Solve the differential equation: d²y/dx² + 2y = x²e^(3x) + e^x cos 2x (10 marks) Solve the initial value problem: d²y/dx² + 4y =…

(a) y″ + 2y = x²e³ˣ + eˣ cos 2x

Complementary function: m² + 2 = 0, so y_c = C₁cos(√2 x) + C₂sin(√2 x).

For x²e³ˣ, try y₁ = e³ˣ(ax² + bx + c). Since (D² + 2)(e³ˣu) = e³ˣ(u″ + 6u′ + 11u), comparing coefficients gives 11a = 1, 12a + 11b = 0, 2a + 6b + 11c = 0, so a = 1/11, b = −12/121, c = 50/1331.

For eˣ cos 2x, take the real part of e^((1+2i)x): (1 + 2i)² + 2 = −1 + 4i, so y₂ = Re[e^((1+2i)x)/(−1 + 4i)] = eˣ(4 sin 2x − cos 2x)/17.

General solution:

y = C₁cos(√2 x) + C₂sin(√2 x) + e³ˣ(x²/11 − 12x/121 + 50/1331) + eˣ(4 sin 2x − cos 2x)/17.

(b) y″ + 4y = e⁻²ˣ sin 2x, y(0) = y′(0) = 0, by Laplace transform

Taking transforms: (s² + 4)Y = 2/((s + 2)² + 4) = 2/(s² + 4s + 8).

Y = 2/[(s² + 4)(s² + 4s + 8)] = (1 − s)/(10(s² + 4)) + (s + 3)/(10(s² + 4s + 8)).

Inverting, the first term gives (1/20)sin 2x − (1/10)cos 2x. For the second, (s + 3) = (s + 2) + 1 over (s + 2)² + 4 gives e⁻²ˣ[(1/10)cos 2x + (1/20)sin 2x].

y = −(1/10)cos 2x + (1/20)sin 2x + e⁻²ˣ[(1/10)cos 2x + (1/20)sin 2x].

(c) Two rods hung from L

Let LM = a and MN = b. Since (LM)² + (MN)² = (LN)², angle LMN is a right angle. Let LM make angle θ with the downward vertical through L, with MN turned back towards that vertical.

Taking L as origin with x horizontal, M = (a sin θ, a cos θ). The centre of gravity of LM is at x = (a/2) sin θ, and that of MN at x = a sin θ − (b/2)cos θ. Weights are proportional to a and b.

In equilibrium the combined centre of gravity lies on the vertical through L, so the moment about it vanishes:

a·(a/2)sin θ + b(a sin θ − (b/2)cos θ) = 0, giving (a²/2 + ab)sin θ = (b²/2)cos θ.

tan θ = b²/(a² + 2ab), that is θ = tan⁻¹[b²/(a(a + 2b))].

(d) Planet stopped in a circular orbit

Let R be the orbit radius and μ = GM. After stopping, the planet falls radially from rest at r = R. Energy conservation gives ½(dr/dt)² = μ(1/r − 1/R), so

t = ∫₀ᴿ dr/√(2μ(1/r − 1/R)) = √(R/2μ) ∫₀ᴿ √(r/(R − r)) dr.

With r = R sin²φ, the integral equals πR/2. So the time of fall is

t = (π/(2√2))√(R³/μ).

The orbital period is T = 2π√(R³/μ). Hence t/T = 1/(4√2), about 0.177.

(e) ∇²[∇·(r⃗/r)]

With r = |r⃗|: ∇·(r⃗/r) = 3/r + r⃗·∇(1/r) = 3/r − 1/r = 2/r.

∇²(2/r) = 0 for r ≠ 0, since 1/r is harmonic. So the expression as printed equals 0, not 2/r⁴; the result 2/r⁴ holds for ∇²[∇·(r⃗/r²)]: here ∇·(r⃗/r²) = 3/r² − 2/r² = 1/r², and ∇²(1/r²) = d²/dr²(r⁻²) + (2/r)d/dr(r⁻²) = 6/r⁴ − 4/r⁴ = 2/r⁴. The printed question probably omits the square.