(a) (i) Find the general and singular solutions of the differential equation : (x² - a²)p² - 2xyp + y² + a² = 0, where p = dy/dx…

(a)(i) General and singular solutions

Write p = dy/dx. The equation is

(x² − a²)p² − 2xy p + y² + a² = 0.

Completing a square gives

(xp − y)² = a²(p² − 1),

so y = xp ± a√(p² − 1).

This is a Clairaut-type equation. Taking p = c constant gives the general solution

y = cx ± a√(c² − 1),

or equivalently

(y − cx)² = a²(c² − 1).

This is the family of straight lines. For real solutions, |c| ≥ 1.

For the singular solution, use the envelope condition. Let

F(x,y,c) = (y − cx)² − a²(c² − 1) = 0.

Then

∂F/∂c = −2x(y − cx) − 2a²c = 0,

so

c(x² − a²) = xy.

Eliminating c from this equation and F = 0 gives

y² = x² − a².

Hence the singular solution is

y² = x² − a², with |x| ≥ |a| for real curves.

Geometrically, the singular curve y² = x² − a² is the envelope of the family of straight lines (y − cx)² = a²(c² − 1); every line of the general solution is tangent to this hyperbola.

(a)(ii) Cauchy–Legendre linear equation

Put t = 3x + 2. Then

d/dx = 3 d/dt, d²/dx² = 9 d²/dt²,

and x = (t − 2)/3. The equation becomes

9t² d²y/dt² + 15t dy/dt − 3y = x² + x + 1.

Now

x² + x + 1 = ((t − 2)/3)² + (t − 2)/3 + 1 = (t² − t + 7)/9.

Thus

9t²y″ + 15t y′ − 3y = (t² − t + 7)/9.

This is a Cauchy–Euler equation. For the homogeneous part, put y = tᵐ. Then

9m(m − 1) + 15m − 3 = 0

⇒ 3m² + 2m − 1 = 0

⇒ m = 1/3, −1.

Hence

y_h = C₁t^(1/3) + C₂t⁻¹.

For a particular integral, try

y_p = At² + Bt + C.

Then

y_p′ = 2At + B, y_p″ = 2A.

Substitution gives

9t²(2A) + 15t(2At + B) − 3(At² + Bt + C) = 45A t² + 12B t − 3C.

Comparing with (t² − t + 7)/9,

45A = 1/9 ⇒ A = 1/405,

12B = −1/9 ⇒ B = −1/108,

−3C = 7/9 ⇒ C = −7/27.

Therefore

y_p = t²/405 − t/108 − 7/27.

Putting t = 3x + 2, the general solution is

y = C₁(3x + 2)^(1/3) + C₂/(3x + 2) + (3x + 2)²/405 − (3x + 2)/108 − 7/27.

The solution is valid on intervals not containing x = −2/3.

(b) Inclinations of a chain of rods

Let each rod have length l and weight W. Number the rods from the top as i = 1, 2, …, n. Let θᵢ be the inclination of the i-th rod to the downward vertical, positive in the direction of the horizontal force P at the lower end Aₙ₊₁.

Use the principle of virtual work. If x is the horizontal displacement of the lower end and y_mid is the vertical downward displacement of the midpoint of a rod, then for equilibrium

P δx + W Σ δy_mid = 0.

For the i-th rod,

∂x/∂θᵢ = l cos θᵢ.

Also, the rods below the i-th rod contribute their weights through the joint, and the midpoint of the i-th rod contributes half its weight. Hence

∂(Σ y_mid)/∂θᵢ = −l(n − i + 1/2) sin θᵢ.

Thus the equilibrium condition is

P l cos θᵢ − W l(n − i + 1/2) sin θᵢ = 0.

Therefore

tan θᵢ = P/[W(n − i + 1/2)].

Equivalently, if the r-th rod is counted from the bottom, r = 1, 2, …, n, then

tan θᵣ = P/[W(r − 1/2)] = 2P/[(2r − 1)W].

So the inclinations are

θᵢ = arctan[2P/((2(n − i + 1) − 1)W)], i = 1, 2, …, n,

where θᵢ is measured from the downward vertical toward the direction of P. The bottom rod is most inclined and the top rod least inclined.

(c) Flux by Gauss’ divergence theorem

By Gauss’ divergence theorem,

∬_S F·n dS = ∭_V (∇·F) dV,

where V is the solid cylinder x² + y² ≤ 4, 0 ≤ z ≤ 1, and n is the outward normal to the closed surface S.

Now

F = x i − y j + (z² − 1) k.

Hence

∇·F = ∂x/∂x + ∂(−y)/∂y + ∂(z² − 1)/∂z = 1 − 1 + 2z = 2z.

Therefore

∬_S F·n dS = ∭_V 2z dV.

Using cylindrical coordinates,

∭_V 2z dV = 2 ∫₀¹ z dz ∬_(x²+y²≤4) dx dy.

The area of the disk x² + y² ≤ 4 is π(2²) = 4π, and

∫₀¹ z dz = 1/2.

Thus

∬_S F·n dS = 2 · (1/2) · 4π = 4π.

Final answer: 4π.