Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 4 pot

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 4 pot

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 4 pot

... 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...

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

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

... 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. 2: The discrete population of bacteria and a continuous population approximation. That is, ... 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) =...

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

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

... 9x 2 = 0 2y 2 − xy + x −3x 3 + c = 0 y = 1 4  x ±  x 2 − 8(c + x − 3x 3 )  2. We consider the differential equation, (2x − 2y)y  + (2x + 4y) = 0. 837 14. 11 Solutions Solution 14. 1 1. y  (x) y(x) = ... equation u  − sin(1/ζ) ζ u = 0. 840 -5 -4 -3 -2 -1 1 -3 -2 -1 1 Figure 14. 11: The function 2(1 − x) e x −1. Solution 14. 11 1. We consider the differential equation, (4y −x...

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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx

... for r > 0, it must be a global minimum. The cup has a radius of 4 3 √ π cm and a height of 4 3 √ π . 106 a x x x x x x∆ 1 2 3 i n-2 n-1 b f( ) ξ 1 Figure 4. 2: Divide -and- Conquer Strategy for ... f  (r), 2πr − 128 r 2 = 0, 2πr 3 − 128 = 0, r = 4 3 √ π . The second derivative of the surface area is f  (r) = 2π + 256 r 3 . Since f  ( 4 3 √ π ) = 6π, r = 4...

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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx

... Integral Formula to evaluate the integrals along C 1 and C 2 .  C (z 3 + z + ı) sin z z 4 + ız 3 dz =  C 1 (z 3 + z + ı) sin z z 3 (z + ı) dz +  C 2 (z 3 + z + ı) sin z z 3 (z + ı) dz = ı2π  (z 3 + ... integrand to s ee that there are singularities at the cube roots of 9. z z 3 − 9 = z  z − 3 √ 9  z − 3 √ 9 e ı2π /3  z − 3 √ 9 e −ı2π /3  Let C 1 , C 2 an...

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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot

... the 656 Hint 13. 39 Hint 13. 40 Definite Integrals Involving Sine and Cosine Hint 13. 41 Hint 13. 42 Hint 13. 43 Hint 13. 44 Make the changes of variables x = sin ξ and then z = e ıξ . Infinite Sums Hint 13. 45 Use ... ıa)     z=a = 1 4a 3 Res  1 z 4 − a 4 , z = −a  = 1 (z − a)(z − ıa)(z + ıa)     z=−a = − 1 4a 3 Res  1 z 4 − a 4 , z = ıa  = 1 (z − a)(z +...

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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 3 pptx

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 3 pptx

... components 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 , 4 3 = 2c 2 , 2η 3 = 2c 1 − 3c 2 c 1 = c 2 , η 3 = − c 1 2 888 We see ... 0     e 2 e 2 e 2 /2 0 e 2 e 2 0 0 e 2     0 3 −2 0 −1 −1 1 −1 −1   e A =   0 2 2 3 1 −1 −5 3 5   e 2 2 859 Thus there are two eigenvectors.   4 3 −2 8 −...

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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 5 pdf

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 5 pdf

... 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 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 /...

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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 6 ppsx

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 6 ppsx

... 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 ... − 1 4 y 4 + c 1 u =  2c 1 − 1 2 y 4  1/2 y  =  2c 1 − 1 2 y 4  1/2 dy (2c 1 − 1 2 y 4 ) 1/2 = dx Integrating gives us the implici...

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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps

... 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  y...

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