Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 1 potx

... y 2 ((x − 1) 2 + y 2 ) 2 = −(x − 1) 2 + y 2 ((x − 1) 2 + y 2 ) 2 and 2( x − 1) y ((x − 1) 2 + y 2 ) 2 = 2( x − 1) y ((x − 1) 2 + y 2 ) 2 The Cauchy-Riemann equations are each identities. The first partial ... 8.4.3 1/ sin (z 2 ) has a second order pole at z = 0 and first order poles at z = (nπ) 1 /2 , n ∈ Z ± . lim z→0 z 2 sin (z 2 ) = lim z→...

Ngày tải lên: 06/08/2014, 01:21

... into two equations on separate domains. y  1 − y 1 = 0, y 1 (0) = 1, for x < 1 y  2 − y 2 = 1, y 2 (1) = y 1 (1) , for x > 1 797 • y  + 3xy  + 2y = x 2 • y  = y  y The degree ... doubles every hour. For the continuous problem, we assume that this i s true for y(t). We write this as an equation: y  (t) = αy(t). 775 1 2 3 4 4 8 12 16 1 2 3 4 4...

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 ... rearranging terms to form exact derivatives. 4yy  − xy  − y + 1 −9x 2 = 0 d dx  2y 2 − xy  + 1 − 9x 2 = 0 2y 2 − xy + x −3x 3 + c = 0 y = 1 4  x ±  x 2 − 8(c...

Ngày tải lên: 06/08/2014, 01:21

... of η to be zero.   4 3 −2 8 −6 −4 −4 3 2     0 0 η 3   = c 1   1 0 2   + c 2   0 2 3   −2η 3 = c 1 , −4η 3 = 2c 2 , 2η 3 = 2c 1 − 3c 2 c 1 = c 2 , η 3 = − c 1 2 888 We see ... one obtained in part (ii). 866 simplify the algebra. (A − 2I)ζ = η   −1 1 1 2 −1 −1 3 2 2     ζ 1 0 ζ 3   =   1 0 1   −ζ 1 + ζ 3 = 1, 2ζ 1 − ζ 3 = 0, 3 1 + 2ζ 3...

Ngày tải lên: 06/08/2014, 01:21

... exp  −  p n−1 (x) dx  . Example 16 .4. 4 Consider the differential equation y  − 3y  + 2y = 0. The Wronskian of the two independent solutions is W (x) = c exp  −  3 dx  = c e 3x . For the choice of solutions ... that e Jt =   e 2t t e 2t 0 0 e 2t 0 0 0 e 3t   8 94 Particular Solutions. Any function, y p , that satisfies the inhomogeneous equation, L[y p ] = f (x), is called a...

Ngày tải lên: 06/08/2014, 01:21

... make this an exact equation. d dx  e x 3 /3 y  = c 1 e x 3 /3 e x 3 /3 y = c 1  e x 3 /3 dx + c 2 y = c 1 e −x 3 /3  e x 3 /3 dx + c 2 e −x 3 /3 9 45 Exercise 17.21 Find the general solution ... real and positive. 944 Method 1. Note that this is an exact equation. d dx (y  − x 2 y) = 0 y  − x 2 y = c 1 d dx  e −x 3 /3 y  = c 1 e −x 3 /3 y = c 1 e x 3 /...

Ngày tải lên: 06/08/2014, 01:21

... x  = dx dy . dy dx = 1 y 3 − xy 2 dx dy = y 3 − xy 2 x  + y 2 x = y 3 Now we have a first order equation for x. d dy  e y 3 /3 x  = y 3 e y 3 /3 x = e −y 3 /3  y 3 e y 3 /3 dy + c e −y 3 /3 Example 18 .3. 2 Consider ... equation of order n − 1 for u  . Writing the derivatives of e u(x) , d dx e u = u  e u d 2 dx 2 e u = (u  + (u  ) 2 ) e u d 3 dx 3 e...

Ngày tải lên: 06/08/2014, 01:21

... y 1/2 exp  − 2y 3/ 2 3  x exp  − 2y 3/ 2 3  = −exp  − 2y 3/ 2 3  + c 1 x = −1 + c 1 exp  2y 3/ 2 3  x + 1 c 1 = exp  2y 3/ 2 3  log  x + 1 c 1  = 2 3 y 3/ 2 y =  3 2 log  x + 1 c 1  2 /3 y =  c + 3 2 log(x ... 3p  )u  exp  − 1 3  p(x) dx  y  =  u  − pu  + 1 3 (p 2 − 3p  )u  + 1 27 (9p  − 9p  − p 3 )u  exp  − 1 3  p(x) dx ...

Ngày tải lên: 06/08/2014, 01:21

... 1) 1 3 x 3 + x  1 3 − 1 2 − 1 3 x 3 + 1 2 x 2  y = 1 6 (x 3 − x). Example 21.7.4 Find the solution to the differential equation y  − y = sin x, that is bounded for all x. The Green function for ... the homogeneous solutions to y p and it will still be a particular solution. For example, η p = − 1 3 sin(2x) − 1 3 sin x = − 2 3 sin  3x 2  cos  x 2  is a particul...

Ngày tải lên: 06/08/2014, 01:21

... the problem for u, (and hence the problem for y). As a check, then general solution for y is y = − 1 3 cos 2x + c 1 cos x + c 2 sin x. 1115 We guess a particular solution of the form y p = t e −t (a ... y 2 = e 3t . We compute the Wronskian of these solutions. W (t) =     e 2t e 3t 2 e 2t 3 e 3t     = e 5t We find a particular solution with variation of parameters. y p = −...

Ngày tải lên: 06/08/2014, 01:21

... u  3 y  3 + u 3 y  3 = u 1 y  1 + u 2 y  2 + u 3 y  3 y  p = u  1 y  1 + u 1 y  1 + u  2 y  2 + u 2 y  2 + u  3 y  3 + u 3 y  3 Substituting the expressions for ... u 3 y  3 ) + p 0 (u 1 y 1 + u 2 y 2 + u 3 y 3 ) = f(x) u  1 y  1 + u  2 y  2 + u  3 y  3 + u 1 L[y 1 ] + u 2 L[y 2 ] + u 3 L[y 3 ] = f(x) u  1 y  1...

Ngày tải lên: 06/08/2014, 01:21

... second equation. c 1 r 2 1 + 1 r 2 (1 − c 1 r 1 )r 2 2 = 1 c 1 (r 2 1 − r 1 r 2 ) = 1 − r 2 11 81 0.2 0 .4 0.6 0.8 1 1.2 0.3 0 .4 0.5 0.6 0.7 0.8 0.9 Figure 23.2: Plot of the solution and approximations. Recall ... that a n =   n/2 j=0  − 4j 2 −2j +1 (2j+2)(2j +1)  for even n, 0 for odd n. 11 93 c 1 = 1 − r 2 r 2 1 − r 1 r 2 = 1 − 1 √ 5 2...

Ngày tải lên: 06/08/2014, 01:21

... − t 3 6 + O(t 5 ). Thus the Laplace transform of sin t has the behavior L[sin t] ∼ 1 s 2 − 1 s 4 + O(s −6 ) as s → +∞. 1 48 8 We use integration by parts and utilize the homogeneous boundary conditions. [v(pv  )  ] b a − ... ≤ √ 6 Thus the smallest zero of J 0 (x) is less than or equal to √ 6 ≈ 2 .44 94. (The smallest zero of J 0 (x) is approximately 2 .40 483 .) Solution 29.9 We as...

Ngày tải lên: 06/08/2014, 01:21

... take the Laplace transform of the differential equation and solve for ˆy(s). s 2 ˆy −sy(0) − y  (0) + 4sˆy −4y(0) + 4 y = 4 s + 1 s 2 ˆy −2s + 3 + 4sˆy −8 + 4 y = 4 s + 1 ˆy = 4 (s + 1)(s + 2) 2 + 2s ... define the Fourier transform. Define the two functions f + (x) =      1 for x > 0 1/2 for x = 0 0 for x < 0 , f − (x) =      0 for x > 0 1/2 for x = 0 1...

Ngày tải lên: 06/08/2014, 01:21

... integration by parts we can evaluate the integral.  1 0 (1 − τ) n τ z 1 dτ =  (1 − τ) n τ z z  1 0 −  1 0 −n (1 − τ) n 1 τ z z dτ = n z  1 0 (1 − τ) n 1 τ z dτ = n(n − 1) z(z + 1)  1 0 (1 − τ) n−2 τ z +1 dτ = n(n ... 1)  1 0 (1 − τ) n−2 τ z +1 dτ = n(n − 1) ··· (1) z(z + 1) ···(z + n 1)  1 0 τ z+n 1 dτ = n(n − 1) ··· (1) z(z + 1) ···(z + n 1...

Ngày tải lên: 06/08/2014, 01:21