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

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps

... values. For instance, (1 2 ) 1/ 2 = 1 1/2 = 1 and  1 1/2  2 = ( 1) 2 = 1. Example 6. 6.2 Consider 2 1/ 5 , (1 + ı) 1/ 3 and (2 + ı) 5 /6 . 2 1/ 5 = 5 √ 2 e ı2πk/5 , for k = 0, 1, 2, 3, 4 19 9 Example 6. 5 .1 ... ellipse. 18 7 6. 8 Hints Complex Numbers Hint 6 .1 Hint 6. 2 Hint 6. 3 Hint 6. 4 Hint 6. 5 Hint 6. 6 Hint 6. 7 The Complex Plane Hint 6. 8...

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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

... z z 3  z=−ı + 2 2!  d 2 dz 2 (z 3 + z + ı) sin z z + ı  z=0 = 2 (−ı sinh(1)) + ıπ  2  3z 2 + 1 z + ı − z 3 + z + ı (z + ı) 2  cos z +  6z z + ı − 2( 3z 2 + 1) (z + ı) 2 + 2( z 3 + z + ı) (z ... formula.  C z z 2 + 1 dz =  C 1 /2 z −ı dz +  C 1 /2 z + ı dz = 1 2 2 + 1 2 2 = 2 3.  C z 2 + 1 z dz =  C  z + 1 z  dz =  C z dz +  C 1 z dz = 0 + 2 = 2...

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

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

... function. Hint 12. 22 cos z = −cos(z − π) sin z = −sin(z − π) Hint 12. 23 CONTINUE Hint 12. 24 CONTINUE Hint 12. 25 Hint 12. 26 Hint 12. 27 Hint 12. 28 Hint 12. 29 Hint 12. 30 CONTINUE 581 Solution 12. 22 cos z ... polynomial. 2 6 12 20 4 6 8 2 2 We s ee that the polynomial is second order. p(n) = an 2 + bn + c. We solve for the coefficients. a + b + c = 2 4a + 2b + c = 6 9...

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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

... 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...

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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

... 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...

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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

... 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...

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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 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 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 /...

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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 ... 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...

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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 ...

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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 8 ppsx

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 8 ppsx

... 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...

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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 9 doc

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 9 doc

... 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 = −...

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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 10 ppt

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 10 ppt

... 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...

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Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 6 pps

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 6 pps

... = ∞  n=1 2(−1) n+1 n sin(nx) for x ∈ (−π . . . π). We apply Parseval’s theorem for this series to find the value of  ∞ n=1 1 n 2 . ∞  n=1 4 n 2 = 1 π  π −π x 2 dx ∞  n=1 4 n 2 = 2π 2 3 ∞  n=1 1 n 2 = π 2 6 13 74 ... no value of a for which both cos a and sin a vanish, the system is not orthogonal for any interval of length π. 2. First note that  π 0 cos nx dx = 0 for...

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Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps

... + 4 3 P 2 (cos θ) exp  − 6a 2 R 2 t  Ψ(θ, t) = − 1 3 +  2 cos 2 θ − 2 3  exp  − 6a 2 R 2 t  Solution 37.34 Since we have homogeneous boundary conditions at x = 0 and x = 1, we will expand the solution ... − Y  Y = −λ We have differential equations for X and Y . X  + λX = 0, X(0) = X(1) = 0 Y  − λY = 0, Y (0) = 0 The eigenvalues and orthonormal eigenfunctions for X are λ n...

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