(a) Obtain the solution of the initial-value problem dy/dx - 2xy = 2, y(0) = 1 in the form y = eˣ²[1 + √π erf(x)]. (10…

(a) Initial-value problem

The equation is linear: dy/dx - 2xy = 2. Integrating factor: μ = e^(∫ -2x dx) = e^(-x²).

Multiply the equation by e^(-x²): d/dx [y e^(-x²)] = 2 e^(-x²).

Integrate: y e^(-x²) = ∫ 2 e^(-x²) dx + C.

Since erf(x) = (2/√π) ∫₀ˣ e^(-t²) dt, we have ∫ 2 e^(-x²) dx = √π erf(x) + C.

Thus y e^(-x²) = √π erf(x) + C, so y = e^(x²)[√π erf(x) + C].

Using y(0)=1 and erf(0)=0: 1 = e⁰[0 + C] = C. Hence C = 1.

Final answer: y = e^(x²)[1 + √π erf(x)].

(b) Integral of f(t)/t and its value

Given L{f(t); p} = F(p), where F(p) = ∫₀^∞ e^(-pt) f(t) dt.

Using the standard Laplace-transform property for division by t: L{f(t)/t; p} = ∫_p^∞ F(x) dx.

Proof: ∫_p^∞ F(x) dx = ∫_p^∞ ∫₀^∞ e^(-xt) f(t) dt dx. Interchanging the order of integration, = ∫₀^∞ f(t) [∫_p^∞ e^(-xt) dx] dt = ∫₀^∞ f(t) e^(-pt)/t dt = L{f(t)/t; p}.

Putting p = 0 gives ∫₀^∞ f(t)/t dt = ∫₀^∞ F(x) dx.

Now take f(t) = e^(-t) - e^(-3t). Then F(x) = 1/(x+1) - 1/(x+3).

Therefore, ∫₀^∞ (e^(-t) - e^(-3t))/t dt = ∫₀^∞ [1/(x+1) - 1/(x+3)] dx = [ln((x+1)/(x+3))]₀^∞ = 0 - ln(1/3) = ln 3.

Final answer: ln 3.

(c) Beam resting on a wall and a cylinder

Let the vertical wall be x = 0. In cross-section, the cylinder is a circle of radius a with centre C = (a, 0). Let the beam touch the wall at A and the cylinder at B. Take the beam sloping downward from the wall to the cylinder. Let B = (l/√2, y). Then A = (0, y + l/√2).

Let r = a/l. Since friction is neglected, the reaction at A is horizontal, and the reaction at B is along the radius CB. The weight W acts vertically through the midpoint M.

For equilibrium of three forces, their lines of action must be concurrent. The weight acts through M = (l/(2√2), y + l/(2√2)). The wall reaction is horizontal through A, so the intersection point is Q = (l/(2√2), y + l/√2).

Since Q lies on CB, vectors BQ and CB are collinear: (B - Q) × (C - B) = 0.

Here B - Q = (l/(2√2), -l/√2), C - B = (a - l/√2, -y).

Thus (l/(2√2))(-y) - (-l/√2)(a - l/√2) = 0, which simplifies to y = 2a - √2 l.

Also B lies on the circle: (l/√2 - a)² + y² = a².

Substitute y = 2a - √2 l. Dividing throughout by l² and writing r = a/l, (1/√2 - r)² + (2r - √2)² = r².

Let d = r - 1/√2. Then d² + 4d² = r², so 5d² = r².

Since r > 1/√2, we take √5(r - 1/√2) = r. Hence r(√5 - 1) = √5/√2, so r = √5/[√2(√5 - 1)] = (5 + √5)/(4√2).

Therefore, a/l = (√5 + 5)/(4√2).

Now let R_c be the cylinder reaction. Its direction is along C → B. The unit normal is (-î + 2ĵ)/√5. Let R_c = k(-î + 2ĵ)/√5.

Vertical equilibrium: 2k/√5 = W, so k = W√5/2. Thus the cylinder reaction has magnitude R_c = W√5/2, with horizontal component W/2 toward the wall and vertical component W upward.

Horizontal equilibrium gives the wall reaction R_w: R_w - W/2 = 0, so R_w = W/2.

Final answers: cylinder reaction = W√5/2, directed along (-î + 2ĵ)/√5; wall reaction = W/2, directed away from the wall.

(d) Return time in simple harmonic motion

Let the SHM be x = A sin(ωt + α), where ω = 2π/T.

At t = 0, the particle is at P, so x = p and velocity is v along OP. Hence p = A sin α, v = Aω cos α. Therefore tan α = pω/v = 2πp/(vT).

The particle next returns to P with velocity reversed when the phase becomes π - α. Hence the time elapsed is τ = (π - 2α)/ω = T/2 - (T/π) tan⁻¹(2πp/(vT)).

If τ = T/2, then T/2 = T/2 - (T/π) tan⁻¹(2πp/(vT)), so tan⁻¹(2πp/(vT)) = 0. Thus 2πp/(vT) = 0, giving p = 0.

Final answers: time = T/2 - (T/π) tan⁻¹(2πp/(vT)); when time = T/2, p = 0.

(e) Derivative of the vector function

Use the vector triple product identity: ā × (b̄ × c̄) = (ā·c̄)b̄ - (ā·b̄)c̄.

Given ā = sinθ î + cosθ ĵ + θ k̂, b̄ = cosθ î - sinθ ĵ - 3k̂, c̄ = 2î + 3ĵ - 3k̂.

Compute: ā·b̄ = sinθ cosθ - cosθ sinθ - 3θ = -3θ, ā·c̄ = 2 sinθ + 3 cosθ - 3θ.

Therefore, V(θ) = ā × (b̄ × c̄) = (2 sinθ + 3 cosθ - 3θ)b̄ + 3θ c̄.

Differentiate with respect to θ: V'(θ) = (2 cosθ - 3 sinθ - 3)b̄ + (2 sinθ + 3 cosθ - 3θ)(-sinθ î - cosθ ĵ) + 3(2î + 3ĵ - 3k̂).

At θ = π/2: sinθ = 1, cosθ = 0, b̄ = -ĵ - 3k̂. Thus V'(π/2) = -6(-ĵ - 3k̂) + (2 - 3π/2)(-î) + 6î + 9ĵ - 9k̂ = (4 + 3π/2)î + 15ĵ + 9k̂.

At θ = π: sinθ = 0, cosθ = -1, b̄ = -î - 3k̂. Thus V'(π) = -5(-î - 3k̂) + (-3 - 3π)ĵ + 6î + 9ĵ - 9k̂ = 11î + (6 - 3π)ĵ + 6k̂.

Final answers:

• At θ = π/2: V' = (4 + 3π/2)î + 15ĵ + 9k̂
• At θ = π: V' = 11î + (6 - 3π)ĵ + 6k̂