Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 2 potx

... 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

... 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 equation y  = y − y 2 is separable. y  y −y 2 = ... αy(t). 775 1 2 3 4 4 8 12 16 1 2 3 4 4 8 12 16 Figure 14.1: The p opulation of bacteria. 1 2 3 4 32 64 96 128 1 2 3 4 32 64 96 128 Figure 14...

Ngày tải lên: 06/08/2014, 01:21

... = 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 + ... = 1 2  ln   (x − ξ) 2 + (y − η) 2  − ln   (x + ξ) 2 + (y − η) 2  −ln   (x − ξ) 2 + (y + η) 2  + ln   (x + ξ) 2 + (y...

Ngày tải lên: 06/08/2014, 01:21

... k a 1 a 2 a 3 b 1 b 2 b 3       = i     a 2 a 3 b 2 b 3     − j     a 1 a 3 b 1 b 3     + k     a 1 a 2 b 1 b 2     = (a 2 b 3 − a 3 b 2 )i − (a 1 b 3 − a 3 b 1 )j ... (b 1 i + b 2 j + b 3 k) = a 1 b 2 k + a 1 b 3 (−j) + a 2 b 1 (−k) + a 2 b 3 i + a 3 b 1 j + a 3 b 2 (−i) = (a 2 b 3 − a 3 b 2 )i − (a...

Ngày tải lên: 06/08/2014, 01:21

... 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   2 3 = c 1 , −4η 3 = 2c 2 , 2 3 = 2c 1 − 3c 2 c 1 = c 2 , η 3 = − c 1 2 888 We see that we ... λ           856 e At =   2 e −2t +3 e −t − e −2t + e −t − e −2t + e −t 5 e −2t −8 e −t +3 e t 2 15 e −2t −16 e −t + 13 e t 12 15 e −2t −16 e −t + e...

Ngày tải lên: 06/08/2014, 01:21

... u  1 (x 0 ) u 2 (x 0 ) u  2 (x 0 )  =  c 11 c 12 c 21 c 22  y 1 (x 0 ) y  1 (x 0 ) y 2 (x 0 ) y  2 (x 0 )  =  1 0 0 1   c 11 c 12 c 21 c 22  =  y 1 (x 0 ) y  1 (x 0 ) y 2 (x 0 ) y  2 (x 0 )  −1 If ... u 2 }.  u 1 u 2  =  c 11 c 12 c 21 c 22  y 1 y 2  For {u 1 , u 2 } to satisfy the relations that define a fundamental set, it must s...

Ngày tải lên: 06/08/2014, 01:21

... exp(  x 2 dx) to 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 /3 945 Exercise 17 .21 Find the ... of (a) x 2 y  − 2xy  + 2y = 0, (b) x 2 y  − 2y = 0, (c) x 2 y  − xy  + y = 0. Hint, Solution 954 Noting that e (2+ 3i)x = e 2x [cos(3x) + ı sin(3x)] e...

Ngày tải lên: 06/08/2014, 01:21

... c 1 e √ 2x +c 2 e − √ 2x u  = 2 √ 2 c 1 e √ 2x −c 2 e − √ 2x c 1 e √ 2x +c 2 e − √ 2x u = 2  c 1 √ 2 e √ 2x −c 2 √ 2 e − √ 2x c 1 e √ 2x +c 2 e − √ 2x dx + c 3 u = 2 log  c 1 e √ 2x +c 2 e − √ 2x  + ... 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 Exa...

Ngày tải lên: 06/08/2014, 01:21

... exp  − 2y 3/ 2 3  = 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 ... p n 2 (x)y (n 2) + ···+ p 0 (x)y = 0 into the form u (n) + a n 2 (x)u (n 2) + a n 3 (x)u (n 3) + ···+ a 0 (x)u = 0. 10 23 This is...

Ngày tải lên: 06/08/2014, 01:21

... equation. u  1 y 1 + 2u  1 y  1 + u 1 y  1 + u  2 y 2 + 2u  2 y  2 + u 2 y  2 + p(u  1 y 1 + u 1 y  1 + u  2 y 2 + u 2 y  2 ) + q(u 1 y 1 + u 2 y 2 ) = f u  1 y 1 + 2u  1 y  1 + u  2 y 2 + 2u  2 y  2 + ... y p = e t (at 2 + 4at + 2a) e t 2( at 2 + 2at) e t +at 2 e t = e t 2a e t = e t a = 1 2 A particular solution is y p = t 2...

Ngày tải lên: 06/08/2014, 01:21