1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

484 Answers to Odd-Numbered Exercises

ony=

√

3 ( 1 −x),φ=sin( 4 nπx)+sin( 2 nπ( 1 − 2 x))−sin 2nπ.

23. For a sextant,φn=J 3 n(λr)sin( 3 nθ)andJ 3 n(λ)=0. Thusλ 1 = 6 .380,
    which is less than

√

16 π^2 / 3 = 7. 255.

25.b=−1.

27. Sincey(x)=hJ 0 (kx)/J 0 (ka),wherek=ω/√gU, the solution cannot
    have this form ifJ 0 (ka)=0.
    29.u(r,t)=R(t)T(t);R(r)=r−mJm(λr),wherem=(n− 2 )/2;T(t)=
    acos(λct)+bsin(λct).

31.b=DrL

2

DzR^2

,ρ=UL

Dz

.

33.a 0 = 12 .77,a 1 =− 4 .88.

35. tan(λ)=Dλ/(D+λ^2 ),λ=π(D=0), 3.173 (D=1), 4.132 (D=10).

Chapter 6

Section 6.1

1. c. s

(^2) + 2 ω 2
s(s^2 + 4 ω^2 )
;d.ωcos(φ)−ssin(φ)
s^2 +ω^2

;

e. e

2

s− 2

;f.^2 ω

2

s(s^2 + 4 ω^2 )

.

3. a.e

−as

s ;b.

e−as−e−bs

s ;c.

1 −e−as

s^2.

5. a.e−tsinh(t);b.e−tcos(t);c.e−at

sin(

√

b^2 −a^2 t)

√

b^2 −a^2

.

7. a.e

at−ebt

a−b ;b.

t

2 asinh(at);d.

t^2 eat

2 ;

e.f(t)=

{ 1 , 0 <t<1,

0 , 1 <t.

9. a. [sin(ωt)−ωtcos(ωt)]/ 2 ω^2 ;
   b. See Table 2;
   c. See 7c;
   d. [cos(ωt)−ωtsin(ωt)]/2.