Page 824 - Advanced_Engineering_Mathematics o'neil

804 Answers to Selected Problems

13. The functions are linearly independent, since neither is a constant multiple of the other on [−1,1]. The differential
equation in linear form is
2 2

y − y + y = 0,
x x 2
which is undefined at 0, so the theorem does not apply.
15. We know that the initial value problem
y + py + qy = 0; y(x 0 ) = y (x 0 ) = 0

has the unique solution y = 0 in some interval about x 0 .If ϕ(x 0 ) = ϕ (x 0 ) = 0, then ϕ is also a solution of this

problem, hence ϕ would have to be identically zero, which is a contradiction.
Section 2.2 The Constant Coefficient Case
1. y = c 2 e −2x + c 2 e 3x 3. y = e −3x (c 1 + c 2 x)
5. y = e −5x (c 1 cos(x) + c 2 sin(x))
√ √
7. y = e −3x/2 [c 1 cos(3 7x/2) + c 2 sin(3 7x/2)]
7x
9. y = e (c 1 + c 2 x)
11. y = 5 − 2e −3x 13. y(x) = 0
9 3(x−2)
5 −4(x−2)
15. y = e + e
7 7
17. y = e x−1 (29 − 17x)
√ √
5
19. y = e (x+2)/2 cos( 15(x + 2)/2) + √ sin( 15(x + 2)/2)
15
x
ax
ax
21. (a) ϕ(x) = e (c 1 + c 2 x) (b) ϕ (x) = e (c 1 e + c 2 e − x )
ax
(c) lim →0 ϕ (x) = e (c 1 + c 2 )
= ϕ(x) in general.
23. The characteristic equation has roots
√ √
2
2
−a + a − 4b −a − a − 4b
λ 1 = ,λ 2 = .
2 2
Now take cases. If a − 4b > 0, then λ 1 and λ 2 are both negative, so the solution c 1 e 1 x + c 2 e 2 x decays to zero as
2
λ
λ
2
2
x →∞.If a = 4b, the solution e −ax/2 (c 1 + c 2 x) → 0as x →∞ because a > 0. If a − 4b < 0, the solution is a linear
combination of sines and cosines, multiplied by e −ax/2 , and this solution goes to zero as x →∞.
Section 2.3 The Nonhomogeneous Equation
1. y = c 1 cos(x) + c 2 sin(x) − cos(x)ln|sec(x) + tan(x)|
4
3. y = c 1 cos(3x) + c 2 sin(3x) + 4x sin(3x) + cos(3x)ln|cos(3x)|
3
x
2x
−x
2x
5. y = c 1 e + c 2 e − e cos(e )
2x
2
7. y = c 1 e + c 2 e −x − x + x − 4
2
x
9. y = e [c 1 cos(3x) + c 2 sin(3x)]+ 2x + x − 1
2x
4x
11. y = c 1 e + c 2 e + e x
x
2x
13. y = c 1 e + c 2 e ++3cos(x) + sin(x)
2x
1 2x
1 3x
15. y = e [c 1 cos(3x) + c 2 sin(3x)]+ e − e
3 2
7
1
7 2x
3 −2x
2x
17. y = e − e − xe − x
4 4 4 4
1 −x
3 −2x
19. y = e − 19 e −6x + e + 7
8 120 5 12
4x
21. y = 2e + 2e −2x − 2e −x − e 2x
2
23. y = 4e −x − sin (x) − 2
Section 2.4 Spring Motion
√ √
10
1. With y(0) = 5and y (0) = 0, y = e −2t 5cosh( 2t) + √ sinh( 2t) .

2
√
5
With y(0) = 0and y (0) = 5, y = √ e −2t sinh( 2t).

2
These functions are graphed in Figure A.1.
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October 14, 2010 17:50 THM/NEIL Page-804 27410_25_Ans_p801-866