Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 10 potx

... 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

... ≤ √ 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 assume that 0 < l < π. Recall that the ... y  1 (l) = y  2 (l) c 1 l = c 2 (π − l), c 1 = −c 2 146 4 Figure 29.1: x and −tan(x). To solve the inhomogeneous problem, we expand the solution and the inhomogeneity in a series of...

Ngày tải lên: 06/08/2014, 01:21

... circle |z +1 /4| = 1 /4. Thus the current image is the left half plane minus a circle: (z) < 0 and     z + 1 4     > 1 4 . The magnification by 2 yields (z) < 0 and     z ... |z + ı /4| = 1 /4. Thus the current image is the region between two circles:    z + ı 2    < 1 2 and    z + ı 4    > 1 4 . The magnification by 2 yields |z + ı| <...

Ngày tải lên: 06/08/2014, 01:21

... +  − π 4 120 + π 4 36 − π 4 12 + π 4 24  z 4 + ···  = ··· − π 4 45 1 z + ··· Thus the residue at z = 0 is −π 4 /45 . Summing the residues, −1  n=−∞ 1 n 4 − π 4 45 + ∞  n=1 1 n 4 = 0. ∞  n=1 1 n 4 = π 4 90 Solution ... cot(πz)/z 4 . π cos(πz) z 4 sin(πz) = π z 4  1 − π 2 2 z 2 + π 4 24 z 4 − ···  1 πz  1 +  π 2 6 z 2 − π 4 120 z 4 + ···  +  π...

Ngày tải lên: 06/08/2014, 01:21

... integrating factor for the Equation 14. 4. The proof of this is left as an exercise for the reader. (See Exercise 14. 2.) Result 14. 4 .4 Homogeneous Coefficient Differential Equations. If P(x, y) and Q(x, y) are ... 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 = −2 For positive A, the solution is bounded at the origin only for c = 0. For A = 0, there are no bounded solutions. For negative A, the solution is bounded there for any value of c and thus ... a  1/(α+1) 822 1 2 3 4 1 1 2 3 4 1 Figure 14. 10: The Solution as α → 0 and α → ∞ In the limit as α → ∞ we have, lim α→∞ y(x) = lim α→∞ 1 α − 1  e −x +(α − 2) e −αx  = lim α→∞...

Ngày tải lên: 06/08/2014, 01:21

... function. Solution 11 64 1 2 3 4 5 -0 .4 -0.3 -0.2 -0.1 1 2 3 4 5 -0.5 -0 .4 -0.3 -0.2 -0.1 Figure 21.8: G(x; 1) and G(x; −1) The solutions that respectively satisfy the left and right boundary conditions ... =  ∞ 0 G(x|ξ)ξ 2 dξ =  x 0  1 2 x − ξ 2 2x  ξ 2 dξ = 1 6 x 4 − 1 10 x 4 = x 4 15 . Now to solve the homogeneous differential equation with inhomogeneous boundary co...

Ngày tải lên: 06/08/2014, 01:21

... 1)c n+2 + c n−2 = 0, for n ≥ 2 c n +4 = − c n (n + 4) (n + 3) For our first solution we have the difference equation a 0 = 1, a 1 = 0, a 2 = 0, a 3 = 0, a n +4 = − a n (n + 4) (n + 3) . For our second solution, b 0 = ... polynomial are α 1 = 1 +  1 − 3 /4 2 = 3 4 , α 2 = 1 −  1 − 3 /4 2 = 1 4 . Thus our two series solutions will be of the form w 1 = z 3 /4 ∞  n=0 a n z n , w...

Ngày tải lên: 06/08/2014, 01:21

... =  x 3 3  1 −1 = 2 3  1 −1 P 2 (x)P 2 (x) dx =  1 −1 1 4  9x 4 − 6x 2 + 1  dx =  1 4  9x 5 5 − 2x 3 + x  1 −1 = 2 5  1 −1 P 3 (x)P 3 (x) dx =  1 −1 1 4  25x 6 − 30x 4 + 9x 2  dx =  1 4  25x 7 7 − 6x 5 + 3x 3  1 −1 = 2 7 Solution ... equation is α 2 + 1 4 α − 1 8 = 0  α + 1 2  α − 1 4  = 0. Since the roots are distinct and do not differ by an integer, the...

Ngày tải lên: 06/08/2014, 01:21

... 21) π 4 P 4 (x) = 105 8π 4 [(315 − 30π 2 )x 4 + ( 24 2 − 270)x 2 + (27 − 2π 2 )] The cosine and this polynomial are plotted in the second graph in Figure 25.1. The le ast squares fit method uses information ... x 3 , x 4 } is independent, but not orthogonal in the interval [−1, 1]. Using Gramm-Schmidt orthogo- 12 84 1 2 3 4 5 6 -60 -40 -20 Figure 24. 3: log(error in approxi...

Ngày tải lên: 06/08/2014, 01:21