(a) Solve the differential equation (x + 2)(d^2y)/(dx²) - (2x + 5)dy/dx + 2y = (1 + x) e^x by the method of variation of…

(a)

The homogeneous equation is (x+2)y'' - (2x+5)y' + 2y = 0. It is satisfied by y1 = e^(2x) and y2 = 2x+5. Their Wronskian is W = e^(2x)·2 - 2e^(2x)(2x+5) = -4(x+2)e^(2x).

Let y = u e^(2x) + v(2x+5). By variation of parameters, u' e^(2x) + v'(2x+5) = 0, 2u' e^(2x) + 2v' = (1+x)e^x/(x+2).

Solving these, u' = (1+x)(2x+5)e^(-x)/[4(x+2)^2], v' = -(1+x)e^x/[4(x+2)^2].

A convenient integration gives u = -(2x+3)e^(-x)/[4(x+2)], v = -e^x/[4(x+2)]. Differentiating these gives exactly the above u', v'. Hence y_p = u e^(2x) + v(2x+5) = -(2x+3)e^x/[4(x+2)] - (2x+5)e^x/[4(x+2)] = -e^x.

Therefore the general solution, valid for x ≠ -2, is y = C1 e^(2x) + C2(2x+5) - e^x.

(b)

∇·F = ∂/∂x(x^2-yz) + ∂/∂y(y^2-zx) + ∂/∂z(z^2-xy) = 2x + 2y + 2z.

Thus ∭_V ∇·F dV = ∫_0^a ∫_0^b ∫_0^c 2(x+y+z) dz dy dx = abc(a+b+c).

Now compute the surface flux over the six faces. x = 0: n = -i, F·n = yz. Flux = (b^2/2)(c^2/2) = b^2c^2/4. x = a: n = i, F·n = a^2 - yz. Flux = a^2bc - b^2c^2/4. Sum over x-faces = a^2bc.

y = 0: n = -j, F·n = zx. Flux = a^2c^2/4. y = b: n = j, F·n = b^2 - zx. Flux = ab^2c - a^2c^2/4. Sum over y-faces = ab^2c.

z = 0: n = -k, F·n = xy. Flux = a^2b^2/4. z = c: n = k, F·n = c^2 - xy. Flux = abc^2 - a^2b^2/4. Sum over z-faces = abc^2.

Therefore ∬_S F·n dS = a^2bc + ab^2c + abc^2 = abc(a+b+c). This equals the volume integral, so Gauss’s divergence theorem is verified.

(c)

Let a be the radius and let θ be measured from the lowest point. The height above the lowest point is h = a(1 - cos θ). By conservation of energy, v^2 = u^2 - 2ga(1 - cos θ).

The radial equation, with N as the inward normal reaction, is m v^2/a = N - mg cos θ. Hence N = m(v^2/a + g cos θ) = m[(u^2 - 2ag)/a + 3g cos θ].

(i) If u^2 ≤ 2ag, then at the extreme position v = 0, so u^2 = 2ga(1 - cos θ). Since u^2 ≤ 2ag, this gives cos θ ≥ 0, i.e. θ ≤ π/2. Thus the particle cannot cross the horizontal diameter; it oscillates about the lowest point in the lower half.

(ii) For complete circular motion, contact must be maintained throughout. The minimum reaction occurs at the highest point θ = π: N_top = m[(u^2 - 2ag)/a - 3g] = m(u^2 - 5ag)/a. For N_top ≥ 0, we require u^2 ≥ 5ag. Then N ≥ 0 everywhere, so the particle executes complete circular motion.

(iii) If 2ag < u^2 < 5ag, then at θ = π/2, N > 0, while at θ = π, N < 0. Hence N becomes zero for some θ0 in (π/2, π). Setting N = 0, (u^2 - 2ag)/a + 3g cos θ0 = 0, so cos θ0 = -(u^2 - 2ag)/(3ag).

At this point the particle leaves the curve tangentially. If α is the acute angle made by the tangent with the horizontal, then cos α = |cos θ0| = (u^2 - 2ag)/(3ag). Thus the particle leaves in a tangential direction making the stated angle α with the horizontal.