(a) A solid sphere rests inside a fixed rough and hemispherical bowl of twice its radius. If a large amount of weight…

(a)

Let the sphere radius be r m and the bowl radius be 2r m. Let O be the bowl centre and let the line OG from O to the sphere centre G make angle θ rad with the downward vertical. Then G = (r sin θ, −r cos θ) m. For rolling without slipping inside a bowl of twice radius, the rotation angle of the sphere is −θ. The material point A initially at the highest point of the sphere has vector GA = (0, r) m. After rolling, GA = (r sin θ, r cos θ) m, so A = G + GA = (2r sin θ, 0) m. Thus the attached weight moves on a fixed horizontal line through O, so its potential energy is constant. Total potential energy: V(θ) = −Mgr cos θ + constant. Hence dV/dθ = Mgr sin θ, so θ = 0 is equilibrium. Also d²V/dθ² = Mgr cos θ, and at θ = 0 this equals Mgr > 0. By the potential-energy criterion, the equilibrium is stable for any attached weight. Final: stable.

(b)

Green’s theorem states ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy. Here P = xy + y², Q = x², and R is bounded by y = x² and y = x. They meet at x = 0 and x = 1. Positively orient C as y = x² from x = 0 to 1, then y = x from x = 1 to 0.

On y = x², dy = 2x dx: ∫_0^1 [(x³ + x⁴) + x²·2x] dx = ∫_0^1 (3x³ + x⁴) dx = 3/4 + 1/5 = 19/20.

On y = x, dy = dx, x goes from 1 to 0: ∫_1^0 [(x² + x²) + x²] dx = ∫_1^0 3x² dx = −1. Therefore ∮_C (P dx + Q dy) = 19/20 − 1 = −1/20.

Now ∂Q/∂x − ∂P/∂y = 2x − (x + 2y) = x − 2y. Thus ∬_R (x − 2y) dx dy = ∫_0^1 ∫_x²^x (x − 2y) dy dx = ∫_0^1 [xy − y²]_x²^x dx = ∫_0^1 (x⁴ − x³) dx = 1/5 − 1/4 = −1/20. Both sides agree. Final: Green’s theorem is verified; common value = −1/20.

(c)(i)

Let p = dy/dx and u = x + y. Then du/dx = 1 + p. The equation becomes (1 + p)³ = (27/(8a))u(1 − p)³. Put r = (1 + p)/(1 − p). Then r³ = 27u/(8a), so r = (3/2)(u/a)^(1/3). Also p = (r − 1)/(r + 1), hence du/dx = 1 + p = 2r/(r + 1). Thus dx/du = (r + 1)/(2r) = 1/2 + 1/(2r). Substituting r, dx/du = 1/2 + a^(1/3)/(3u^(1/3)). Integrating, x = u/2 + (a^(1/3)/2)u^(2/3) + C. So 2x = u + a^(1/3)u^(2/3) + 2C. Since u = x + y, x − y − 2C = a^(1/3)(x + y)^(2/3). Let c = 2C. General solution: (x − y − c)³ = a(x + y)².

For the singular solution, differentiate the family with respect to c: −3(x − y − c)² = 0 ⇒ x − y − c = 0. Using this in the family gives a(x + y)² = 0, hence x + y = 0. Singular solution: y = −x.

(c)(ii)

Let θ = x d/dx. Then θ³ y = x³ y''' + 3x² y'' + x y'. So the equation becomes (θ³ + 1)y = x log x.

Homogeneous part: for y = x^m, θy = m x^m, so m³ + 1 = 0. Thus m = −1 and m = (1 ± i√3)/2. Hence y_c = C1/x + √x [C2 cos((√3/2) log x) + C3 sin((√3/2) log x)].

For particular integral, put t = log x, so x = e^t and θ = d/dt. The equation becomes (D³ + 1)y = e^t t. Try y_p = e^t(At² + Bt + C). Then (D³ + 1)(e^t v) = e^t(v''' + 3v'' + 3v' + 2v). Thus v''' + 3v'' + 3v' + 2v = t. With v = At² + Bt + C, comparing coefficients gives A = 0, B = 1/2, C = −3/4. So y_p = x[(1/2) log x − 3/4]. Therefore the complete solution is y = C1/x + √x [C2 cos((√3/2) log x) + C3 sin((√3/2) log x)] + (x/2) log x − 3x/4, x > 0.