Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 2 potx

... = 1 4π ln  ((x 2 − y 2 − ξ 2 + η 2 ) 2 + (2xy − 2 η) 2 ) ((x 2 − y 2 − ξ 2 + η 2 ) 2 + (2xy + 2 η) 2 )  u = 1 4π ln  ((x − ξ) 2 + (y − η) 2 ) ((x + ξ) 2 + (y + η) 2 ) ((x + ξ) 2 + (y − η) 2 ) ((x − ξ) 2 + ... sin  nπct L  . 20 28 0.05 0.1 0.15 0 .2 0 .25 0 0 .2 0.4 0 .6 0.8 1 0 1 2 0.05 0.1 0.15 0 .2 0 .25 0.05 0.1 0.15 0 .2 0 .25 0 0 ....

Ngày tải lên: 06/08/2014, 01:21

... A = 0, 2 ln t + 1 2 t 2 + c, A = 0 − 1 2 t 2 + ln t + c, A = 2 y =      1 A + t 2 A +2 + ct −A , A = 2 ln t + 1 2 t 2 + c, A = 0 − 1 2 + t 2 ln t + ct 2 , A = 2 For positive A, the ... are bounded at the origin are: y =      1 A + t 2 A +2 , A > 0 1 A + t 2 A +2 + ct −A , A < 0, A = 2 − 1 2 + t 2 ln t + ct 2 , A = 2 Equations in the...

Ngày tải lên: 06/08/2014, 01:21

... a and b are orthogonal, (perpendicular), or one of a and b are zero. 26 1 2 1 2 2 4 6 8 10 1 2 Figure 1.11: Plots of f(x) = p(x)/q(x). 1 .6 Hints Hint 1.1 area = constant ×diameter 2 . Hint 1 .2 A ... = 2x 2 x 2 + βx + χ Now we use the fact that f(x) is even to conclude that q(x) is even and thus β = 0. f(x) = 2x 2 x 2 + χ Finally, we use that f(1) = 1 to determin...

Ngày tải lên: 06/08/2014, 01:21

... a 2 z) = O(z 2 ) −a 0 z −1 +  a 2 + a 0 6  z + a 0 z −1 + a 1 + a 2 z = O(z 2 ) a 1 +  2a 2 + a 0 6  z = O(z 2 ) a 0 is arbitrary. Equating powers of z, z 0 : a 1 = 0. z 1 : 2a 2 + a 0 6 = 0. Thus ... 1/x. 8 02 and integrate to find the solution. dy dx = xy 2 y 2 dy = x dx  y 2 dy =  x dx + c −y −1 = x 2 2 + c y = − 1 x 2 /2 + c Example 14.4 .2 The eq...

Ngày tải lên: 06/08/2014, 01:21

... cases for even and odd n separately. a 2n = − a 2n 2 2n = a 2n−4 (2n)(2n − 2) = (−1) n a 0 (2n)(2n − 2) ···4 · 2 = (−1) n a 0  n m=1 2m , n ≥ 0 a 2n+1 = − a 2n−1 2n + 1 = a 2n−3 (2n + 1)(2n − 1) = ... 2t 3 d dt . 124 0 Solution 23 .2 1. First we write the differential equation in the standard form.  1 − x 2  y  − 2xy  + α(α + 1)y = 0 (23 .2) y  − 2x 1 − x 2 y  + α(α...

Ngày tải lên: 06/08/2014, 01:21

... denominator. s 2 + s − 1 (s − 2) (s − ı)(s + ı) We expand the function in partial fractions and then invert each term. s 2 + s − 1 (s − 2) (s − ı)(s + ı) = 1 s − 2 − ı /2 s − ı + ı /2 s + ı s 2 + s − 1 (s − 2) (s ... µ n x  1 460 L −1  1 s 2 1 s − 1  = e t −t − 1. Example 31.3 .2 We can find the inverse Laplace transform of s 2 + s − 1 s 3 − 2s 2 + s − 2 with the ai...

Ngày tải lên: 06/08/2014, 01:21

... transforms, F c [ e −2x ] = 2/ (π(ω 2 + 4)). −ω 2 ˆy c (ω) − 1 π y  (0) − ˆy c (ω) = 2 π(ω 2 + 4) ˆy c (ω) = − 2 π(ω 2 + 4)(ω 2 + 1) = 2 π  1/3 ω 2 + 1 − 1/3 ω 2 + 4  = 1 3 2/ π ω 2 + 4 − 2 3 1/π ω 2 + ... 1)(s + 2) 2 + 2s + 5 (s + 2) 2 ˆy = 4 s + 1 − 2 s + 2 − 3 (s + 2) 2 We take the inverse Laplace transform to determine the solution. y = 4 e −t −...

Ngày tải lên: 06/08/2014, 01:21

... + ζ 2 2(2n 2) + ζ 4 2 4·(2n 2) (2n−4) + ··· + ζ n 2 4···n·(2n 2) ···(2n−n)  for even n 2 n−1 n! ζ n+1  1 + ζ 2 2(2n 2) + ζ 4 2 4·(2n 2) (2n−4) + ··· + ζ n−1 2 4···(n−1)·(2n 2) ···(2n−(n−1))  for ... J 3 /2 (z), J 3 /2 (z) = 1 /2 z J 1 /2 (z) − J  1 /2 (z) = 1 /2 z  2 π  1 /2 z −1 /2 sin z −  − 1 2  2 π  1 /2 z −3 /2 sin z −  2 π  1 /2 z −1 /2 co...

Ngày tải lên: 06/08/2014, 01:21

... J −1 /2 (z)  z ζJ 1 /2 (ζ) 2/ πζ dζ = π 2 J 1 /2 (z)  z ζ 2 J −1 /2 (ζ) dζ − π 2 J −1 /2 (z)  z ζ 2 J 1 /2 (ζ) dζ. Thus the general solution is y = c 1 J 1 /2 (z) + c 2 J −1 /2 (z) + π 2 J 1 /2 (z)  z ζ 2 J −1 /2 (ζ) ... I  ν (z) + ν z I ν (z). I −1 /2 (z) = I  1 /2 (z) + 1 2z I 1 /2 (z) =  2 π z −1 /2 cosh(z) − 1 2  2 π z −3 /2 sinh(z) + 1 2z  2...

Ngày tải lên: 06/08/2014, 01:21

... −ϑ) − ∞  n,m=1  j  2n,m a  2 g nm 2j  2n,m √ π a  j  2 2n,m − 4n 2 |J 2n (j  2n,m )| J 2n  j  2n,m r a  sin(2nθ) = ∞  n,m=1 2j  2n,m √ π a  j  2 2n,m − 4n 2 |J 2n (j  2n,m )| J 2n  j  2n,m ρ a  sin(2nϑ) 2j  2n,m √ π ... 4n 2 |J 2n (j  2n,m )| J 2n  j  2n,m ρ a  sin(2nϑ) 2j  2n,m √ π a  j  2 2n,m − 4n 2 |J 2n (j  2n,m )| J 2n  j  2n,m r a  sin(2nθ...

Ngày tải lên: 06/08/2014, 01:21