Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 1 potx
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
... y 2 ((x − 1) 2 + y 2 ) 2 =−(x − 1) 2 + y 2 ((x − 1) 2 + y 2 ) 2 and 2( x − 1) y((x − 1) 2 + y 2 ) 2 = 2( x − 1) y((x − 1) 2 + y 2 ) 2 The Cauchy-Riemann equations are each identities. The first partial ... 8.4.3 1/ sin (z 2 ) has a second order pole at z = 0 and first order poles at z = (nπ) 1 /2 , n ∈ Z±.limz→0z 2 sin (z 2 )= limz→02z2z cos (z 2 )= limz→0 2 2 cos (z 2 ) − 4z 2 sin (z 2 )= ... L’Hospital’s rule. 2. limz→ı 1 + z 2 2 + 2z6=2z 12 z5z=ı= 1 6limz→ıπsinh(z)ez +1 =cosh(z)ezz=ıπ= 1 399 8.7 HintsComplex DerivativesHint 8 .1 Hint 8 .2 Start with the Cauchy-Riemann...
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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 y2− y2= 1, y2 (1) = y 1 (1) , for x > 1 797 • y+ 3xy+ 2y = x2• y= yyThe degree ... doubles everyhour. 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 3 448 12 16 1 2 3 448 12 16 Figure 14 .1: The p opulation ... =y(x)x.P (1, u) + Q (1, u)u + xdudx= 0This equation is separable.P (1, u) + uQ (1, u) + xQ (1, u)dudx= 0 1 x+Q (1, u)P (1, u) + uQ (1, u)dudx= 0ln |x| + 1 u + P (1, u)/Q (1, u)du...
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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 =1A+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 = 0ddx2y 2 − xy+ 1 − 9x 2 = 02y 2 − xy + x −3x 3 + c = 0y =14x ±x 2 − 8(c + x − 3x 3 ) 2. We consider the ... = x0≡ 2 01The matrix has the distinct eigenvalues λ1= 1, λ 2 = 2, λ 3 = 3. The corresponding eigenvectors arex1=0 2 1, x 2 =110, x 3 = 2 21.849...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 3 pptx
... of η to be zero.4 3 −28 −6 −4−4 3 200η 3 = c1102+ c202 3 −2η 3 = c1, −4η 3 = 2c2, 2η 3 = 2c1− 3c2c1= c2, η 3 = −c12888 We see ... one obtained in part (ii).866 simplify the algebra.(A − 2I)ζ = η−1 1 12 −1 −1 3 2 2ζ10ζ 3 =101−ζ1+ ζ 3 = 1, 2ζ1− ζ 3 = 0, 3 1+ 2ζ 3 = 1ζ =102A ... −λ 3 + 3 2− 3 + 1 = −(λ − 1) 3 λ = 1 is an eigenvalue of multiplicity 3. The rank of the null space of A − I is 2. (The second and thirdrows are multiples of the first.)A − I =4 3 −28...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 4 pot
... exp−pn−1(x) dx.Example 16 .4. 4 Consider the differential equationy− 3y+ 2y = 0.The Wronskian of the two independent solutions isW (x) = c exp− 3 dx= ce3x. For the choice of solutions ... thateJt=e2tte2t00e2t00 0e3t8 94 Particular Solutions. Any function, yp, that satisfies the inhomogeneous equation, L[yp] = f (x), is called aparticular solution or particular integral of ... R(x)y = 0. (16 .4) An exact equation can be written in the form:ddx[a(x)y+ b(x)y] = 0.If Equation 16 .4 is exact, then we can write it in the form:ddx[P (x)y+ f(x)y] = 0 for some function...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 5 pdf
... make this an exact equation.ddxex 3 /3 y= c1ex 3 /3 ex 3 /3 y = c1ex 3 /3 dx + c2y = c1e−x 3 /3 ex 3 /3 dx + c2e−x 3 /3 9 45 Exercise 17.21Find the general solution ... real and positive.944 Method 1. Note that this is an exact equation.ddx(y− x2y) = 0y− x2y = c1ddxe−x 3 /3 y= c1e−x 3 /3 y = c1ex 3 /3 e−x 3 /3 dx + c2ex 3 /3 Method ... Solution 954 Noting thate(2+3i)x=e2x[cos(3x) + ı sin(3x)]e(2−3i)x=e2x[cos(3x) − ı sin(3x)],we can write the two linearly independent solutionsy1=e2xcos(3x), y2=e2xsin(3x).Solution...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 6 ppsx
... x=dxdy.dydx=1y 3 − xy2dxdy= y 3 − xy2x+ y2x = y 3 Now we have a first order equation for x.ddyey 3 /3 x= y 3 ey 3 /3 x =e−y 3 /3 y 3 ey 3 /3 dy + ce−y 3 /3 Example 18 .3. 2 Consider ... equation of ordern − 1 for u. Writing the derivatives ofeu(x),ddxeu= ueud2dx2eu= (u+ (u)2)eud 3 dx 3 eu= (u+ 3uu+ (u) 3 )eu.997 Finally ... α)q(t)Thus we obtain a linear equation for u which when solved will give us an implicit solution for y.1008 Example 18 .6 .3 Consider the linear equation in Example 18 .6. 1y+ p(x)y+ q(x)y = 0.Under...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 7 pps
... y1/2exp−2y 3/ 2 3 x exp−2y 3/ 2 3 = −exp−2y 3/ 2 3 + c1x = −1 + c1exp2y 3/ 2 3 x + 1c1= exp2y 3/ 2 3 logx + 1c1=2 3 y 3/ 2y = 3 2logx + 1c12 /3 y =c + 3 2log(x ... 3p)uexp−1 3 p(x) dxy=u− pu+1 3 (p2− 3p)u+1 27 (9p− 9p− p 3 )uexp−1 3 p(x) dxyields the differential equationu+1 3 (3q − 3p− p2)u+1 27 (27r ... 0into the formu(n)+ an−2(x)u(n−2)+ an 3 (x)u(n 3) + ···+ a0(x)u = 0.10 23 This is a first order differential equation for x(y).x− y1/2x = y1/2ddyx exp−2y 3/ 2 3 = y1/2exp−2y 3/ 2 3 x...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 8 ppsx
... 1)1 3 x 3 + x1 3 −12−1 3 x 3 +12x2y =16(x 3 − x).Example 21.7.4 Find the solution to the differential equationy− y = sin x,that is bounded for all x.The Green function for ... the homogeneous solutions to yp and it will still be a particular solution. For example,ηp= −1 3 sin(2x) −1 3 sin x= −2 3 sin3x2cosx2is a particular solution.21.2 Method ... the form of a particular solution. This form will contain some unknown parameters. We substitute this forminto the differential equation to determine the parameters and thus determine a particular...
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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 isy = −1 3 cos 2x + c1cos x + c2sin x.1115 We guess a particular solution of the formyp= te−t(a ... y2=e3t.We compute the Wronskian of these solutions.W (t) =e2te3t2e2t 3 e3t=e5tWe find a particular solution with variation of parameters.yp= −e2t2ete3te5tdt ... c2e−tsin(2t).11 29 The solution for v isv =1e − cos 1 − sin 1(e + sin 1 − 3) cos x + (2e − cos 1 − 3) sin x + (3 − cos 1 − 2 sin 1)ex.Now we find the Green function for the problem...
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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 10 ppt
... u 3 y 3 + u 3 y 3 = u1y1+ u2y2+ u 3 y 3 yp= u1y1+ u1y1+ u2y2+ u2y2+ u 3 y 3 + u 3 y 3 Substituting the expressions for ... u 3 y 3 )+ p0(u1y1+ u2y2+ u 3 y 3 ) = f(x)u1y1+ u2y2+ u 3 y 3 + u1L[y1] + u2L[y2] + u 3 L[y 3 ] = f(x)u1y1+ u2y2+ u 3 y 3 = ... u2y2+ u 3 y 3 = 0Differentiating the expression for yp,yp= u1y1+ u1y1+ u2y2+ u2y2+ u 3 y 3 + u 3 y 3 = u1y1+ u2y2+ u 3 y 3 yp= u1y1+...
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Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx
... second equation.c 1 r2 1 + 1 r2 (1 − c 1 r 1 )r22= 1 c 1 (r2 1 − r 1 r2) = 1 − r2 11 81 0.2 0 .4 0.6 0.8 1 1.20.30 .4 0.50.60.70.80.9Figure 23.2: Plot of the solution and approximations.Recall ... thatan=n/2j=0−4j2−2j +1 (2j+2)(2j +1) for even n,0 for odd n. 11 93 c 1 = 1 − r2r2 1 − r 1 r2= 1 − 1 √52 1+ √52√5= 1+ √52 1+ √52√5= 1 √5Substitute this result into the equation for ... − 1) + 2n + n(n − 1) + 1 4 an(α) + 1 4 an 1 (α) = 0an(α) = −n(n + 1) 4 +α(α − 1) 4 + 1 16an 1 (α).The first few an’s area0, −α(α − 1) +9 16 a0,α(α − 1) +25 16 α(α...
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Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 8 potx
... −t36+ O(t5).Thus the Laplace transform of sin t has the behaviorL[sin t] ∼1s2−1s 4 + O(s−6) as s → +∞. 1 48 8 We use integration by parts and utilize the homogeneous boundary conditions.[v(pv)]ba− ... ≤√6Thus the smallest zero of J0(x) is less than or equal to√6 ≈ 2 .44 94. (The smallest zero of J0(x) is approximately2 .40 483 .)Solution 29.9We assume that 0 < l < π.Recall that the ... inverse Laplace transform for t > 0. 1 48 5 • te2tis of exponential order α for any α > 2.ãet2is not of exponential order for any .ã tnis of exponential order for any > 0.ã t2does...
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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).s2ˆy −sy(0) − y(0) + 4sˆy −4y(0) + 4 y = 4 s + 1s2ˆy −2s + 3 + 4sˆy −8 + 4 y = 4 s + 1ˆy = 4 (s + 1)(s + 2)2+2s ... define the Fourier transform. Define the two functionsf+(x) =1 for x > 01/2 for x = 00 for x < 0, f−(x) =0 for x > 01/2 for x = 01 for x < 0.1550 This ... Transforms of DerivativesCosine Transform. Using integration by parts we can find the Fourier cosine transform of derivatives. Let y be afunction for which the Fourier cosine transform of y and...
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Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 1 pdf
... integration by parts we can evaluate the integral. 1 0 (1 − τ)nτz 1 dτ = (1 − τ)nτzz 1 0− 1 0−n (1 − τ)n 1 τzzdτ=nz 1 0 (1 − τ)n 1 τzdτ=n(n − 1) z(z + 1) 1 0 (1 − τ)n−2τz +1 dτ=n(n ... 1) 1 0 (1 − τ)n−2τz +1 dτ=n(n − 1) ··· (1) z(z + 1) ···(z + n 1) 1 0τz+n 1 dτ=n(n − 1) ··· (1) z(z + 1) ···(z + n 1) τz+nz + n 1 0=n!z(z + 1) ···(z + n) 16 10 For x > 0 we close ... limn→∞nzn!z(z + 1) ···(z + n)= 1 zlimn→∞ (1) (2) ···(n)(z + 1) (z + 2) ···(z + n)nz= 1 zlimn→∞ 1 (1 + z) (1 + z/2) ··· (1 + z/n)nz= 1 zlimn→∞ 1 (1 + z) (1 + z/2) ··· (1 + z/n)2z3z···nz 1 z2z···(n...
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