(a) Solve x² d²y/dx² − 3x dy/dx + y = (sin(log x) + 1)/x log x. (15 marks) (b) (i) A battleship is steaming ahead with velocity…

(a) Cauchy–Euler equation The right-hand side is read as [(sin(log x) + 1)/x]·log x. Put x = e^z (z = log x) and D = d/dz, so x d/dx = D and x² d²/dx² = D(D − 1). The equation becomes [D(D − 1) − 3D + 1]y = (D² − 4D + 1)y = e^−z(z sin z + z). CF: m² − 4m + 1 = 0 ⇒ m = 2 ± √3, so CF = c₁x²+√3 + c₂x²−√3. PI: using 1/f(D)[e^azg(z)] = e^az·1/f(D + a) g(z) with a = −1, f(D − 1) = (D − 1)² − 4(D − 1) + 1 = D² − 6D + 6. PI = e^−z·1/(D² − 6D + 6) (z + z sin z). Term z: 1/(6(1 − D + D²/6)) z = (1/6)(1 + D + …)z = (z + 1)/6. Term z sin z = Im(z e^iz): 1/(D² − 6D + 6)(z e^iz) = e^iz·1/(D² + (2i − 6)D + (5 − 6i)) z = e^iz·[1/(5 − 6i)]·[z − (2i − 6)/(5 − 6i)] = e^iz[(5 + 6i)z/61 + (54 + 382i)/3721], since 1/(5 − 6i) = (5 + 6i)/61 and (2i − 6)(5 + 6i)/61 = −(42 + 26i)/61. Imaginary part: (5z/61 + 54/3721) sin z + (6z/61 + 382/3721) cos z. Returning to x (e^−z = 1/x, z = log x): y = c₁x²+√3 + c₂x²−√3 + (log x + 1)/(6x) + (1/x)[(5 log x/61 + 54/3721) sin(log x) + (6 log x/61 + 382/3721) cos(log x)].

(b)(i) Range of a gun firing backwards from a moving ship Relative to the ground, the shot has horizontal velocity v cos α − u (backwards; the ship moves forward with u) and vertical velocity v sin α (u is horizontal and does not affect it). Time of flight T = 2v sin α/g. Range R = (v cos α − u)·T = (2v/g) sin α (v cos α − u). For maximum range: dR/dα = (2v/g)[cos α(v cos α − u) − v sin²α] = (2v/g)(v cos 2α − u cos α) = 0 ⇒ v(2cos²α − 1) = u cos α ⇒ 2v cos²α − u cos α − v = 0 ⇒ cos α = [u ± √(u² + 8v²)]/(4v). The minus sign gives cos α < 0 (obtuse elevation), so it is rejected. At this root dR/dα changes from positive to negative, so R is a maximum. α = cos⁻¹[(u + √(u² + 8v²))/(4v)] (α = 45° when u = 0, as a check).

(b)(ii) Chain partly on a rough table Let w be the weight per unit length, x the length on the table, so the hanging part has length s = l − x. The hanging part is a catenary whose tangent is horizontal where the chain meets the table, so that point is the vertex; there the tension is horizontal, T₀ = wc (c the parameter). The part on the table is straight, with normal reaction N = wx. On the point of slipping, T₀ = μN ⇒ wc = μwx ⇒ c = μx. Measured from the directrix, the vertex has ordinate c and the fixed end has ordinate y = c + h. For the catenary, s² = y² − c²: (l − x)² = (c + h)² − c² = 2ch + h² = 2μxh + h². ⇒ x² − 2(l + μh)x + (l² − h²) = 0 ⇒ x = (l + μh) ± √[(l + μh)² − l² + h²] = (l + μh) ± √[(μ² + 1)h² + 2μlh]. The plus sign gives x > l, which is impossible. Hence length on the table = (l + μh) − √[(μ² + 1)h² + 2μlh].

(c) Frenet–Serret quantities for y² = 4ax Parametrise: r = at² î + 2at ĵ (z = 0). dr/dt = 2at î + 2a ĵ, ds/dt = |dr/dt| = 2a√(1 + t²). T̂ = (dr/dt)/(ds/dt) = (t î + ĵ)/√(1 + t²). dT̂/dt = (î − t ĵ)/(1 + t²)³/2, so dT̂/ds = (dT̂/dt)/(ds/dt) = (î − t ĵ)/[2a(1 + t²)²]. κ = |dT̂/ds| = 1/[2a(1 + t²)³/2] (radius of curvature ρ = 2a(1 + t²)³/2 = 2(a + x)³/2/√a; ρ = 2a at the vertex). N̂ = (1/κ) dT̂/ds = (î − t ĵ)/√(1 + t²). B̂ = T̂ × N̂ = [(t î + ĵ) × (î − t ĵ)]/(1 + t²) = (−t² − 1)k̂/(1 + t²) = −k̂. B̂ is constant, so dB̂/ds = 0; from dB̂/ds = −τN̂, τ = 0 (as expected for a plane curve).