1000 Solved Problems in Modern Physics

44 1 Mathematical Physics NowΓ(n)= ∫∞ 0 x n− (^1) e−xdx, putx=y (^2) ,dx= 2 ydy, so that Γ(n)= 2 ∫∞ 0 y^2 n−^1 e−y 2 dy Γ(1/2)= ...

1.3 Solutions 45 HXi=λiXi (1) NowX ̄ι′HXi=X ̄′ιλiXi=λiX ̄ι′Xi (2) is real and non-zero. Similarly the conjugate transpose X ̄′ιH ...

46 1 Mathematical Physics Fig. 1.10bReflection about a line passing through origin at 45 ◦ Fig. 1.10cElongating a vector in the ...

1.3 Solutions 47 1.31 A= ⎛ ⎝ 6 − 22 − 23 − 1 2 − 13 ⎞ ⎠ In Problem 1.30, the characteristic roots are found to beλ= 2 , 2 ,8. Wi ...

48 1 Mathematical Physics on settingk=1 andλ 1 = 5 − 2 C 11 + 2 C 21 =0(2) 4 C 11 − 4 C 21 =0(3) ThusC 21 =C 11 =a= 1 The substi ...

1.3 Solutions 49 h=− f(v) f′(v) ;f(v)=f(− 2 .1) f′(v)= dy dx |v;f′(v)= 10. 23 h=− 0. 039 10. 23 =− 0. 0038 To a first approximat ...

50 1 Mathematical Physics Differentiating, limn=∞ ( −2(n^2 +n1) ) =∞∞ Differentiating again, limn=∞ ( −^22 ) =−1(=L) ∣ ∣ ∣ ∣ 1 L ...

1.3 Solutions 51 1.38 f(a+x)=sin(a+x) Putx= 0 f(a)=sina f′(a)=cosa f′′(a)=−sina f′′′(a)=−cosa Substitute in f(x)=f(a)+ (x−a) 1! ...

52 1 Mathematical Physics =− 1 5 ln(2x+1)+ 1 5 ln(x−2)+C = 1 5 ln ( x− 2 2 x+ 1 ) +C 1.42r^2 =a^2 sin 2θ Elementary area dA= 1 2 ...

1.3 Solutions 53 1.45 (a) ∫ tan^6 xsec^4 xdx= ∫ tan^6 x(tan^2 x+1) sec^2 xdx = ∫ (tanx)^8 sec^2 xdx+ ∫ tan^6 xsec^2 xdx = ∫ (tan ...

54 1 Mathematical Physics 1.48x^2 /^3 +y^2 /^3 =a^2 /^3 (1) The arcABgenerates only one half of the surface. Sx 2 = 2 π ∫b a y [ ...

1.3 Solutions 55 = 2 ∫ ydx = 2 ∫ 2 1 1 x dx=2ln2 = 1 .386 units Fig. 1.14Area enclosed between the curvesy= 1 /x andy=− 1 /xand ...

56 1 Mathematical Physics Fig. 1.15aArea bounded by ABGFA (see the text, Prob 1.53a ) Fig. 1.15bVolume of the cylinder plus the ...

1.3 Solutions 57 1.3.8 OrdinaryDifferentialEquations...................... 1.55 dy dx = x^3 +y^3 3 xy^2 The equation is homogeno ...

58 1 Mathematical Physics 1.58 d^2 y dx^2 +m^2 y=cosbx (1) Replacing the right-hand member by zero, d^2 y dx^2 +m^2 y= 0. (2) So ...

1.3 Solutions 59 The complete solution is y=U+V (4) whereV=C 3 x+C 4 (5) In order thatVbe a particular solution of (1), substitu ...

60 1 Mathematical Physics The equation of motion for mass 2 is mx ̈ 2 +k(2x 2 −x 1 )=0(2) The two Eqs. (1) and (2) are coupled e ...

1.3 Solutions 61 Fig. 1.16Two modes of Oscillation If initiallyx 1 =x 2 , the masses oscillate in phase with frequencyω 1 (sym- ...

62 1 Mathematical Physics dy dx +Py=Q (2) dy dx + y(x+1) x =9(3) Lety=Uz (4) dy dx = Udz dx + zdU dx (5) Substituting (4) and (5 ...

1.3 Solutions 63 Differentiating (1) twice d^4 y dx^4 + d^3 y dx^3 − 2 d^2 y dx^2 =8 cosh 2x (3) Multiply (1) by (4) and subtrac ...