Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 9 docx

... transform of ˆy(s) by first finding its partial fraction expansion. ˆy(s) = s/3 s 2 + 1 − s/3 s 2 + 4 + s s 2 + 1 = − s/3 s 2 + 4 + 4s/3 s 2 + 1 y(t) = − 1 3 cos(2t) + 4 3 cos(t) Example 31 .4. 3 ... transform methods and show that i 1 = E 0 2R + E 0 2ωL e −αt sin(ωt) where α = R 2L and ω 2 = 2 LC − α 2 . Hint, Solution Exercise 31.23 Solve the initial value problem, y  + 4y  + 4...

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

... = 1 6 − 1 π 2 ∞  n=1 1 n 2 ∞  n=1 1 n 2 = π 2 6 141 8 29. 5 Hints Hint 29. 1 Hint 29. 2 Hint 29. 3 Hint 29 .4 Write the problem in Sturm-Liouville form to show that the eigenfunctions are orthogonal ... πx) 2 dx = π 3 /3 π 5 /30 = 10 π 2 ≈ 1.013 142 9 29 .4 Exercises Exercise 29. 1 Find the eigenvalues and eigenfunctions of y  + 2αy  + λy = 0, y(a) = y(b) = 0, where a < b....

Ngày tải lên: 06/08/2014, 01:21

... ı  1 /4 =  2 e −ıπ/6  1 /4 = 4 √ 2 e −ıπ/ 24 1 1 /4 = 4 √ 2 e ı(πn/2−π/ 24) , n = 0, 1, 2, 3 1 ı /4 = e (ı /4) log 1 = e (ı /4) (ı2πn) = e −πn/2 , n ∈ Z 308 7.11 Hints Cartesian and Modulus-Argument Form Hint ... coordinates.    e z 2    =    e (x+ıy) 2    =    e x 2 −y 2 +ı2xy    = e x 2 −y 2 3 04 25 50 75 100 -1 1 Figure 7 .43 : The values of ı ı . See Figu...

Ngày tải lên: 06/08/2014, 01:21

... are, Res  1 z 4 + 1 , e ıπ /4  = lim z→ e ıπ /4 z − e ıπ /4 z 4 + 1 = lim z→ e ıπ /4 1 4z 3 = 1 4 e −ı3π /4 = −1 − ı 4 √ 2 , Res  1 z 4 + 1 , e ı3π /4  = 1 4( e ı3π /4 ) 3 = 1 4 e −ıπ /4 = 1 − ı 4 √ 2 , We evaluate ... ζ j ) for j = 1, . . . , n. 7 04 we can apply Result 13 .4. 1.  ∞ −∞ 1 x 4 + 1 dx = ı2π  Res  1 z 4 + 1 , e ıπ /4  + Res  1 z 4 + 1 ,...

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 ... v +  b a G(x|ξ)f(ξ) dξ. 1 099 A particular solution is y p = t 3 6 e t +4. The general solution of the differential equation is y = c 1 e t +c 2 t e t + t 3 6 e t +4. We use the in...

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

... 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 approximation) In Figure 24. 4 we see ... 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 square...

Ngày tải lên: 06/08/2014, 01:21

... the formula to obtain information about the eigenvalues before we solve a problem. Example 27 .4. 2 Consider the self-adjoint eigenvalue problem −y  = λy, y(0) = y(π) = 0. 1322 Example 27 .4. 1 ... eigenfunction φ. Green’s formula states φ|L[φ] − L[φ]|φ = 0 φ|λφ − λφ|φ = 0 (λ − λ)φ|φ = 0 Since φ ≡ 0, φ|φ > 0. Thus λ = λ and λ is real. 13 19 27.7 Hints Hint 27.1 1327 2. F...

Ngày tải lên: 06/08/2014, 01:21

... solution and discuss in as much detail as possible what goes wrong. 13 64 Note that this formula is valid for m = 0, 1, 2, . . Similarly, we can multiply by sin(mx) and integrate to solve for b m . ... ∼ ∞  n=1 oddn 4 n  x −π sin(nt) dt = ∞  n=1 oddn 4 n  − 1 n cos(nt)  x −π = ∞  n=1 oddn 4 n 2 (−cos(nx) + (−1) n ) = 4 ∞  n=1 oddn −1 n 2 − 4 ∞  n=1 oddn cos(nx) n 2...

Ngày tải lên: 06/08/2014, 01:21