(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 the Dirac delta function. (15 marks)

(b) Using Gauss divergence theorem, evaluate the integral ∬_S (y²î + xz³ĵ + (z-1)²k̂)

Question

(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 the Dirac delta function. (15 marks)

(b) Using Gauss divergence theorem, evaluate the integral ∬_S (y²î + xz³ĵ + (z-1)²k̂) · n̂ dS over the region bounded by the cylinder x² + y² = 16 and the planes z = 1 and z = 5. (15 marks)

(c) A particle moves with a central acceleration μ(3/r³ + d²/r⁵) being projected from a distance d at an angle 45° with a velocity equal to that in a circle at the same distance. Prove that the time it takes to reach the centre of force is d²/√(2μ) (2 - π/2). (20 marks)

UPSC Answer Check · Accepted Answer

(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).

(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…

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