 vanish. Then W(y1fy2)(c) = y1(c)y/2(c) - y1 (c)y2 (c) = 0. Hence, by Theorem 3.3.3 the functions y1 and y2 cannot form a fundamental set.

26. Suppose that y1 and y2 have a point of inflection at t0

//

and either p(t0) n 0 or q(t0) n 0. Since y1(t0) = 0 and

y 2 (t0) = 0 it follows from the D.E. that p(t0)y1(t0) +

q(t0)y1(t0) = 0 and p(t0)y'2(t0) + q(t0)y2(t0) = 0. If p(t0) = 0 and q(t0) n 0 then y1(t0) = y2(t0) = 0, and W(y1,y2)(t0) = 0 so the solutions cannot form a fundamental set. If p(t0) n 0 and q(t0) = 0 then

y1(t0) = y2(t0) = 0 and W(y1,y2)(t0) = 0, so again the solutions cannot form a fundamental set. If p(t0) n 0

and q(t0) = 0 then y1(t0) = q(t0)y1(t0)/p(t0) and

y'2(t0) = q(t0)y2(t0)/p(t0) and thus

W(y1,y2)(t0) = y1(t0)y2(t0) - y1(t0)y2(t0)

= y1(t0)[q(t0)y2(t0)/p(t0)] -[q(t0)y1(t0)/p(t0)]y2(t0)

= 0.

2

27. Let -1 < t0, t1 < 1 and t0 n t1. If y1 = t and y2 = t

2

are linearly dependent then c1t1 + c2t1 = 0 and 2

c1t0 + c2t0 = 0 have a solution for c1 and c2 such that c1 and c2 are not both zero. But this system of equations has a non-zero solution only if t1 = 0 or t0 = 0 or t1 = t0. Hence, the only set c1 and c2 that satisfies the system for every choice of t0 and t1 in -1 < t < 1 is

2

c1 = c2 = 0. Therefore t and t are linearly independent

Section 3.4

43

22

on -1 < t < 1. Next, W(t,t ) = t clearly vanishes at

22 t = 0. Since W(t,t ) vanishes at t = 0, but t and t are

linearly independent on -1 < t < 1, it follows that t and

2

t cannot be solutions of Eq.(7) on -1 < t < 1. To show

2

that the functions y1 = t and y2 = t are solutions of

2

t y" - 2ty' + 2y = 0, substitute each of them in the equation. Clearly, they are solutions. There is no contradiction to Theorem 3.3.3 since p(t) = -2/t and q(t) = 2/t are discontinuous at t = 0, and hence the theorem does not apply on the interval -1 < t < 1.

28. On 0 < t < 1, f(t) = t3 and g(t) = t3. Hence there are

nonzero constants, c1 = 1 and c2 = -1, such that

c1f(t) + c2g(t) = 0 for each t in (0,1). On -1 < t < 0,

f(t) = -t3 and g(t) = t3; thus c1 = c2 = 1 defines

constants such that c1f(t) + c2g(t) = 0 for each t in (-1,0). Thus f and g are linearly dependent on

0 < t < 1 and on -1 < t < 0. We will show that f(t) and

g(t) are linearly independent on -1 < t < 1 by demonstrating that it is impossible to find constants c1

and c2, not both zero, such that c1f(t) + c2g(t) = 0 for all t in (-1,1). Assume that there are two such nonzero constants and choose two points t0 and t1 in -1 < t < 1

such that t0 < 0 and t1 > 0. Then -c1t^ + c2t^ = 0 and

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