Page 871 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 851

and
2
∂ y 1
2
2

= (c f (x + ct) + c f (x − ct))
∂t 2 2
5. The wave equation is
2
2
2
∂ z ∂ z ∂ z
= c 2 +
∂t 2 ∂x 2 ∂y 2
for 0 < x < a,0 < y < b; boundary conditions are
z(0, y,t) = z(a, y,t) = z(x,0,t) = z(x,b,t) = 0
for t ≥ 0; initial conditions are
∂z
z(x, y,0) = f (x, y), (x, y,0) = g(x, y).
∂t
Section 16.2 Wave Motion on an Interval

16sin(nπ/2) − 8nπcos(nπ/2) nπx nπct
∞

1. y(x,t) = n=1 sin sin
3
n π c 2 2
3
108
∞

3. y(x,t) = n=1 sin((2n − 1)πx/3)sin(2(2n − 1)πt/3)
4
(2n − 1) π 4
∞ 24(−1) n+1 √

5. y(x,t) = n=1 sin((2n − 1)x/2)cos((2n − 1) 2t)
2
(2n − 1) π
−32
∞

7. y(x,t) = sin((2n − 1)πx/2)cos(3(2n − 1)πt/2)
n=1 3 3
(2n − 1) π
4
∞

+ n=1 cos(nπ/4) − cos(nπ/2) sin(nπx/2)sin(3nπt/2)
n π 2
2
3
9. Let Y(x,t) = y(x,t) + h(x) and substitute into the problem to choose h(x) = (x − 4x)/9. The problem for Y is
2
2
∂ Y ∂ Y
= 3
∂t 2 ∂x 2
Y(0,t) = Y(2,t) = 0,
1 ∂Y
3
Y(x,0) = (x − 4x), (x,0) = 0,
9 ∂t
with solution
∞ n
32(−1) √

Y(x,t) = sin(nπx/2)cos(nπ 3t/2).
3
3n π 3
n=1
11. Let Y(x,t) = y(x,t) + h(x) and find that h(x) = cos(x) − 1. The problem for Y is
2
2
∂ Y ∂ Y
=
∂t 2 ∂x 2
Y(0,t) = Y(2π,t) = 0,
∂Y
Y(x,0) = cos(x) − 1, (x,0) = 0,
∂t
with solution
∞
16 1
Y(x,t) = sin((2n − 1)x/2)cos((2n − 1)t/2).
2
π (2n − 1)[(2n − 1) − 4]
n=1

1
13. u(x,t) = e −At/2 ∞ C n sin(nπx/L) AL n cos(r n t/2L) + sin(r n t/2L) ,
r
n=1
where
2 L
C n = f (ξ)sin(nπξ/L)dξ
L 0
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October 14, 2010 17:50 THM/NEIL Page-851 27410_25_Ans_p801-866