(a) A cable of weight w per unit length and length 2l hangs from two points P and Q in the same horizontal line. Show that the…

(a) Span of a tightly stretched cable

Let S be the span. For a uniform cable, the catenary is y = c cosh(x/c), where c = T₀/w. Put u = S/(2c). Then the length is 2l = 2c sinh u, so 2l = S(sinh u/u). The middle sag is h = c(cosh u − 1), so h = S(cosh u − 1)/(2u).

For a tightly stretched cable u is small. Use expansions: sinh u/u = 1 + u²/6 + … (cosh u − 1)/(2u) = u/4 + u³/48 + … Hence u ≈ 4h/S. Therefore 2l ≈ S[1 + (4h/S)²/6] = S + 8h²/(3S). Thus S + 8h²/(3S) = 2l, so S² − 2lS + 8h²/3 = 0. Taking the larger root, S = l + √(l² − 8h²/3) = l + l(1 − 4h²/(3l²) + …) ≈ 2l − 4h²/(3l) = 2l(1 − 2h²/(3l²)). This proves the required span formula for h ≪ l. The weight w cancels; all lengths are in the same unit.

(b) Variation of parameters

The reduced equation is (x² − 1)y″ − 2xy′ + 2y = 0. Given y₁ = x. Let y = xv. Then y′ = v + xv′, y″ = 2v′ + xv″. Substitution gives x(x² − 1)v″ − 2v′ = 0. Put p = v′. Then dp/p = 2 dx/[x(x² − 1)] = (−2/x + 1/(x−1) + 1/(x+1)) dx. Integrating, p = C(x² − 1)/x² = C(1 − 1/x²). So v = C(x + 1/x) + C₂, and hence a second independent solution is y₂ = x(x + 1/x) = x² + 1. Thus y_h = c₁x + c₂(x² + 1).

Write the equation in standard form: y″ − (2x/(x² − 1))y′ + (2/(x² − 1))y = x² − 1. Here R = x² − 1, y₁ = x, y₂ = x² + 1. The Wronskian is W = y₁y₂′ − y₁′y₂ = x(2x) − 1(x² + 1) = x² − 1. By variation of parameters, y_p = −y₁∫ y₂R/W dx + y₂∫ y₁R/W dx = −x∫(x² + 1)dx + (x² + 1)∫x dx = −x(x³/3 + x) + (x² + 1)(x²/2) = x⁴/6 − x²/2. Therefore y = c₁x + c₂(x² + 1) + x⁴/6 − x²/2, valid for x ≠ ±1.

(c) Verification of Green’s theorem

Green’s theorem states ∮_C (P dx + Q dy) = ∬_D (∂Q/∂x − ∂P/∂y)dA for positively oriented C. Here P = 3x² − 8y², Q = 4y − 6xy. So ∂Q/∂x = −6y, ∂P/∂y = −16y, and ∂Q/∂x − ∂P/∂y = 10y. The region D is the triangle with vertices (0,0), (1,0), (0,1). Hence ∬_D 10y dA = 10∫₀¹∫₀^(1−x) y dy dx = 5∫₀¹(1 − x)² dx = 5/3.

Now compute the line integral along C in the positive direction.

• C₁: y = 0, x from 0 to 1. ∫ 3x² dx = 1.

• C₂: x + y = 1 from (1,0) to (0,1). Let x = 1 − t, y = t, t from 0 to 1. Then dx = −dt, dy = dt, P = 3(1 − t)² − 8t² = 3 − 6t − 5t², Q = 4t − 6(1 − t)t = −2t + 6t². Thus ∫(P dx + Q dy) = ∫₀¹(Q − P)dt = ∫₀¹(4t + 11t² − 3)dt = 8/3.

• C₃: x = 0, y from 1 to 0. ∫ 4y dy = −2.

Therefore ∮_C (P dx + Q dy) = 1 + 8/3 − 2 = 5/3. This equals the double integral, so Green’s theorem is verified.