(a)(i) Find the solution of the differential equation : dy/dx=-(2xy³+2)/(3x²y²+8e^(4y)) 10 (a)(ii) Reduce the equation…

(a)(i) Exact differential equation

Write the equation as (3x²y²+8e^(4y))dy+(2xy³+2)dx=0. Let M=2xy³+2, N=3x²y²+8e^(4y). Then ∂M/∂y=6xy²=∂N/∂x, so the equation is exact. Hence F_x=M and F_y=N. Integrating F_x with respect to x, F=x²y³+2x+g(y). Then F_y=3x²y²+g'(y)=N, so g'(y)=8e^(4y) and g(y)=2e^(4y). Thus x²y³+2x+2e^(4y)=C.

(a)(ii) Reduction to Clairaut's form, solution and singular solution

Let p = dy/dx. With u = y and v = xy, dv/du = (y + xp)/p. Call this P. Then y + xp = Pp, so xp + y = Pp.

The given equation can be written as x²p² + 2xyp + y²p + y² = (xp + y)² + y²p = 0. Substituting xp + y = Pp and y = u, P²p² + u²p = 0, so p(P²p + u²) = 0. For p ≠ 0, P²p = −u², i.e. p = −u²/P².

Also xp = Pp − u gives x = P − u/p, and with x = v/u, v/u = P − u/p = P + P²/u, so v = uP + P², which is Clairaut's form in (u, v) with slope P.

Solution: the general solution of Clairaut's equation is P = a (constant), so v = au + a², that is xy = ay + a², or y = a²/(x − a).

Singular solution: differentiate v = uP + P² with respect to u: P = P + (u + 2P) dP/du, so (u + 2P) dP/du = 0. The factor dP/du = 0 gives the general solution; the factor u + 2P = 0 gives P = −u/2, hence v = −u²/2 + u²/4 = −u²/4, that is xy = −y²/4, so y = 0 or y + 4x = 0.

The line y + 4x = 0 is the envelope of the family xy = ay + a², and it contains no arbitrary constant, so it is a singular solution. Direct check with y = −4x, p = −4: x²(16) + (−4x)(2x − 4x)(−4) + 16x² = 16x² − 32x² + 16x² = 0.

(b) Hemisphere supported by a string against a smooth wall

Let R be the radius (m) and W the weight (N); all coordinates are in metres. Work in the vertical plane through the centre O, the contact point C with the wall, and the string. Since the curved surface touches the smooth vertical wall at C, OC is horizontal. Take C=(0,0), the wall x=0, x horizontal away from the wall, z upward; then O=(R,0). Let A be the attached rim point in this plane on the wall side. If the base diameter makes angle φ with the vertical, A=(R(1−sinφ), Rcosφ). For a solid hemisphere, the centre of mass G is on the normal to the base at OG=3R/8 towards the curved surface. Since C lies on the curved surface, the horizontal component of OG towards the wall is (3R/8)cosφ, so G_x=R−(3R/8)cosφ. The forces are W downward at G, the smooth-wall reaction H (N) horizontal at C, and tension T (N) along the string. Vertical equilibrium gives Tcosθ=W. Taking moments about C, H has no moment. Equating the moment of W to that of T, Tcosθ A_x+Tsinθ A_z=WG_x. Using Tcosθ=W, A_x+A_z tanθ=G_x. Substitute the coordinates: R(1−sinφ)+Rcosφ tanθ=R−(3R/8)cosφ. Cancel R and divide by cosφ: tanθ=tanφ−3/8. Therefore tanφ−tanθ=3/8. The result is dimensionless.

(c) Curve whose tangent makes a constant angle with a fixed line

Let s be arc length, T the unit tangent, N the principal normal, B the binormal, κ curvature and τ torsion (both m⁻¹). Prime denotes d/ds. The Frenet-Serret formulas are T'=κN, N'=−κT+τB, B'=−τN. The radius of curvature is ρ=1/κ (m) and the radius of torsion is σ=1/τ (m), so σ/ρ=κ/τ. Suppose T makes a constant angle θ with a fixed unit vector a. Then T·a=cosθ. Differentiating, κN·a=0, so N·a=0. Differentiating N·a=0, (−κT+τB)·a=0, so τB·a=κcosθ. Since a is a unit vector orthogonal to N, a=cosθT+(a·B)B, and (a·B)²=1−cos²θ=sin²θ. Thus a·B=±sinθ and κ/τ=±tanθ. Hence σ/ρ=±tanθ, i.e. the ratio is proportional to tanθ. Conversely, suppose σ/ρ=κ/τ=c is constant. Define a=(T+cB)/√(1+c²). Then a'=(κN−cτN)/√(1+c²)=0, so a is a fixed unit vector. Also T·a=(1+cT·B)/√(1+c²)=1/√(1+c²), which is constant. Therefore T makes a constant angle θ with the fixed direction a, where cosθ=1/√(1+c²) and tanθ=|c|. The proof assumes κ>0 and τ≠0; plane or straight limiting cases follow by continuity.