1000 Solved Problems in Modern Physics

64 1 Mathematical Physics 1.67 (a)y′− 2 y x = 1 x^3 (1) Lety=px,y′=p+xp′ Then (1) becomes xp′−p= 1 /x^3 Nowddx (p x ) =xpx− 2 p ...

1.3 Solutions 65 d^4 y dx^4 + 4d^2 y dx^2 =−8 cos(2x)(2) Multiply (1) by 4 and add to (2), d^4 y dx^4 + 8d^2 y dx^2 + 16 y= 0 D^ ...

66 1 Mathematical Physics 1.70 (i)d (^2) y dx^2 − 4dy dx + 4 y= 8 x^2 − 4 x−4(1) Replace the RHS member by zero to get the auxil ...

1.3 Solutions 67 1.3.9 LaplaceTransforms 1.72 dNA(t) dt =−λANA(t)(1) dNB(t) dt =−λBNB(t)+λANA(t)(2) Applying Laplace transform t ...

68 1 Mathematical Physics = λAλBNA^0 (λB−λA) [ 1 λA { 1 s − 1 s+λA } − 1 λB { 1 s − 1 s+λB }] =N 10 [ 1 s − λB (λB−λA) 1 (s+λA) ...

1.3 Solutions 69 H ′′ n−^2 ξH ′ n+^2 nHn=^0 1.76 Jn(x)= ∑ k (−1)k (x 2 )n+ 2 k k!Γ(n+k+1) (a) Differentiate d dx [xnJn(x)]=Jn(x) ...

70 1 Mathematical Physics = √ 2 πx [ 1 − x^2 2! + x^4 4! −··· ] = √ 2 πx cosx 1.79 The normalization of Legendre polynomials can ...

1.3 Solutions 71 1.80 Legendre’s equation is (1−x^2 ) ∂^2 Pn(x) ∂x^2 − 2 x ∂Pn(x) ∂x +n(n+1)Pn(x)=0(1) Putx=cosθ, Eq. (1) then b ...

72 1 Mathematical Physics (b) Differentiate with respect tox ∂T ∂x =s(1− 2 sx+s^2 )− (^32) = ∑ (1− 2 sx+s^2 )−^1 plsl+^1 = ∑ pl′ ...

1.3 Solutions 73 ∮ c dz (z−a)n = 2 πiifn= 1 =0ifn> 1 Method 2 By Cauchy’s integral formula f(n)(a)= n! 2 πi ∮ c f(z) (z−a)n+^ ...

74 1 Mathematical Physics Residue at exp(πi/4)=limz→exp(π 4 i) { z−exp ( πi 4 ) 1 z^4 + 1 } = 1 4 z^3 = 1 4 exp ( − 3 πi 4 ) Res ...

1.3 Solutions 75 But in (2),F=F(y′). Hence ∂F ∂y = 0 ∂F ∂y′ = ∂ ∂y′ (1+y′^2 ) (^12) =y′(1+y′^2 )−^1 /^2 d dx [ y′(1+y′^2 )−^1 /^ ...

76 1 Mathematical Physics F= √ 1 +y′^2 y = y′^2 √ 1 +y′^2 +C Simplifying√^1 y(1+y′^2 ) =Cwhere we have used (3) Ory(1+y′^2 )=con ...

1.3 Solutions 77 The volume: V=π ∫a 0 y^2 dx Therefore, dropping off the constant factors K=y^2 +λy(1+y′^2 )^1 /^2 which must sa ...

78 1 Mathematical Physics (c)<x^2 >= ∑ x^2 e−mmx x! = ∑ [x(x−1)+x] e−mmx x! = ∑∞ x= 0 e−mmx (x−2)! + ∑∞ x= 0 xe−m mx x! =e ...

1.3 Solutions 79 Mx(t)=Eext= ∑∞ x= 0 extB(x) = ∑∞ x= 0 ( N r ) (pet)r(1−p)N−r =(pet+ 1 −p)N ∑(N r ) prqN−r =(pet+ 1 −p)N μ^0 n= ...

80 1 Mathematical Physics N!→ √ 2 πNNNe−N x!→ √ 2 πxxxe−x (N−x)!→ √ 2 π(N−x)(N−x)N−xe−N+x B(x)→f(x)= √ N 2 πx(N−x) pxqN−xNN xx(N ...

1.3 Solutions 81 (b)B(x)= N!pxqN−x x!(N−x)! = N(N−1)...(N−x−1)px(1−p)N−x x! = N(N−1)...(N−x+1)(Np)x(1−p)N−x Nxx! = mx x! ( 1 − 1 ...

82 1 Mathematical Physics Normal equation gives λ= ∑ tnyn ∑ 6 t^2 n ∑ n= 1 tnyn= 1 ×ln(1/ 0 .835)+ 2 ×ln(1/ 0 .695) + 3 ×ln(1/ 0 ...

1.3 Solutions 83 Two limiting cases (a)t 2 =∞. We find that the number of intervals greater than any duration is Ne−atin whichat ...