(a) Using Laplace transform, solve the initial value problem y'' + 2y' + 5y = δ(t-2), y(0) = 0, y'(0) = 0 where δ(t-2) denotes…

(a) Let Y(s) = L{y(t)}. By the Laplace transform of derivatives and the zero initial conditions y(0)=y′(0)=0, L{y″} = s²Y, L{y′} = sY. Also L{δ(t−2)} = exp(−2s). Hence s²Y + 2sY + 5Y = exp(−2s), so Y = exp(−2s)/(s² + 2s + 5) = exp(−2s)/((s+1)² + 4). Now the standard inverse Laplace transform gives L⁻¹{1/((s+1)² + 4)} = (1/2) exp(−t) sin 2t. Using the second shifting theorem, L⁻¹{exp(−2s)F(s)} = f(t−2)u(t−2), where u is the unit step function. Therefore y(t) = (1/2) exp(−(t−2)) sin(2(t−2)) u(t−2), that is, y(t)=0 for t<2 and y(t)=(1/2) exp(−(t−2)) sin(2(t−2)) for t≥2.

(b) Let F = y²î + xz³ĵ + (z−1)²k̂. By the Gauss divergence theorem, for the outward normal n̂, ∬ F·n̂ dS = ∭ div F dV, where V is the solid cylinder x² + y² ≤ 16, 1 ≤ z ≤ 5. Now div F = ∂(y²)/∂x + ∂(xz³)/∂y + ∂((z−1)²)/∂z = 0 + 0 + 2(z−1) = 2z − 2. Use cylindrical coordinates x = r cosθ, y = r sinθ, z = z, with dV = r dr dθ dz. The limits are 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π, 1 ≤ z ≤ 5. Hence ∭ (2z−2) dV = (∫ dθ from 0 to 2π)(∫ r dr from 0 to 4)(∫ (2z−2) dz from 1 to 5) = (2π)(8)(16) = 256π. Thus the required surface integral is 256π for outward normal. For inward normal, the sign is −256π.

(c) Let r be the radial distance and θ the angular coordinate. For a central acceleration μ(3/r³ + d²/r⁵) directed toward the centre, the equations of motion in polar form are r″ − r(θ′)² = −μ(3/r³ + d²/r⁵), r²θ′ = h, where h is the constant angular momentum per unit mass.

At distance r = d, the circular speed v_c satisfies v_c²/d = μ(3/d³ + d²/d⁵) = μ(3/d³ + 1/d³) = 4μ/d³. Hence v_c = 2√μ/d. Since the particle is projected at 45°, the angular momentum is h = d v_c sin45° = d(2√μ/d)(1/√2) = √(2μ). Thus h² = 2μ. The initial radial speed is v_r = v_c cos45° = √(2μ)/d, directed inward.

Substituting h² = 2μ in the radial equation gives r″ = h²/r³ − μ(3/r³ + d²/r⁵) = 2μ/r³ − 3μ/r³ − μd²/r⁵ = −μ/r³ − μd²/r⁵. Multiply by r′ and integrate: (1/2)(r′)² = μ/(2r²) + μd²/(4r⁴) + C. At r = d, (r′)² = 2μ/d², so μ/d² = μ/(2d²) + μ/(4d²) + C, which gives C = μ/(4d²). Therefore (r′)² = μ/r² + μd²/(2r⁴) + μ/(2d²). Put u = r/d. Then (r′)² = μ/(2d²)(1 + 1/u²)². Since the motion is inward,

|r′| = √μ/(d√2)(1 + 1/u²). The time T to reach the centre r = 0 from r = d is T = ∫ (r from 0 to d) dr/|r′| = d∫ (u from 0 to 1) du/[√μ/(d√2)(1 + 1/u²)] = d²√2/√μ ∫ (u from 0 to 1) u²/(1 + u²) du. Now ∫ u²/(1 + u²) du = ∫ (1 − 1/(1 + u²)) du = u − arctan u. Thus ∫ from 0 to 1 u²/(1 + u²) du = 1 − π/4. Therefore T = d²√2/√μ (1 − π/4) = d²/√(2μ)(2 − π/2). Hence the time taken to reach the centre of force is d²/√(2μ) (2 − π/2).