(a) By eliminating the arbitrary functions f and g from z = f(x² - y) + g(x² + y), form partial differential equation. (10…

(a) Partial differential equation from z = f(x² − y) + g(x² + y)

Let u = x² − y and v = x² + y, so z = f(u) + g(v). Write p = ∂z/∂x, q = ∂z/∂y, r = ∂²z/∂x², t = ∂²z/∂y².

First derivatives: p = 2x[f′(u) + g′(v)], q = −f′(u) + g′(v).

Second derivatives: r = 2[f′(u) + g′(v)] + 4x²[f″(u) + g″(v)], t = f″(u) + g″(v).

From the first equation, f′ + g′ = p/(2x), so r = p/x + 4x² t.

Multiplying by x, the arbitrary functions are eliminated: x ∂²z/∂x² − ∂z/∂x − 4x³ ∂²z/∂y² = 0.

(b) Euler's method for dy/dx = (y² − x)/(y² + x), y(0) = 1

The method is yₙ₊₁ = yₙ + h f(xₙ, yₙ) with h = 0.1.

n = 0: x₀ = 0, y₀ = 1, f = 1/1 = 1. y₁ = 1 + 0.1(1) = 1.1000.

n = 1: x₁ = 0.1, y₁ = 1.1, y₁² = 1.21, f = (1.21 − 0.1)/(1.21 + 0.1) = 1.11/1.31 = 0.847328. y₂ = 1.1 + 0.0847328 = 1.184733.

n = 2: x₂ = 0.2, y₂² = 1.403592, f = (1.403592 − 0.2)/(1.403592 + 0.2) = 1.203592/1.603592 = 0.750560. y₃ = 1.184733 + 0.075056 = 1.259789.

n = 3: x₃ = 0.3, y₃² = 1.587068, f = (1.587068 − 0.3)/(1.587068 + 0.3) = 1.287068/1.887068 = 0.682046. y₄ = 1.259789 + 0.068205 = 1.327993.

So y(0.4) ≈ 1.3280, correct to 4 decimal places.

(c) Binary arithmetic in octal and hexadecimal

Part (i): convert each octal digit to three bits. (634.235)₈ = 110 011 100 . 010 011 101 (132.223)₈ = 001 011 010 . 010 010 011

Binary subtraction: 110011100.010011101 − 001011010.010010011 = 101000010.000001010

Regrouping in threes, 101 000 010 . 000 001 010 gives the answer 502.012 in base 8, written (502.012)₈.

Part (ii): convert each hexadecimal digit to four bits. (7AB.432)₁₆ = 0111 1010 1011 . 0100 0011 0010 (5CA.D61)₁₆ = 0101 1100 1010 . 1101 0110 0001

Binary subtraction: 0111 1010 1011 . 0100 0011 0010 − 0101 1100 1010 . 1101 0110 0001 = 0001 1110 0000 . 0110 1101 0001

Regrouping in fours, 0001 1110 0000 . 0110 1101 0001 gives the answer 1E0.6D1 in base 16, written (1E0.6D1)₁₆.

(d) Hamiltonian and Hamilton's equations of the planet

With T = ½m(ṙ² + r²θ̇²) and V = GMm(1/(2a) − 1/r), the Lagrangian is L = T − V. The generalised momenta are pᵣ = ∂L/∂ṙ = mṙ, p_θ = ∂L/∂θ̇ = mr²θ̇, so ṙ = pᵣ/m and θ̇ = p_θ/(mr²).

The Hamiltonian is H = pᵣṙ + p_θθ̇ − L = T + V, since T is quadratic in the velocities. Therefore H = pᵣ²/(2m) + p_θ²/(2mr²) + GMm(1/(2a) − 1/r).

Hamilton's equations: ṙ = ∂H/∂pᵣ = pᵣ/m, θ̇ = ∂H/∂p_θ = p_θ/(mr²), ṗᵣ = −∂H/∂r = p_θ²/(mr³) − GMm/r², ṗ_θ = −∂H/∂θ = 0.

Thus p_θ is constant, which is conservation of angular momentum. The constant GMm/(2a) does not affect the equations of motion.

(e) Streamlines of a source and two sinks

The complex potential of a source or sink at z₀ is proportional to log(z − z₀), with opposite signs for source and sink. Streamlines do not depend on the sign convention or on the overall constant. With the source of strength 2m at z = 2 and sinks of strength m at z = 2 ± i,

W = (m/2π)[2 log(z − 2) − log(z − 2 − i) − log(z − 2 + i)] = (m/2π) log[(z − 2)²/((z − 2)² + 1)],

because (z − 2 − i)(z − 2 + i) = (z − 2)² + 1. The total source strength 2m − m − m is zero, so there is no net flow at infinity.

Streamlines are the curves ψ = Im W = constant, that is arg[(z − 2)²/((z − 2)² + 1)] = constant.

In Cartesian form, with z = x + iy, 2 tan⁻¹(y/(x − 2)) − tan⁻¹((y − 1)/(x − 2)) − tan⁻¹((y + 1)/(x − 2)) = constant, with the inverse tangents taken as the arguments of x − 2 + iy, x − 2 + i(y − 1) and x − 2 + i(y + 1). The real axis y = 0 and the line x = 2 are among the streamlines, by symmetry.