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8 Oct 2015: x = A cos ωt B sin ωt a cos(ωt ϵ) Peiωt Qeiωt, (2). ... The solution isx = A cos ωt B sin ωt. Initially, x = a and ẋ = 0, so.

8 Oct 2015: The other end,B, is fixed. The particle moves in a horizontal circle a distance d below B. ... cos θ,. where cos θ = d/l. 2 A particle rests at the top of a smooth sphere.

18 Nov 2015: 2. z2h5 B. 2. θ2h5 = 0,. (curiously A α/6, but B β/80). ... t1/4. h(π) =1 cos L. t1/4. (K ((π L) cos L sin L.

23 Feb 2015: weak. charged current interaction). Given R(θ) =. (cos θ sin θsin θ cos θ. ), ... 4 cos θ. (2a2D v2a2D2. )},. where v = 2 sin2 θW 12,a = 12.

31 Dec 2015: ex. ez. ) a2 y(ez. ey eyez). (16). The contour of the elliptical cross section can be parameterized as x = a cos , y = b sin , with0   2. ... a sin e. x. b cos eypa2sin2 b2cos2. , n =b cos e.

9 Nov 2015: 87 –. while the jump condition gives. B( cos(1 ξ)) A cos ξ = 1. ... The generalsolution of this homogeneous equation c1 cos x c2 sin x and the inhomogeneous boundary.

10 Oct 2015: H′3 H′′′3 = G1. 1. Pn1 (1 cos x)n1, H3 Sx. ... H′5 H′′′5 = 0, H5 = A5 B5 sin x C5 cos x.

4 Nov 2015: n 1)(n (n 1) cos x). (2n 1)(2n 1). xπ. 1. ... H′3 H′′′3 = G1. 1. Pn1 (1 cos x)n1, H3 Sx.

26 Oct 2015: When λ 0 then X(x) = a sin(λx) b cos(. λx) solves X′′ = λX. ... λx B cos. λx, but if we impose the boundary condition y(1) = y(1) then.

4 Aug 2015: where. Vn = (An Bny)eny sin nξ (Cn Dny)eny cos nξ.(11). ... 3), is given by. rs = (δ ρ cos θ )x ρ sin θ y z z, δ = b sin k(z ct ),(18).