Page 875 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 855

where
−16L
d n = .
2
(2n − 1)π[(2n − 1) − 4]
2 4
2
5. u(x,t) = π − ∞ cos(nx)e −4n 2 t
n=1
3 n 2
−6
1 1 − e (−1) n
−6
7. u(x,t) = (1 − e ) + 12 ∞ cos(nπx/6)e −n 2 π 2 t/18
6 n=1 36 + n π 2
2
4B
∞ 2 2 2

9. u(x,t) = sin((2n − 1)πx/2L)exp(−(2n − 1) π kt/4L )
n=1
(2n − 1)π
11. Substitute e αx+βt v(x,t) into the partial differential equation, and solve for α and β,sothat v t = kv xx .Obtain
2
α =−A/2and β = k(B − A /4).
3x
13. Let u(x,t) = e −3x−9t v(x,t).Then v(0,t) = v(4,t) = 0and v(x,0) = e ,so
2nπ
∞

n
12
v(x,t) = (1 − e (−1) ) sin(nπx/4)e −n 2 π 2 t/16 .
144 + n π 2
2
n=1
2
15. Let u(x,t) = v(x,t) + f (x) and choose f (x) = 3x + 2. We obtain v(0,t) = v(1,t) = 0and u(x,0) = x − f (x).The
solution for v is
∞
4
2
n
2
2
2
v(x,t) = 2 (−1) (1 + 2n π ) − (1 + n π ) sin(nπx)e −16n 2 π 2 t .
3
n π 3
n=1
17. Let u(x,t) = e −At W(x,t).Then w(0,t) = w(9,t) = 0and u(x,0) = 3x.Wefind that
∞ n+1
54(−1)

w(x,t) = sin(nπx/9)e −4n 2 π 2 t/81 .
nπ
n=1

1 1 − (−1) n
∞ 2 −4n 2 t

19. u(x,t) = n=1 −1 + 4n t + e sin(nx)
8π n 5
1 − (−1)
n
∞ −4n 2 t
4
+ n=1 sin(nx)e
π n 3
50 1 − cos(5)(−1) n

∞ 2 2 −n 2 π 2 t/25
21. u(x,t) = −25 + n π t + 25e sin(nπx/5)
n=1 3 3 2 2
n π n π − 25
(−1) − 1
n+1
∞ −n 2 π 2 t/25

+ 500 sin(nπx/5)e
n=1 3 3
n π
2 2 −16n 2 π 2 t/9
n
23. ∞ 27(−1) 16n π t + 9e − 9
u(x,t) = n=1 sin(nπx/3)
5
128 n π 5
∞ 1 − (−1) n

+2K sin(nπx/3)e −16n 2 π 2 t/9
n=1 nπ
Section 17.3 Solutions in an Infinite Medium
1. By separation of variables,
1 ∞ 8
u(x,t) = cos(ωx)e −ω 2 kt dω
π 0 16 + ω 2
By Fourier transform,
1 ∞ −4|ξ| −(x−ξ) 2 /4kt
u(x,t) = √ e e dξ
2 πkt −∞
3.
∞

u(x,t) = (a ω cos(ωx) + b ω sin(ωx))e −ω 2 kt dω,
0
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October 14, 2010 17:50 THM/NEIL Page-855 27410_25_Ans_p801-866