(a) Solve the partial differential equation
(2D² - 5DD' + 2D'²) z = 24 (y - x) + sin (y - x)
where D ≡ ∂/∂x , D' ≡ ∂/∂y , z = z(x, y). (15 marks)

(b) Evaluate the integral ∫₀⁰.8 f(x) dx,
where f(x) = 0.2 + 25x - 200x² + 675x³ - 900x⁴ + 400x⁵
by usin

Question

(a) Solve the partial differential equation
(2D² - 5DD' + 2D'²) z = 24 (y - x) + sin (y - x)
where D ≡ ∂/∂x , D' ≡ ∂/∂y , z = z(x, y). (15 marks)

(b) Evaluate the integral ∫₀⁰.8 f(x) dx,
where f(x) = 0.2 + 25x - 200x² + 675x³ - 900x⁴ + 400x⁵
by using single application of Simpson's 3/8 rule. (15 marks)

(c) A smooth uniform rod, say OA, of length 2a and mass m is pivoted at one end to a fixed point O. The rod is inclined at an angle θ with the downward vertical line OZ and the plane OAZ makes an angle φ with a fixed vertical plane. A bead of mass λm slides smoothly on the rod and is connected to O by a light elastic string of modulus nmg and natural length a. Show that the kinetic energy T of the system is given by 2T = (4/3) ma² (θ̇² + φ̇² sin θ) + λm (ẋ² + x²θ̇² + x²θ̇² sin² θ), where x is the stretched length of the string, and derive the equations of motion. (20 marks)

UPSC Answer Check · Accepted Answer

(a) Linear PDE with constant coefficients
Auxiliary equation (D = mD′): 2m² − 5m + 2 = 0 ⇒ m = 2, 1/2, so 2D² − 5DD′ + 2D′² = (D − 2D′)(2D − D′).
(D − 2D′)z = 0 gives z = φ₁(y + 2x); (2D − D′)z = 0 gives z = φ₂(2y + x).
CF = φ₁(y + 2x) + φ₂(2y + x).
PI for 24(y − x): try z = k(y − x)³. Then zₓₓ = 6k(y − x), z_xy = −6k(y − x), z_yy = 6k(y − x), so
F(D, D′)z = [2(6k) − 5(−6k) + 2(6k)](y − x) = 54k(y − x) = 24(y − x) ⇒ k = 4/9.
PI₁ = (4/9)(y − x)³.
PI for sin(y − x) (a = −1, b = 1): D² → −a² = −1, DD′ → −ab = 1, D′² → −b² = −1, so F = 2(−1) − 5(1) + 2(−1) = −9 and
PI₂ = −(1/9) sin(y − x).
z = φ₁(y + 2x) + φ₂(2y + x) + (4/9)(y − x)³ − (1/9) sin(y − x).

(b) Simpson's 3/8 rule
A single application uses n = 3 subintervals: h = 0.8/3 = 0.266667, nodes 0, 0.266667, 0.533333, 0.8.
f(0) = 0.2
f(0.266667) = 1.432724
f(0.533333) = 3.487177
f(0.8) = 0.232
I ≈ (3h/8)[f₀ + 3f₁ + 3f₂ + f₃] = 0.1 × [0.2 + 4.298173 + 10.461531 + 0.232] = 0.1 × 15.191704
I ≈ 1.5192.
(The exact value is 1.640533; the true error of about 7.4% arises because the 3/8 rule is exact only for polynomials of degree ≤ 3, while f is of degree 5.)

(c) Rod and bead: kinetic energy and equations of motion
Rod: moment of inertia about a perpendicular axis through O is I_O = (1/3)m(2a)² = (4/3)ma². The rod's angular velocity has components θ̇ (in the plane OAZ) and φ̇ sin θ (perpendicular to it, about the rod's normal), so
2T_rod = (4/3)ma²(θ̇² + φ̇² sin²θ).
Bead: at distance x from O along the rod, in spherical coordinates (x, θ, φ) its speed satisfies v² = ẋ² + x²θ̇² + x²φ̇² sin²θ, so
2T_bead = λm(ẋ² + x²θ̇² + x²φ̇² sin²θ).
Hence 2T = (4/3)ma²(θ̇² + φ̇² sin²θ) + λm(ẋ² + x²θ̇² + x²φ̇² sin²θ). (This is the printed expression with the sin²θ factors read as φ̇² sin²θ, which dimensional consistency requires.)
Potential energy (θ from the downward vertical; string stretched, x > a):
V = −mga cos θ − λmgx cos θ + nmg(x − a)²/(2a).
With L = T − V, Lagrange's equations give:
φ (cyclic): ∂L/∂φ̇ = [(4/3)ma² + λmx²] sin²θ φ̇ = constant (angular momentum about the vertical).
θ: d/dt{[(4/3)ma² + λmx²]θ̇} = [(4/3)ma² + λmx²] φ̇² sin θ cos θ − mg(a + λx) sin θ.
x: λm ẍ = λm x(θ̇² + φ̇² sin²θ) + λmg cos θ − nmg(x − a)/a.
(If the string becomes slack, x < a, the last term is dropped.)

(a) Solve the partial differential equation (2D² - 5DD' + 2D'²) z = 24 (y - x) + sin (y - x) where D ≡ ∂/∂x , D' ≡ ∂/∂y , z =…

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