(a) Solve the differential equation: d³y/dx³ - 3d²y/dx² + 4dy/dx - 2y = eˣ + cos x. (15 marks) (b) When a particle is projected…

(a) Solution of the differential equation

Let L = d³/dx³ - 3d²/dx² + 4d/dx - 2. The characteristic equation is r³ - 3r² + 4r - 2 = 0, i.e. (r - 1)(r² - 2r + 2) = 0. Hence r = 1, 1 ± i. Therefore the complementary function is y_c = C₁eˣ + eˣ(C₂ cos x + C₃ sin x).

For eˣ, since r = 1 is a simple root, take y_p1 = A x eˣ. Here P(r) = r³ - 3r² + 4r - 2, so P(1) = 0 and P′(1) = 1. Hence L(A x eˣ) = A eˣ, giving A = 1. Thus y_p1 = x eˣ.

For cos x, take y_p2 = B cos x + C sin x. Substitution gives L(y_p2) = (B + 3C) cos x + (C - 3B) sin x. Equating with cos x gives B + 3C = 1 and C - 3B = 0. Thus B = 1/10, C = 3/10, so y_p2 = (1/10)(cos x + 3 sin x).

Therefore the general solution is y = C₁eˣ + eˣ(C₂ cos x + C₃ sin x) + x eˣ + (1/10)(cos x + 3 sin x).

(b) Ranges of projection from two heights

Let O₁ be the origin, x horizontal and y vertical upward. For projection from sea level, y = v sinθ t - (1/2)gt², x = v cosθ t. Time of flight is T₁ = 2v sinθ/g, hence R₁ = v cosθ T₁ = v² sin2θ/g = 2v² sinθ cosθ/g.

For projection from O₂ at height h above O₁, y = h + v sinθ t - (1/2)gt². At sea level y = 0, so (1/2)gt² - v sinθ t - h = 0, giving the positive root t = [v sinθ + √(v² sin²θ + 2gh)]/g. Therefore R₂ = v cosθ t = (v² sinθ cosθ/g)[1 + √(1 + 2gh/(v² sin²θ))]. Since v² sinθ cosθ/g = R₁/2, R₂ = (R₁/2)[1 + √(1 + 2gh/(v² sin²θ))]. As h > 0, the square root is greater than 1, so R₂ > R₁. Also R₂ - R₁ = (R₁/2)[√(1 + 2gh/(v² sin²θ)) - 1]. Hence (R₂ - R₁):R₁ = (1/2)[√(1 + 2gh/(v² sin²θ)) - 1]:1. This assumes 0 < θ < π/2 and neglects air resistance.

(c) Surface integral and Gauss divergence theorem

Write F = 3y²z² î + 4z²x² ĵ + z²y² k̂. The surface is the upper hemisphere of 4x² + 4y² + 4z² = 1, i.e. x² + y² + z² = a² with a = 1/2.

Parametrize S by r(φ,θ) = (a sinφ cosθ, a sinφ sinθ, a cosφ), where 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π. Then r_φ × r_θ = a² sinφ (sinφ cosθ î + sinφ sinθ ĵ + cosφ k̂), giving the outward normal. Computing F · (r_φ × r_θ), the terms involving cosθ and sinθ vanish after θ-integration from 0 to 2π. What remains is ∬ₛ F·n dS = a⁶ (∫₀^π/2 sin³φ cos³φ dφ)(∫₀^2π sin²θ dθ). Now ∫₀^π/2 sin³φ cos³φ dφ = 1/12, ∫₀^2π sin²θ dθ = π. Thus ∬ₛ F·n dS = a⁶π/12 = (1/64)π/12 = π/768.

To verify Gauss-Divergence theorem, take V as the upper half-ball bounded by S and the disk D: z = 0, x² + y² ≤ a². Now ∇·F = ∂(3y²z²)/∂x + ∂(4z²x²)/∂y + ∂(z²y²)/∂z = 2y²z. In spherical coordinates, ∭ᵥ 2y²z dV = ∫₀^a∫₀^π/2∫₀^2π 2(ρ² sin²φ sin²θ)(ρ cosφ)(ρ² sinφ) dθ dφ dρ = 2(a⁶/6)(1/4)(π) = a⁶π/12 = π/768. On D, z = 0, so F = 0 and ∬ᴅ F·n dS = 0. Therefore ∬ₛ F·n dS + ∬ᴅ F·n dS = π/768 + 0 = ∭ᵥ ∇·F dV. Hence Gauss-Divergence theorem is verified.