(a) Find the solution of the equation (D² + DD' - 2D'²)z = ysin x, where D ≡ (∂)/(∂ x) and D' ≡ (∂)/(∂ y). (10 marks) (b) Solve…

(a)

Let D = ∂/∂x and D' = ∂/∂y. The operator factors as D² + DD' - 2D'² = (D - D')(D + 2D') = (D + 2D')(D - D').

Complementary function:

• (D - D')z = 0 gives z = f(x + y), since dx/1 = dy/(-1), so x + y = constant.
• (D + 2D')z = 0 gives z = g(y - 2x), since dx/1 = dy/2, so y - 2x = constant.

Hence C.F. = f(x + y) + g(y - 2x).

For P.I., take z_p = -y sin x - cos x. Then z_px = -y cos x + sin x, z_py = -sin x, z_pxx = y sin x + cos x, z_pxy = -cos x, z_pyy = 0.

Therefore D²z_p + DD'z_p - 2D'²z_p = (y sin x + cos x) + (-cos x) - 0 = y sin x.

Thus the complete solution is z = f(x + y) + g(y - 2x) - y sin x - cos x, where f and g are arbitrary twice differentiable functions.

(b)

The system is 10x + 2y + z = 9 2x + 20y - 2z = -44 -2x + 3y + 10z = 22.

For Gauss-Seidel, rearrange as x = (9 - 2y - z)/10 y = (-22 - x + z)/10 z = (22 + 2x - 3y)/10.

The coefficient matrix is strictly diagonally dominant, so Gauss-Seidel converges. Start with x₀ = y₀ = z₀ = 0.

Iteration 1: x₁ = (9 - 0 - 0)/10 = 0.9 y₁ = (-22 - 0.9 + 0)/10 = -2.29 z₁ = (22 + 1.8 + 6.87)/10 = 3.067.

Iteration 2: x₂ = 1.0513 y₂ = -1.99843 z₂ = 3.009789.

Iteration 3: x₃ = 0.998707 y₃ = -1.998892 z₃ = 2.999409.

Iteration 4: x₄ = 0.999837 y₄ = -2.000043 z₄ = 2.999980.

Iteration 5: x₅ = 1.000011 y₅ = -2.000003 z₅ = 3.000003.

Hence the solution is x = 1, y = -2, z = 3.

(c)(i)

Convert (3479)₁₀ to binary by repeated division by 2: 3479 → remainders 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1 from least to most significant. Therefore (3479)₁₀ = (110110010111)₂.

Convert (7AE·9F)₁₆ to decimal: 7 × 16² + A × 16 + E + 9/16 + F/16² = 7 × 256 + 10 × 16 + 14 + 9/16 + 15/256 = 1792 + 160 + 14 + 159/256 = 1966 + 159/256.

Thus (7AE·9F)₁₆ = 1966.62109375.

(c)(ii)

For F(x,y,z) = (x + y + z')(x' + y'), the truth table is:

• (0,0,0) → 1
• (0,0,1) → 0
• (0,1,0) → 1
• (0,1,1) → 1
• (1,0,0) → 1
• (1,0,1) → 1
• (1,1,0) → 0
• (1,1,1) → 0

The minterms where F = 1 are m₀, m₂, m₃, m₄, m₅. Hence the full disjunctive normal form is F = x'y'z' + x'yz' + x'yz + xy'z' + xy'z. Equivalently, F = Σm(0,2,3,4,5). It simplifies to F = x'y + xy' + x'z'.

(d)

The cycloid is x = a(θ - sinθ), y = a(1 + cosθ).

Then dx/dt = a(1 - cosθ) dθ/dt, dy/dt = -a sinθ dθ/dt.

Kinetic energy: T = 1/2 m[(dx/dt)² + (dy/dt)²] = 1/2 m a²[(1 - cosθ)² + sin²θ](dθ/dt)² = 1/2 m a²[2 - 2 cosθ](dθ/dt)² = 2ma² sin²(θ/2)(dθ/dt)².

Potential energy, taking y upward: V = mgy = mga(1 + cosθ) = 2mga cos²(θ/2).

Hence the Lagrangian is L = 2ma² sin²(θ/2)(dθ/dt)² - 2mga cos²(θ/2).

Put u = cos(θ/2). Then sin²(θ/2) = 1 - u² and dθ/dt = -2(du/dt)/√(1 - u²), so (dθ/dt)² = 4(du/dt)²/(1 - u²).

Thus T = 8ma²(du/dt)², V = 2mga u², so L = 8ma²(du/dt)² - 2mga u².

Lagrange’s equation: d/dt(∂L/∂u̇) - ∂L/∂u = 0.

Here ∂L/∂u̇ = 16ma² du/dt, ∂L/∂u = -4mga u.

Therefore 16ma² d²u/dt² + 4mga u = 0, which gives d²u/dt² + g/(4a) u = 0. This is valid in the interior of the motion, where u ≠ ±1.

(e)

Let z = x + iy and c = a/2. Take a source at z = c and an equal sink at z = -c. In free space, W₀(z) = m log((z - c)/(z + c)).

By the Milne-Thomson circle theorem, for the circular boundary |z| = a, W(z) = W₀(z) + W̄₀(a²/z), where W̄₀ is W₀ with all constants conjugated.

Since c and m are real, the image term is m log((a²/z − c)/(a²/z + c)); dropping the constant, this simplifies to W(z) = m log[((z - c)(a² - cz))/((z + c)(a² + cz))].

Let U = (z - c)(a² - cz), V = (z + c)(a² + cz). Then the stream function is ψ = Im W = arg U - arg V.

For a fixed streamline, ψ = constant. Hence Im(U V̄) = tan ψ Re(U V̄).

With c = a/2, direct expansion gives Re(U V̄) = -(a²/4)[(r² - a²/4)(r² - 4a²) - 4a²y²], Im(U V̄) = -(5a³/4)y(r² - a²), where r² = x² + y².

Substituting into Im(U V̄) = tan ψ Re(U V̄), -(5a³/4)y(r² - a²) = tan ψ [-(a²/4){(r² - a²/4)(r² - 4a²) - 4a²y²}].

Simplifying, (r² - a²/4)(r² - 4a²) - 4a²y² = (5a cot ψ) y(r² - a²).

Let k = 5a cot ψ. Therefore the streamlines are (r² - a²/4)(r² - 4a²) - 4a²y² = ky(r² - a²), where k is a constant. The boundary and the x-axis occur as limiting streamlines.