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

**Download**(direct link)

**:**

**427**> 428 429 430 431 432 433 .. 609 >> Next

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

and q(t0) = 0 then y/1(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 Ï 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

2 2

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

c^3+ c2t! = 0. These equations have a nontrivial solution for c1 and c2 only if the determinant of coefficients is

zero. But the determinant of coefficients is -2t^t1 Ï 0 for t0 and t1 as specified. Hence f(t) and g(t) are linearly independent on -1 < t < 1.

Section 3.4, Page 158 1. exp(1+2i) = e^21 = ºº21 = e(cos2 + isin2).

5. Recall that 21-1 = eln(21-1 = e(1-l)ln2

rt

7. As in Section 3.1, we seek solutions of the form y = e . Substituting this into the D.E. yields the characteristic equation r - 2r + 2 = 0, which has the roots r1 = 1 + i

44

Section 3 . 4

11.

14.

18.

22.

23a.

and r2 = 1 - i, using the quadratic formula. Thus l = 1 and m = — and from Eq.(17) the general solution is y = c1etcost + c2efcsint.

The characteristic equation is r + 6r + 13 = 0, which

-6+v-—á .

has the roots r = ------------- = -3+2i. Thus l = -3 and

2

m = 2, so Eq.(17) becomes y = c—e 3tcos2t + c2e 3tsin2t.

2

The characteristic equation is 9r + 9r - 4, which has the real roots -4/3 and 1/3. Thus the solution has the

same form as in Section 3.1, y(t) = c—et/3 + c2e-4t/3.

The characteristic equation is

r + 4r + 5 = 0, which has the roots r—,r2 = -2 + i. Thus

-2t -2t . . , y = c—e cost + c2e sint and

y' = (-2c1+c2)e-2tcost +

-2t

(-c—-2c2)e sint, so that

y(0)= c—= 1 and y'(0) = -2c—+c2= 0,

- 21

or c2 = 2. Hence y = e (cost + 2sint).

«

VI

The characteristic equation is ,e

r2 + 2r + 2 = 0, so r-,, r9 = -1 ± i. 14

¦L A I.C

Since the I.C. are given at 7l/4 in

we want to alter Eq. (17) by oe

letting c-l = e7l^4d1 and c2 = e7l^4d2. 04

Thus, for X = -1 and |l = 1 we ®*

have ó = e_(t_7l/4) (d-LCOSt + d2sint) ; .©*

so y' = -e-(t-p/4)(d—cost + d2sint) +e-(t-p/4)(-d—sint +d2cost). Thus ^/2d—/2 + \J~2d2/2 = 2 and -\J~2 d— = -2 and hence y = \[2 e (t p/4) (cost + sint).

2

The characteristic equation is 3r - r + 2 = 0, which has

³ a/23

the roots r—,r2 = — + -----------1. Thus u(t) =

1 2 6 6

t/6 /23 \p22

e (c—cos-1 + c2sin------------1) and we obtain u(0) = c— = 2

1 6 2 6 1

Section 3.4

45

, 1 óÒ¯¯

and u (0) = — c + -----------c2 = 0. Solving for c2 we find

**427**> 428 429 430 431 432 433 .. 609 >> Next