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Example text

N = N0 eKt N = N0 eKt − (R/K)(eKt − 1) where N0 = number of bacteria at t = 0, KN = rate of increase, R = removal rate. T = 100[1 − (ln r)/(ln 2)] T = 100(2r−1 −1) (a) k = weight divided by terminal speed. 1 cm/sec, so t is of the order of 10−4 sec. 15 S = 21 × 107 (1 + 3t/104) + (1 + 3t/104 )−1/3 , where S = number of pounds of salt, and t is in hours. (λn − λ2 ) etc. (all λ s different). 20 y = Ae−x + Beix x −3x −5x = Ae + Be + Ce = Aeix + Be−ix + Cex + De−x √ = Ae−x + Bex/2 sin 21 x 3 + γ = A + Be2x + Ce−3x = Ae5x + (Bx + C)e−x = Ax + B + (Cx + D)ex + (Ex2 + F x + G)e−2x = ex (A sin x + B cos x) + e−x (C sin x + D cos x) = (A + Bx)e−x + Ce2x + De−2x + E sin(2x + γ) Chapter 8 42 y = (Ax + B) sin x + (Cx + D) cos x + (Ex + F )ex + (Gx + H)e−x θ = θ0 cos ωt, ω = g/l T = 2π√ R/g ∼ = 85 min.

9 Amplitude increases; outward swing takes longer. 7 h(1) n (x) ∼ 0, α=β 1 2 2 jn−1 (α), α = β 1 1 2 1 0 2 παn xN 1/2 (αn x) dx = 2 , where jn (α) = jn (β) = 0. 13 L0 (x) = 1 H3 (x) = 8x3 − 12x H4 (x) = 16x4 − 48x2 + 12 H5 (x) = 32x5 − 160x3 + 120x L1 (x) = 1 − x L2 (x) = (2 − 4x + x2 )/2! L3 (x) = (6 − 18x + 9x2 − x3 )/3! L4 (x) = (24 − 96x + 72x2 − 16x3 + x4 )/4! L5 (x) = (120 − 600x + 600x2 − 200x3 + 25x4 − x5 )/5! Note: The factor 1/n! is omitted in most quantum mechanics books but is included as here in most reference books.

N = 2, ν = v/l 3. n = 3, ν = 3v/(2l), and n = 4, ν = 2v/l, have nearly equal intensity. 4. n = 2, ν = v/l 5, 6, 7, 8. 11 The basis functions for a string pinned at x = 0, free at x = l, and (n+ 1 )πx (n+ 21 )πvt . 12 With bn = 3 3 , odd n, the six solutions on (0, 1) are: n π 1. Temperature in semi-infinite plate: T = bn e−nπy sin nπx 2. Temperature in finite plate of height H: bn sinh nπ(H − y) sin nπx T = sinh (nπH) 2 3. 1-dimensional heat flow: u = bn e−(nπα) t sin nπx ¯h2 n2 π 2 4. Particle in a box: Ψ = bn sin nπx e−iEn t/¯h , En = 2m 5.

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